Class 12

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 10 Vector Algebra Ex 10.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 10 Vector Algebra Ex 10.1

MP Board Class 12th Chemistry Important Questions Chapter 2 Solutions

Solutions Important Questions

Very Short Answer Type Questions

Question 1.
Write formula of van’t Hoff factor ‘i’
Answer:
van’t Hoff factor, i = \(\frac {Observed value of colligative property}{Calculated value of colligative property}\)

Question 2.
Write down van’t Hoff equation. Give formula used for calculating molecular mass with its help.
Answer:
van’t Hoff equation, πV = nRT
or π = \(\frac { nRT}{V}\)
or π = \(\frac { WRT}{MV}\)
or π = \(\frac { WRT}{πV}\)
Where, W is mass of solute, R is Solution constant, T is temperature, n is osmotic pressure and V is volume of Solution.

The normality formula of a solution is the gram equivalent weight of a solute per liter of solution.

Question 3.
Define Normality.
Answer:
Number of gram equivalent of solute present in one litre of Solution: is called normality. It is represented by N.
Normality (N) = \(\frac {Number of gram equivalent of solute}{Volume of Solution: in litres}\)
or N = \(\frac {Mass of solute in grams }{Equivalent mass of solute}\) x \(\frac { 1000}{ Volume of Solution: in ml }\)
∵ [No. of gram equivalent =\(\frac {Mass of solute in grams}{Equivalent mass of solute}\)]
Normality of a Solution changes with temperature as it is based on mass – volume relationship and volume changes with change in temperature.

Question 4.
Differentiate between Molarity and Molality.
Answer:
Differences between Molarity and Molality :
Molarity (M):

• Molarity involves the total volume of Solution.
• In molarity, gram moles of solute are dissolved in 1 litre of Solution:.
• Molarity changes with temperature because volume changes with temperature.

Molality (m):

• Molality involves the mass of solvent.
• In molality, gram moles of solute are dissolved in 1 kg of solvent. Here volume of Solution is not considered.
• Molality is independent of temperature as it takes mass into consideration.

Question 5.
Explain the following term : Parts per million.
Answer:
Parts per million:
When a solute is present in very minute amounts (in traces), the concentration is expressed in parts per million abbreviated as ppm. The parts may be of mass or volume. It is the parts of a component per million parts of the Solution.
i.e., ppm = \(\frac {Mass of component A}{Total mass of Solution:}\) x 10⁶
or ppm = \(\frac {Volume of component A}{Total volume of Solution:}\) x 10⁶
Where, ppm. is the concentration of component A in parts per million.

Question 6.
What role does the molecular interaction play in a Solution: of alcohol and water? (NCERT)
Answer:
Alcohols dissolve in water due to formation of inter – molecular H – bonding with water.

Question 7.
Why do gases always tend to be less soluble in liquids as the temperature is raised? (NCERT)
Answer:
Gas + Liquid ⇌ Dissolved gas   ∆ H = – ve
The dissolution of a gas in a liquid is exothermic process. Therefore in accordance with Le – Chatelier’s principle with increase in temperature, the equilibrium shifts in back-ward direction. Therefore, the solubility of gas in Solution decreases with the rise in temperature.

Question 8.
State Henry’s. (NCERT)
Answer:
Henry’s law:
According to this law, ‘The mass of a gas dissolved per unit volume of a solvent at constant temperature, is proportional to the pressure of the gas with which the solvent is in equilibrium’. Let in unit volume of solvent, mass of the gas dissolved is m and equilibrium pressure be P, then m ∝P or m = KP, where K is a constant.

Question 9.
What are colligative properties?
Answer:
The physical properties of Solution which depend on the number of total particles present in Solution and the ratio of number of particles of solute not on the nature of solute particles are known as colligative properties.

Question 10.
Give an example of Solution: of solid in solid.
Answer:
Mixture of copper and gold is an example of solid in solid.

Question 11.
On dissolving ethanoic acid in benzene, experimental molecular mass of ethanoic acid is generally found to be double. Why?
Answer:
On dissolving ethanoic acid in benzene, it dimerizes due to the formation of hydrogen bond. Thus, its experimental value is generally found to be double.

Question 12.
What is vapour pressure? What is the effect of temperature on it?
Answer:
Pressure exerted by the vapours on the surface of a liquid at the state of equilibrium is known as vapour pressure. On increasing the temperature vapour pressure increases.

Question 13.
What type of deviation is represented by acetone and CS₂ Solution?
Answer:
Positive deviation.

Osmotic pressure calculator … The factor i is also called the dissociation factor or the van’t Hoff factor.

Question 14.
Why is CaCl₂ used for clearing off snow from the roads?
Answer:
On adding CaCl₂, freezing point of water decreases. Therefore, CaCl₂ is used for clearing off snow from the roads.

Question 15.
Based on solute – solvent interactions, arrange the following in order of increasing solubility in n – octane and explain :
Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer:
For solubility we know Tike dissolves like’, n – octane is a non – polar solvent, hence non-polar compounds will be more soluble.
KCl < CH₃OH < CH₃CN < Cyclohexane.

Question 16.
What is transition temperature?
Answer:
The temperature at which the nature of solubility changes (i.e., first it increases, then decreases) is known as transition temperature. Solubility of sodium sulphate in water increases upto 32.4, then it starts decreasing. Thus, 32.4°C is the transition temperature of sodium sulphate.

Question 17.
Define molarity and molality. (NCERT)
Answer:
Molarity “is defined as number pf gram mole§ of solute dissolved in a litre of Solution. It is denoted by M.
Molality (M) = \(\frac {Mass of solute in gram per litre}{ Molecular mass of solute}\)

Molality is defined as number of moles of solute present in a kilogram (1000 gram) of solvent. It is denoted by m.
Molality (m) = \(\frac {Mass of solute in kg of solvent}{Gram molecular mass of solute}\)

Solutions Short Answer Type Questions

Question 1.
Write down Raoult’s law.
Answer:
The vapour pressure of a solution containing non – volatile solute is directly pro – portional to the mole fraction of the solute.
Mathematically,
\(\frac { { X }_{ A } }{ { X }_{ A }+{ X }_{ B } } \) = \(\frac { P{ A }^{ 0 } }{ { P }_{ A }+{ P }_{ B } } \)
Where, P⁰_A = Vapour pressure of pure solvent, P_A = Vapour pressure of solvent in solution, X_B = Mole fraction of solute.

Question 2.
What is Azeotropic mixture? They are of how many types? (MP 2018)
Answer:
Azeotropic mixture is the mixture of liquids which boil at one temperature with-out any change in composition. For example, at the composition of 95 6% alcohol and 4.4% water. It form an azeotropic mixture which boils at 78.13°C. Components of this mixture cannot be separated fully by fractional distillation.

They are of two types :
1. Low boiling azeotropic mixture:
Such Solutions which represent positive deviation towards Raoulfs law i.e. their vapour pressure is high thus their boiling point is low are known as low boiling azeotropic mixture.

Example:

• CS₂ + Acetone
• C₂ H₅ OH + n – hexane.

2. High boiling azeotropic mixture:
Such Solutions which represent negative deviation towards Raoult’s law i.e. their vapour pressure is low thus their boiling point is high are known as high boiling azeotropic mixture.

Example:

• Acetone + Chloroform
• Ether + Chloroform.

Question 3.
What are ideal and non – ideal Solutions? Explain with example.
Answer:
Ideal Solutions:
Ideal Solutions are those Solutions in which Raoult’s law can be applied completely for all concentrations of the Solutions and at all temperatures.

Condition for ideal Solutions are following :

1. P_A = P_A⁰ X _A and P_B= P⁰_B X B
2. ∆V_mixing = 0 and
3. ∆H_mixing = 0 and

Example:
C₂H₅Br + C₂H₅Cl, C₆H₆ + C₆H₅CH₃, CCl₄ + SiCl₄etc.

Non – ideal Solutions:
Solutions in which Raoult’s law cannot be applied completely for all concentrations and temperatures are called non – ideal Solutions.
For these Solutions:

1. P_A ≠ P⁰_A X_A and p_B ≠ P⁰_BX_B 
2. ∆V_mixing ≠ 0 and
3. ∆H_mixing ≠ 0 and

Example:
Benzene + Acetone, CHCl₃ + HNO₃ etc.

Question 4.
Establish van’t Hoff Solution: equation. (MP 2018)
Answer:
Osmotic pressure of dilute Solution of a non – volatile solute is proportional to absolute temperature of the Solution at constant concentration. This is known as van’t Hoff law.
π ∝ T .. (1)
Derivation : Osmotic pressure n of a Solution: is directly proportional to a molar concentration.
π ∝ C
By eqn.(l) and eqn.(2) π ∝ C (T is constant) .. (2)
or π = RCT .. (3)
Where, R = Gas constant
or
C = \(\frac {1}{V}\)
π = \(\frac {RT}{V}\) .. (4)
πV = RT .. (5)
This is known as van’t Hoff Ideal Solution equation.

Question 5.
Define the following:

1. Molal elevation boiling point constant.
2. Molal freezing point depression constant. (MP 2018)

Answer:
1. Molal elevation boiling point constant : Molal elevation constant can be defined as “The elevation in boiling point of the Solution in which 1 gm of solute is dissolved in 1000 gm of solvent.”
∴ Elevation in boiling point
∆ T_b ∝ m
∆ T_b = K_bm
m= 1, ∆ T_b = K_b
Where K_b = Molal boiling point elevation constant.

2. Molal freezing point depression constant : Molal depression constant may be defined as “The depression in freezing point for 1 molal Solution: i.e., Solution: in which 1gm mole of solute is dissolved in 1000 gm of solvent.”
∴ Depression in freezing point
∆ T_f ∝ m
∆ T_f = k_f
If m = 1, ∆ T_f = k_f
Where K_f = Molal freezing point depression constant.

Question 6.
What is Raoult’s law? Establish its mathematical expression.
Or,
What is Raoult’s law? How can molar mass of a non – volatile solute be determined with its help?
Answer:
Raoult’s law:
For a Solution in which solute is non – volatile, the Raoulf s law may be stated as following :
“At any constant temperature, vapour pressure of solvent collected above the Solution of non – volatile solute, is directly proportional to the mole fraction of solute.” If a non – volatile solute is added to a volatile solvent, the vapour pressure of the solvent decreases. The vapour pressure of the solvent is directly proportional to its mole fraction. As the solute is non – volatile, the vapour pressure of the Solution (P) will be equal to the vapour pressure of the solvent (PA).
P = P_A ∝ X_A
or P_A=KX_A .. (1)
Where, K = Proportionality constant.
Apply eqn. (1) for pure solvent if X_A= 1 and P_A = P⁰_A
Then, P⁰_A = K x 1 .. (2)
Where, p⁰_A= Vapour pressure of pure solvent.
Putting the value of K from eqn. (2) in eqn. (1),
P_A=P⁰_AX_A .. (3)
Where, X_A is the mole fraction of the solvent in the Solution.
The mole fraction of solute is represented as XB.
So, X_A+X_B=1
or X_A= 1 – X_B .. (4)
From eqns. (3) and (4),
P_A= P⁰_A(1-X_B) = P⁰ – P⁰_AX_B
X_B = \(\frac { { P }_{ A\quad }^{ 0 }-\quad { P }_{ A } }{ { P }_{ A }^{ 0 } } \)
P⁰_A is lowering in vapour pressure and \(\frac { { P }_{ A\quad }^{ 0 }-\quad { P }_{ A } }{ { P }_{ A }^{ 0 } } \) is relative lowering in vapour pressure.

On the basis of equation (5) Raoult’s law can be defined as “The relative lowering in vapour pressure of a Solution containing non – volatile solute is equal to mole fraction of solute”.

Question 7.
Give relation between elevation in boiling point and molecular mass of solute.
Answer:
Relation between elevation in boiling point and molecular mass of solute:
Suppose, W_B gram of non – volatile solute dissolve in WA gram of solvent and the molecular mass of non – volatile solute is M_B gram. Then, molality, m will be

Question 8.
What is van’t Hoff factor? How does it equation?
Answer:
van’t Hoff Equation : van’t Hoff equation

Question 9.
What are constant boiling mixture? Write three differences in Ideal solution and Non – ideal Solutions. (MP 2011,15,16)
Answer:
Constant boiling mixtures or azeotropic mixture. A Solution which distils with out change in composition is called azeotropic mixture.
Differences between Ideal and Non – ideal Solution:

Question 10.
Define the following :

1. Reverse osmosis
2. Isotonic Solution
3. Semi – permeable membrane.

Answer:
1. Reverse osmosis:
It may be noted that if a pressure higher than the osmotic pressure is applied on the Solution the solvent will flow from the Solution into the pure solvent through the SPM. The process is called reverse osmosis.

2. Isotonic Solutions:
The Solution:s which have same osmotic pressure are called as isotonic Solution:s. For example, 0.91% Solution of NaCl called as saline water is isotonic with human blood corpuscles. That is why medicines are mixed with saline water before intravenous injections.

Osmotic pressure is a colligative property and depends on the number of solute particles in a Solution So, the isotonic Solution:s must have same number of solute particles in a given volume of Solution Consequently, the isotonic Solution:s are those which have same molality, if that the solute neither associate nor dissociate in the Solution.

3. Semipermeable membrane:
These are the membrane which allow the movement of the solvent molecules through them. The membrane appear to be continuous sheets or films. But they have very tiny holes or pores which are semimicro in nature. Through the holes, only the molecules of solvent can pass while those of bigger solute molecules cannot pass through.

Solutions Long Answer Type Questions

Question 1.
Write five differences in Solution: having Positive deviation and Negative deviation.
Answer:
Differences between Positive deviation and Negative deviation :

Question 2.
Explain in brief Berkeley and Hartley’s method of osmotic pressure measurement and state its uses.
Answer:
Berkeley and Hartley’s method:
In this method, pressure is applied over the Solution to stop the flow of solvent. This pressure is equivalent to osmotic pressure.
In this method, the apparatus consists of a strong vessel made up of steel in which porous pot is fitted. In the porous pot, copper ferrocyanide semipermeable membrane is deposited. The porous pot is fitted with a capillary tube on one side and a water reservoir on the other side. A piston and pressure gauge are fitted to the steel vessel.

The porous pot and steel vessel are filled with water and Solution respectively. Osmosis takes place and water moves into the steel vessel from the porous pot through the semipermeable membrane. This is shown by fall in water level in the capillary tube. This flow of water is stopped by applying external pressure on the Solution with the help of piston.

This method has the following advantages :

1. It takes comparatively lesser time to determine osmotic pressure.
2. Concentration of Solution: does not change, hence better results are obtained.
3. As high pressure is not exerted over semipermeable membrane, it does not break.
4. High osmotic pressure can be measured.

Question 3.
What is molal freezing point depression constant? Derive the formula to establish relation between molal freezing point depression constant and molecular mass of solute.
Or,
What is molal freezing point depression constant? Show that depression in freezing point is a colligative property. How can molecular mass of solute be determined from depression in freezing point?
Answer:
Molal freezing point depression constant is equal to depression in freezing point of the Solution when 1 gm mole is dissolved in 1000 gm of solvent. It is represented by i.e.,
K_f = \(\frac {depression in freezing point}{Number of moles}\)
or ∆ T_f = K_f x m
As K_f is molal freezing point depression constant, then
∆ T_f ∝m
Where, m = Number of moles dissolved in 1000 gm solvent.
Thus, depression in freezing point is proportional to molality of Solution. Molality is directly proportional to number or molecules of solute substance. Therefore, depression in freezing point is a colligative property.
Calculation of molecular mass of solute:
By determination of depression in freezing point, the Molecular mass of non – volatile solute can be determined.
For a Solution: of non – volatile solute,
∆T_f = K_f x m .. (1)
Let WB gram non – volatile solute is dissolved in w_A gram solvent and molecular mass of solute is M_B.

Question 4.
What is elevation in boiling points? How addition of a non – volatile solute elevates the boiling point of a solvent? Explain it with the help of graph diagram.
Answer:
The vapour pressure of the Solution:
containing a non – volatile solute is always less than that of pure solvent. Therefore, the Solution has to be heated to higher temperature so that its vapour pressure become equal to the atmospheric pressure. Thus, the boiling point of Solution (T_b)is always higher than the boiling point of solvent (T_b⁰). The difference T_b – T_b⁰ is called elevation in boiling point.

If we plot graph between temperature and vapour pressure of a pure solvent and its Solution, then following curve is obtained. Curve AB gives the vapour pressure for the pure solvent and the curve CD gives the vapour pressure of the Solution: at different temperature. At temperature T_b⁰ the vapour pressure of the solvent becomes equal to the atmospheric pressure hence it boils at T_b⁰.

Now, by the addition of non – volatile solute, lowering of vapour pressure of the Solution takes place. And to increase the vapour pressure of the Solution to become equal to atmospheric pressure, the temperature rises. Hence, at T_b the Solution boils. Thus, the boiling point is now elevated from T_b⁰ to T_b. The rise in temperature that results by the addition of a non – volatile solute in a solvent is termed as elevation in boiling point. It is represented by ∆T_b.
So, elevation in boiling point (∆T_b) = T_b – T_b⁰.

Question 5.
What are Non – Ideal Solution? How many types of are they? Explain with giving examples.
Answer:
Non – ideal Solutions:
Solutions in which Raoult’s law cannot be applied completely for all concentrations and temperatures are called non – ideal Solutions.
For these Solutions:

1. P_A ≠ P⁰_A X_A and p_B ≠ P⁰_BX_B,
2. ∆V_mixmg ≠ 0 and
3. ∆H_mixmg ≠ 0 and

Non – ideal Solution are of two types :

1. Solutions showing positive deviations:
For such Solutions the total vapour pressure will be greater than the corresponding vapour pressure according to the Raoult’s law. The boiling point of such Solutions are lowered. Because interactions between the mol-ecules of Solution: is less than the interactions between pure solvents and solutes. Thus the vapour pressure greater than expected. In these type of Solution formed enthalpy and volume increases.
Characteristics:

• P_A > P⁰_A X_A and P_B > P⁰_B X_B
• ∆V_mixlng > 0
• ∆H_mixlng > 0

Example:
Solution of Cyclohexane and Ethanol:
In Ethanol, its molecules are held together by hydrogen bond

On adding cyclohexane, the molecules tend to occupy the space between ethyl alcohol molecules. As a result the attractive forces between alcohol molecules becomes less. Thus, formation of such Solution: is slight increase in vapour pressure, endothermic and an increase in volume.

2. Solutions showing negative deviations:
For such Solutions the total vapour pressure becomes less than expected according to the Raoult’s law. Because interaction between molecules of Solution is greater than interaction between the pure solvent or solute molecules. Thus the vapour pressure of Solution is less. In these type of Solutions formed enthalpy and volume decreases.
Characteristics:

• P_A < P⁰_A X_A and P_B < P⁰_B X_B
• ∆V_mixlng < 0
• ∆H_mixlng < 0

Example:
Solution of Acetone and Chloroform:
When acetone and chloroform are mixed, there are new attractive forces due to intermolecular hydrogen bonding. These force become stronger. Thus the formation of such Solution:s is an exothermic and the vapour pressure of Solution is less and decrease in volume.

Solutions Numerical Questions

Question 1.
Calculate the mass of urea (NH₂ CONH₂ ) required in making 2.5 kg of 0.25 molal aqueous Solution. (NCERT)
Solution:
0.25 molal aqueous Solution means that
Moles of urea = 0.25 mole
Mass of solvent (water) = 1 kg = 1000 g
Molar mass of urea = 14+ 2 +12+ 16 +14+ 2 = 60 g mol^-1
∴ 0 .25 mole of urea = 60 x 0.25 mole = 15 g
Total mass of the Solution = 1000 + 15 g
= 1015 g = 1.015 g
∵ 1.015 kg of Solution contain urea = 15 g
∴ 2.5 kg of Solution will require urea = \(\frac {1.5}{1.015}\) x 2.5 kg = 37 g.

Question 2.
Henry’s law constant for CO₂ in water is 1.67 x 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 ml of soda water when packed under 2.5 atm CO₂ pressure at 298 K. (NCERT)
Solution:
According to Henry’s law,
P = K_HX .. (1)
P = 2.5 atm = 2.5 x 101325 Pa, KH= 1.67 x 10⁸ Pa
Putting these values in equation (1), we get

Question 3.
Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_fK kg mol^-1. (NCERT)
Solution:
We know that ∆T_f = K_f x \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}\) .. (1)
Given ∆T_f = 1.5, K_f = 3.9 K kg mol^-1, W_A = 75g, MB (ascorbic acid, C₆H₈O6)
= 6 x 12 + 8 x l + 16 x 6 = 176
Putting, these values in equation (1),
1.5 = 3.9 x \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{176 \times 75}\)
W_B = 5.077g.

Question 4.
Calculate the osmotic pressure in pascals exerted by a Solution prepared by dissolving 1.0 g of polymer of molar mass 1,85,000 in 450 ml of water at 37°C. (NCERT)
Solution:
π = CRT
= \(\frac {n}{V}\) RT
Here, number of moles of solute dissolved (n)
= \(\frac {1.0}{185,000}\) mol^-1
= \(\frac {1}{185,000}\) g mol^-1
v = 450mL = 0.450 L
T = 37°C = 37 + 273 = 310 K
R = 8.314 K Pa LK^-1 mol^-1
= 8.314 x 103 Pa LK^-1 mol^-1
Substituting these values we get,

Question 5.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous Solution. What should be the molarity of such a sample of the acid if the density of the Solution is 1.504 g ml^-1? (NCERT)
Solution:
68% nitric acid by mass means that
Mass of nitric acid = 68 g
Mass of Solution = 100 g
Molar mass of HNO₃ = 63 g mol^-1
∴ 68 g HNO₃ = \(\frac {68}{63}\) mole = 1.079 mole
Density of solution = 1.504 g mL^-1
∴ Volume of solution = \(\frac {100}{1.504}\)mL = 66.5 mL = 0 0665 L
Molarity of the solution = \(\frac {Moles of the solute}{Volume of Solution in L}\)
= \(\frac {1.079}{0.0665}\) M = 16.23 M.

Question 6.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293K when 25g of glucose is dissolved in 450g of water. (NCERT)
Solution:
\(\frac { { p }^{ 0 }-{ p }^{ 0 } }{ { p }^{ 0 } } \) = X₂ = \(\frac { { W }_{ B }{ M }_{ A } }{ { M }_{ B }{ W }_{ A } } \)
or \(\frac { 17.535-{ P }_{ s } }{ 17.535 }\) = \(\frac { 25×18 }{ 180×450 }\)
or \(\frac { 17.535-{ P }_{ s } }{ 17.535 }\) = 5.56×10^-3
17.535 – p_s = 0.0975
P_s = 17.438 mm Hg.

Question 7.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10⁵mm Hg. Calculate the solubility of methane in benzene at 298K under 760 mm Hg. (NCERT)
Solution:
Here, KH = 4.27 x 10⁵ mm, P = 760 mm
Applying Hemy’s law, P = K_HX
\(\mathrm{X}=\frac{\mathrm{P}}{\mathrm{K}_{\mathrm{H}}}=\frac{760}{4 \cdot 27 \times 10^{5}}\) =1.78 x 10^-3
∴ Mole fraction of methane in benzene = 1.78 x 10^-3

Question 8.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene. (NCERT)
Solution:
A → Benzene (C₆H₆); B —> Toluene (C₇H₈)
Number of moles of benzene n_A = \(\frac {80}{78}\)
Number of moles of toluene = n_B = \(\frac {100}{92}\)
X_A = \(\frac { { n }_{ A } }{ { n }_{ A }+{ n }_{ B } } \quad\) = \(\frac {1.026}{1.026+1.087}\) = 0.486
X_B = 1 – X_A = 1 – 0.486 = 0.514
Given P°_A = 50.71 mm Hg and P°_B = 32.06 mm Hg
We know, P = P_A + P_B
= P°_AX_A + P°_AX_B
= (50.71 x 0.486) + (32.06 x 0.514)
= 24.65+ 16.48 = 41-13
Mole fraction of components in vapour phase may be calculated by using Dalton’s law,
P_A = y_A x P
y_A = \(\frac { { P }_{ A } }{ p } \quad \) = \(\frac {24.65}{41.13}\) = 0.60

Question 9.
Determine the osmotic pressure of 5% glucose Solution at 25°C. Molecular mass of glucose = 180, R = 0.0821 litre atmosphere. (MP 2013,16)
Solution:
∴5 gm glucose is dissolved in 100 ml.
∴ 180 gm glucose will be dissolved in \(\frac {100}{5}\) x 180
= 3600 ml = 3.6 litre.
We know that,
PV = RT
P x 3.6 = 0.0821 x (25 + 273) = 298
or P = \(\frac {0.52×12.5×1000}{0.63×170}\)
= 6.80 atmospheres.

Question 10.
12.5 gm of urea dissolved in 170 gm of water. The elevation in boiling point was found to be 0.63 K. If KA for water = 0.52 Km^–1, calculate the molecular mass of urea. (MP 2010)
Solution:
Formula: M_B\(\frac { { K }_{ b }x{ W }_{ B }x1000 }{ ∆{ T }_{ b }x{ W }_{ A } }\)
Given: W_A = 170gm,W_B= 12.5gm, ∆T_B=0.63K, K_b=0.52Km^-1
M_B = \(\frac {0.52×12.5×1000}{0.63×170}\)
M_B = 60.7 gm mol^-1

Question 11.
If 6.84 gram sucrose is dissolved in 100 ml. Solution then what will be its osmotic pressure at 20°C? (R = 0.082 litre atmosphere K^-1mol^-1) (Molecular mass of sucrose M_B = 342)
Solution:
Given : Weight of solute (W_B) = 6.84 gm
Molecular mass of solute (M_B) = 342
Volume of Solution: (V) =\(\frac {100}{1000}\) =0.1 litre
Temperature (T) = 20° C = 273 + 20 = 293 K
Solution constant R = 0.082 litre atmosphere K^-1 mol^-1
Osmotic pressure π = \(\frac { { W }_{ B }RT }{ { M }_{ B }V } \)
= \(\frac {6.84×0.082×293}{342×0.1}\)
= 4.8 atmosphere.

MP Board Class 12th Chemistry Important Questions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1

MP Board Class 12th Physics Important Questions Chapter 13 Nuclei

Nuclei Important Questions 

Nuclei Objective Type Questions

Question 1.
Choose the correct answer of the following:

Half-life formula is the time required for the amount of something to fall to half its initial value.

Question 1.
The relation between half life time Tm and decay constant of a radioactive material is:

Answer:

Question 2.
The decayed part of a radioactive sample in two half lives will be :
(a) One fourth
(b) Half
(c) Three fourth
(d) Full part
Answer:
(c) Three fourth

Question 3.
The α – particle is a :
(a) Hydrogen nucleus
(b) Deuterium nucleus
(c) Helium nucleus
(d) Tritium.
Answer:
(c) Helium nucleus

Question 4.
The value of A and Z in the nuclear reaction \(_{ 238 }^{ 92 }{ U }\) → \(_{ A }^{ Z }{ Th }\) + \(_{ 4 }^{ 2}{ He }\) will be :
(a) 234,94
(b) 234,90
(c) 238,94
(d) 238,90.
Answer:
(b) 234,90

Question 5.
The source of energy in the sun is :
(a) Nuclear fission
(b) Nuclear fusion
(c) Chemical reaction
(d) Photo electric reaction.
Answer:
(b) Nuclear fusion

Question 6.
For the nuclear fusion which of the following is suitable :
(a) Heavy nucleus
(b) Lighter nucleus
(c) Atom bomb
(d) Radio active decay.
Answer:
(b) Lighter nucleus

Question 7.
In the symbol of nucleus \(_{ Z }^{ A }{ X }\) there are :
(a) Z – neutrons, (A – Z) protons
(b) Z – protons, (A – Z) neutrons
(c) Z – protons, A neutrons
(d) A protons. Z neutrons.
Answer:
(b) Z – protons, (A – Z) neutrons

Question 8.
In the gamma ray emission of a nucleus :
(a) Only proton number varies
(b) Number of neutrons and protons both varies
(c) There is no change in number of protons and neutrons
(d) Only number of neutrons varies.
Answer:
(c) There is no change in number of protons and neutrons

Question 2.
Fill in the blanks :

1. In the controlled chain reaction the fast moving neutrons are slowed down by using moderator. These neutrons are called …………..
2. The Einstein’s mass energy equivalence relation is …………..
3. The unstable nucleus obtained by artificial radioactivity is called …………..
4. The mass of neutrons in a nucleus is nearly equal to mass of …………..
5. In the process of ………….. the lighter nuclei together makes a heavy nucleus.
6. In nuclear reactor the heavy water is used as …………..
7. The SI unit of radio activity is …………..
8. In the range mass number A = 30 to A = 170 the value of ………….. is nearly constant.

Answer:

1. Thermal neutrons
2. E = me²
3. Radio active isotopes
4. Protons
5. Fusion
6. Moderator
7. Becquerel (Bq)
8. Per nucleon binding energy.

Question 3.
Match the columns :

Answer:

1. (d)
2. (e)
3. (a)
4. (c)
5. (b)

Question 4.
Write the answer in one word/sentence :

1. Write the equation showing the decay of free neutron.
2. The energy distribution of P – rays is continuous. Why?
3. In the α and β particles, which one has more ionization power?
4. The ionization power of α – particles is high. Why?
5. The penetrating power of γ(gamma) – rays is very high. Why?
6. Write an equation showing nuclear fusion.
7. Which part of the electromagnetic spectrum has highest penetrating power?
8. What is order of size of nucleus?
9. Does the density of all nuclei remain same? If yes then write its order.
10. How is nuclear radius related to atomic mass?

Answer:

1. ₀n¹ → ₁H¹ + _-1B⁰ + \(\overline { ν }\)
2. Because during the β – decay anti – neutrino is emitted
3. α – particles
4. Because α – particles have large mass and large size and their velocity is low
5. Because their velocity is very high and they do not have any charge,
6. ₁H² + ₁H² → ₂He⁴ + Q
7. γ – rays
8. 10^-15 m
9. Yes, 10¹⁷ per cubic metre
10. R = R₀ A^1/3

Nuclei Very Short Answer Type Questions

Question 1.
What will be the ratio of radii of two nucleus of mass number A₁ and A₂?
Solution:
From the formula

Question 2.
What will be the energy equivalent to 10 milligram?
Solution:
E = mc²
= 10 x 10^-6 x (3 x 10⁸)²
= 9 x 10⁹ joule.

Question 3.
What are thermal neutrons?
Answer:
In the process of controlled nuclear fission, the fast moving neutrons are slowed down with the help of moderation. These slow stray neutrons are called as thermal neutrons.

Question 4.
Does the ratio of neutron and proton after emission of α – particle in a nucleus increase, decrease or remain constant?
Answer:
The ratio can increase, decrease or can remain same. It depends upon the nature of nucleus.

Question 5.
Nuclear fusion is not possible in laboratory. Why?
Answer:
For nuclear fusion very high temperature (≈ 10⁷K) and very high pressure is required which is not possible in laboratory.

Question 6.
Is a free neutron a stable particle?
Answer:
No, free neutron is decayed in a proton, an electron and an antineutrino (\(\overline { ν }\)).
₀n¹ → ₁P¹ + _-1e⁰ + \(\overline { ν }\)

Question 7.
In heavy nucleus the number of neutrons is more than number of protons. Explain why?
Answer:
In a nucleus there is a coulomb’s repulsive force between protons in addition to a strong attraction nuclear force. On the other hand in case of neutrons, within the nucleus, there is only a short range attractive nuclear force. For heavy nuclei to be stable the repulsive force must be less. This is possible only if the number of neutrons is more than number of protons.

Question 8.
If a radioactive substance that is capable to emit a, p and y rays is kept on a piece of paper, then which rays has the maximum possibility to be stoped?
Answer:
As the mass of α – particles is more and its penetrating power is least, so it will be stopped.

Question 9.
Why is heavy water used as moderator in a nuclear reactor?
Answer:
Heavy water contains protons (of mass nearly equal to the mass of neutrons). Fast moving neutrons undergo elastic collision with these slow moving neutrons and thus get slowed down. Hence heavy water is used as moderator.

Question 10.
An atom expressed by _YX^A emits n α – particles. How many proton will remain in it? What will be the new atomic number of new atom?
Answer:
In an α – particles there are 2 protons. If n α – particles are emitted then 2n proton will be reduced.
∴ No. of remaining protons = Z – 2n
Atomic number = Y – 2n.

Question 11.
Penetrating power of β – particles is more than that of α – particles but its penetrating power is less. Why?
Answer:
β – particles have small size and high velocity so its penetrating power is more than α – particles. Due to small size and high velocity the probability of collision with gas molecules decreases and hence ionization power becomes less.

Question 12.
The nuclear fusion could not be used as an experimental and controlled source of energy till now. Why?
Answer:
The nuclear fusion takes place at very high temperature. Due to such very high temperature it is not possible to control it.

Question 13.
Why is a neutron preferred as a bombarding particle in nuclear fission?
Answer:
Neutron is a neutral particle i.e., it has no charge. So it is neither repelled nor attracted by the nucleus and hence can penetrate deep into the nucleus to cause a nuclear reaction.

Question 14.
β – rays are emitted from nucleus. They are made of fast moving electron though there are no electrons in the nucleus. Why it is so?
Answer:
Nucleus has no electron in it. Actually due to decay of neutron into proton and electron, the β – rays (particle) are produced.

Question 15.
Justify that gamma rays have more penetrating power and α – particles have more ionizing power.
Answer:
Gamma rays are electromagnetic waves. Its speed is equal to the speed of light. So they can penetrate deeply into the matter i.e., their penetrating power is high, α – particles has high mass and posses greater kinetic energy. So when they collide with any atom, it transfer its energy to orbital electrons and they are ejected. Hence α – particles has high ionizing power.

Question 16.
The mass number of two nucleus are 1 : 2. What is the ratio of their nuclear density?
Answer:
1 : 1 because nuclear density does not depend on mass number.

Question 17.
What is effect of temperature and pressure or radio activity?
Answer:
No, effect.

Question 18.
Due to the emission of α – particles does the ratio of neutron to proton decrease, increase or remains the same?
Answer:
The ratio \(\frac {n}{p}\) increases, because number of protons decreases by 2.

Nuclei Short Answer Type Questions

Question 1.
Which particles remains inside the nucleus? What will be the number of neutrons in the nucleus _zX^A?
Answer:
The neutrons and protons remains inside the nucleus. In the nucleus _zX^A number of neutrons be A – Z.

Question 2.
Define decay constant and mean life of a radioactive substance write their units.
Answer:
Decay Constant:
The decay constant is defined as the reciprocal of that time interval during which the number of active nuclei in a given sample of a radioactive substance reduces to times of the initial value of number of nuclei.
If T is half life time and X is decay constant then
T = \(\frac {0.6931}{λ}\)
Its unit is “per second”.

Mean life:
The reciprocal of decay constant is called mean life, its unit is ‘second’.

Question 3.
What is radioactivity? Write the names of the rays emitted from radio – active substances.
Answer:
Radioactivity is that property by virtue of which the nucleus of a heavy element disintegrates itself with the emission of radiation without being forced by any external agent to do so.

Following rays are emitted from a radioactive substance :

1. α – rays
2. β – rays
3. γ – rays.

Question 4.
A radioactive nucleus decays as given below :

If the mass number of A₂ is 176 and atomic number is x then what is the atomic number and mass number of A₁ and A? Are they isotopes or isobars?
Answer:
The mass number of A₂ = 176,
atomic number of A₂ = 71.
Due to decay of α – particle mass number decreases by 4 and atomic number decreases by 2.
∴ Mass number of A₁ = 176 + 4 = 180 and
its atomic number = 71 + 2 = 73
Similarly
The mass number of A = 180 + 1 = 181 and
atomic number of A = 73 + 0 = 73
Since, the atomic number of A and A₁ are same so they are isotopes.

Question 5.
Define decay constant of a radioactive substance. Write its relation with half life?
Answer:
The decay constant is defined as the reciprocal of that time interval during which the number of active nuclei in a given sample of a radioactive substance reduces to times of the initial value of number of nuclei.
If T is half life time and X is decay constant then
T = \(\frac {0.6931}{λ}\).

Question 6.
What is nuclear fission? Give an example.
Answer:
When a heavier nucleus is bombarded with neutrons then it get splitted into two lighter nuclei. This process is called as nuclear fission.
A huge amount of energy is released in this process.
₉₂U²³⁵ + ₀n¹ → ₅₆Ba¹⁴¹ + ₃₆kr⁹² + 3 ₀n¹ + 200 MeV.

Question 7.
What is nuclear fusion? Give an example.
Answer:
When two lighter nuclei fused together to form a heavier nucleus then a huge amount of energy is released in this process. This process is called nuclear fusion.
₁H² + ₁H² → ₂He⁴+ 24MeV.

Question 8.
A radioactive element with mass number 218 and atomic number 84 emits a β – particle. What would be the mass number and atomic number after decay?
Answer:
Since, after the emission of β – particle there is no change in the mass number of element but its atomic number is increased by 1.
Therefore the mass number will be 218 and atomic number will be 84 + 1 = 85.

Question 9.
The ratio of radii of two nuclei are 1 : 2. Find out the ratio of their mass number.
Solution:
Radius of nucleus

Question 10.
The nuclear fusion process is difficult as compared to nuclear fission. Why?
Answer:
For nuclear fusion a high temperature (≈ 10⁷ K) is required. It is very difficult to achieve this temperature, so nuclear fusion is difficult.

Question 11.
Write the important properties of nuclear forces.
Answer:

1. Nuclear forces are attraction force
2. These forces does not depend on charge
3. These forces are short range forces
4. They are strong forces
5. The nuclear forces are not central forces.

Question 12.
Write properties of nuclear forces. Prove that the density of nucleus is independent of mass number.
Answer:
Properties of nuclear forces :

1. Nuclear forces are attraction force
2. These forces does not depend on charge
3. These forces are short range forces
4. They are strong forces
5. The nuclear forces are not central forces.

Let the mass number of a nucleus is A and radius is r then volume of nucleus be

Thus, it is clear that material density of a nucleus is independent of its mass number.

Question 13.
Write difference between nuclear fusion and nuclear fission.
Answer:
Difference between nuclear fusion and nuclear fission :
Nuclear fission:

• In this prosses a heavy nucleus splitted into two lighter nuclei.
• This process is possible at normal temperature.
• Per fission released energy is very high (200 MeV)

Nuclear fusion:

• In this process two lighter nuclei fused together to form a heavy nucleus.
• This process is possible at very high temperature.
• Per fusion released energy is comparatively less (24 MeV).

Nuclei Long Answer Type Questions

Question 1.
Define ‘electron volt’ and ‘Atomic mass unit”. Find out the energy equivalent to mass of a proton in joule.
Answer:
Electron volt:
The energy acquired by an electron when it is being accelerated by a potential difference of 1 volt is called ‘electron volt’.
1 electron volt = 1.6 x 10^-19J.

Atomic mass unit:
An atomic mass unit is defined as (\(\frac {1}{12}\)) of the mass of 1 atom of carbon – 12.
1 atomic mass unit = 1.66 x 10^-27kg
(1 a.m.u. or only 1u)
The mass of proton = 1.673 x 10^-27 x (3 x 10⁸ )²
= 15.05 x 10^-11J
= 1.505 x 10^-10J
∴ The energy equivalent mass of 1 proton is 1.505 x 10^-10J

Question 2.
Draw the graph showing the variation of binding energy per nucleon with the mass number. Write important conclusions drawn from it. (CBSE 1994)
Or
Draw the graph showing the variation of binding energy per nucleon with mass number. Highlight the region where the nuclei are more stable. (CBSE 1996)
Or
Draw the graph showing the variation of binding energy per nucleon with the mass number. Highlight the region where the nuclear fusion takes place. (CBSE 1996)
Answer:
From the binding energy curve, it follows that:

1. The average binding energy/ nucleon for nearly all elements is 8 MeV

2. The maximum value of binding energy per nucleon is 8.79 MeV and it is for Fe. This explains the large abundance of Fe in nature

3. Since for nuclei of intermediate mass numbers, the value of the binding energy per nucleon is around the maximum value, these nuclei are the most stable

4. The binding energy/ nucleon decreases for nuclei of mass number more than 56 and its least value is 7.6 MeV.

5. Below mass number 28, there are peaks in the curve corresponding to those nuclei whose mass numbers are multiples of four. These nuclei contain an equal number of protons and neutrons. Though peaks occur corresponding to \(_{ 4 }^{ 2 }{ He}\), \(_{ 8 }^{ 4 }{ Be }\), \(_{ 12 }^{ 6 }{ C }\), \(_{ 16 }^{ 8 }{ O }\) and \(_{ 20}^{ 10 }{ Ne }\), we have shown only the first peak corresponding to \(_{ 4 }^{ 2 }{ He }\) Obviously, these nuclei have more binding energy per nucleon than their neighbours. This explains the stability of an α – particle (\(_{ 4 }^{ 2 }{ U }\))

6. The smallest value of the binding energy per nucleon is in the case of a deuteron (\(_{ 2 }^{ 1 }{ U }\)) and its value (2.2 MeV)/2 = 1.1 MeV. This is confirmed by the fact that a photon whose energy is 2.2 MeV or more, can split a deuteron into a free neutron and a free proton. This phenomenon is called the photodisintegration of a deuteron.

Question 3.
Calculate binding energy per nucleon of ₂₆Fe⁵⁶.
Given : m(\(_{ 26 }^{ 56 }{ Fe }\)) = 55.934939 a.m.u., m (proton ) = 1.007825 a.m.u., m (neutron) = 1.008665 a.m.u. (CBSE 1993,95, 2000 Supp.)
Solution:
Number of protons in = \(_{ 26 }^{ 56 }{ He }\) is 26 and number of neutrons is 56 – 26 = 30.
∴ Mass of 26 protons = 1 0077825 x 26
= 26.202345 a.m.u.
Mass of 30 neutrons = 1.008665 x 30
= 30.25995 a.m.u.
Total mass of nucleons = 26.202345 + 30.25995
= 56.4623 a.m.u.
Mass of ₂₆Fe⁵⁶ = 55.934939.a.m.u.,
∴ Mass defect ∆m = 56.4623 – 55.934939
= 0.527361 a. m. u.
∴ Binding energy of ₂₆Fe⁵⁶ = 0.527361 x 931
= 490.973091 MeV
Binding energy per nucleon = \(\frac {490.973091}{56}\)
= 8.7674 MeV.

Question 4.
Define the term half-life period and decay constant of a radioactive sample. Derive a relation between these terms.
Answer:
The half – life of a radioactive substance is the time it takes half of a given number of radioactive nuclei to decay. (CBSE 1995, 2001)
The SI unit of Tm is second (s).
Example:
Half life of radium is 1600 years. It means that if 1 gm of radium is taken then after 1600 years \(\frac {1}{2}\) gm of radium will decay.
Let us assume that at time t = 0, the number of radioactive nuclei present is N₀ and after time t, the number of radioactive nucleus remaining is N, then according to law of radioactive decay,
N = N₀e^-λt
Where, λ is decay constant.
If the half – life of radioactive sample is T_1/2, then at t = T_1/2
N = \(\frac { { N }_{ 0 } }{ 2 }\)
Substituting in equation (1), we get ,

From equation (2), it is clear that half life of a radioactive sample is inversely proportional to decay constant.

Question 5.
State law of radioactive decay and derive the following formula :
N = N₀ e^-λt, where symbols have their usual meaning.
Answer:
According to the radioactive decay law :
The rate at which a particular decay process occurs in a radioactive sample is proportional to the number of radioactive nuclei present (that is, those nuclei that have not yet decayed).
Let N₀ = Total number of radioactive nuclei present originally at time t = 0,
(t = 0 refers to the time when the radioactive element is freshly separated from its by products), and N = Total number of radioactive nuclei present at any time t.

According to the radioactive decay law,
– \(\frac {dN}{dt}\) ∝ N
(- ve sign indicated that the number of radioactive nuclei decreases with time).
Thus, – \(\frac {dN}{dt}\) = λN
Where A is a constant of proportionality and is called the disintegration constant or decay constant.
Equation (1) can be written as
\(\frac {dN}{dt}\) = – λdt
When t = 0, N = N₀ and when t = t,N = N.
Integrating equation (2) within proper limits,

From equation (3) it is clear that (i) the radioactive decay is an exponential process i.e., the decay is fast initially and later on it becomes slow.

Nuclei Numerical Questions

Question 1.
An imaginary fission reaction is given below :
\(_{ 92 }^{ 236 }{ X }\) → \(_{ a}^{ 141 }{ Y }\) + \(_{ 36 }^{ b }{ Z }\) + 3\(_{ 0 }^{ 1 }{ n }\)
Find out value of a and b.
Solution:
a + 36 + 0 = 92
a = 56.
and 141 + b + 3 = 236
b = 92.

Question 2.
The half life of a radioactive substance is 30 days. What will be the number of radioactive atoms after 90 days?
Solution:
Given : T = 30 days, t = 90 days
n = \(\frac {t}{T}\) = \(\frac {90}{30}\) = 3
Formula : N = N₀
= N₀(\(\frac {1}{2}\))ⁿ
= N₀(\(\frac {1}{2}\))³
= \(\frac { { N }_{ 0 } }{ 8 }\)
Thus, the number of radioactive atoms be \(\frac {1}{8}\) times of their initial value.

Question 3.
How many fission per second takes place if its half life for α – decay is 1.42 x 10¹⁷sec?
Solution:

MP Board Class 12th Physics Important Questions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 Relations and Functions Ex 1.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 1 Relations and Functions Ex 1.2

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 संबंध एवं फलन Ex 1.4 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Book Solutions Chapter 1 संबंध एवं फलन Ex 1.4

प्रश्न 1.
निर्धारित कीजिए कि क्या निम्नलिखित प्रकार से परिभाषित प्रत्येक संक्रिया से एक द्विआधारी संक्रिया प्राप्त होती है या नहीं। उस दशा में जब एक द्विआधारी संक्रिया नहीं है, औचित्य भी बतलाइए।
(i) Z⁺ में, a * b = a – b द्वारा परिभाषित संक्रिया
(ii) Z⁺ में, a* b = ab द्वारा परिभाषित संक्रिया
(iii) R में, संक्रिया *, a* b = ab² द्वारा परिभाषित
(iv) Z⁺ में, संक्रिया *, a* b = |a – b| द्वारा परिभाषित
(v) Z⁺ में, संक्रिया *, a* b = a द्वारा परिभाषित
हल:
(i) Z⁺ में, a* b = a – b द्वारा परिभाषित संक्रिया है
यदि a > b, a * b = a – b ϵ Z⁺
परन्तु यदि a < b, a * b = a – b < 0, Z⁺ में नहीं है।
अत:* संक्रिया द्विआधारी संक्रिया नहीं है।

(ii) Z⁺ पर * संक्रिया, a * b = ab द्वारा परिभाषित है।
यदि a, b ϵ Z⁺ ⇒ a और b दोनों धनात्मक हैं।
a * b = ab भी धनात्मक है।
ab ϵ Z⁺
अतः यह संक्रिया द्विआधारी है।

(iii) R पर * संक्रिया a* b = ab⁺ द्वारा परिभाषित है।
यदि a, b ϵ R, ab² भी R* में है।
अतः यह संक्रिया द्विआधारी है।

(iv) Z⁺ पर * संक्रिया a * b = |a – b| द्वारा परिभाषित है।
यदि a, b ϵ Z⁺ , |a – b | ϵ Z⁺
अंत: यह संक्रिया द्विआधारी है।

(v) Z⁺ पर * संक्रिया a* b = a द्वारा परिभाषित है।
यदि a, b ϵ Z⁺, ∴ a * b = a ϵ Z⁺
अत: यह संक्रिया द्विआधारी है।

प्रश्न 2.
निम्नलिखित परिभाषित प्रत्येक द्विआधारी संक्रिया के लिए निर्धारित कीजिए कि क्या द्वि आधारी क्रमविनिमेय है तथा क्या साहचर्य है।
(i) Z में, a * b = a – b द्वारा परिभाषित
(ii) Q में, a * b = ab + 1 द्वारा परिभाषित
(iii) Q में, a * b = \(\frac{a b}{2}\) द्वारा परिभाषित
(iv) Z⁺ में, a * b = 2^ab द्वारा परिभाषित
(v) Z⁺ में, a* b = a^b द्वारा परिभाषित
(vi) R – { – 1} में, a* b = \(\frac{a}{b+1}\) द्वारा परिभाषित
हल:
(i) Z पर संक्रिया a* b = a – b द्वारा परिभाषित है।
(a) यदि a * b = a – b और b * a = b – a
परन्तु a – b ≠ b – a ⇒ a* b + b * a
∴ यह संक्रिया क्रमविनिमेय नहीं है।

(b) यदि a * (b * c) = a * (b – c) = a * (b – c)
a – (b – c) = a – b + c
(a * b) * c = (a – b) * c = a – b – c
स्पष्ट है कि a * (b * c) (a * b) * c
∴ संक्रिया साहचर्य नहीं है। अतः संक्रिया न तो क्रमविनिमेय है और न ही साहचर्य है।

(ii) Q पर * संक्रिया, a * b = ab + 1 से परिभाषित है।
(a) a * b = ab + 1, b * a = ba + 1 = ab + 1
∴ a * b = b * a
∴ यह संक्रिया क्रमविनिमेय द्विआधारी है।

(b) यदि a * (b * c) = a * (bc + 1) = a (bc + 1) + 1
= abc + a + 1
(a * b) * c= (ab + 1) * c =(ab + 1)c + 1
= abc + c + 1
∴ (a * b) * c ≠ a + (b + c)
∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है। अतः यह संक्रिया क्रमविनिमेय है परन्तु साहचर्य नहीं है।

(iii) Q पर * संक्रिया, a * b = \(\frac{a b}{2}\) द्वारा परिभाषित है।

∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया है।
अतः यह संक्रिया क्रमविनिमेय और साहचर्य दोनों हैं।

(iv) Z⁺ पर * संक्रिया a* b = 2^ab से परिभाषित है।
(a) ∴ a * b = 2^ab, b * a = 2^ba = 2^ab
⇒ a * b = b * a
अतः संक्रिया क्रमविनिमेय संक्रिया है।

(b) a * (b * c) = a * 2^bc = a^a.2^bc
(a * b) * c = 2^ab * c = 2^2ab.c
∴ a * (b * c) ≠ * (a * b) *c
∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है। अतः यह संक्रिया क्रमविनिमेय है परन्तु साहचर्य नहीं है।

(v) Z⁺ पर * संक्रिया, a * b = a^b से परिभाषित है।
(a) a * b = a^b, b * a = b^a
∴ a * b + b* a
अत: यह संक्रिया क्रमविनिमेय नहीं है।
(b) a * (b * c)=a * b^c = a^(bc)
(a * b) * c = ad * c = a^(b)c = a^bc
∴ (a * b) * c * a * (b * c)
∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है।
अत: यह संक्रिया न तो क्रमविनिमेय है और न ही साहचर्य

(vi) R – {-1} पर * संक्रिया, a * b = \(\frac{a}{b+1}\) द्वारा परिभाषित है।

∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है।
अत: यह संक्रिया क्रमविनिमेय है और न ही साहचर्य है।

प्रश्न 3.
समुच्चय {1, 2, 3, 4,5} में a ^ b= निम्नतम {a, b} द्वारा परिभाषित द्विआधारी संक्रिया पर विचार कीजिए। संक्रिया के लिए संक्रिया सारणी लिखिए।
हल:
समुच्चय {1, 2, 3, 4, 5} पर संक्रिया ^ सारणी निम्न है-

प्रश्न 4.
समुच्चय {1, 2, 3, 4, 5} में, निम्नलिखित संक्रिया सारणी (सारणी 1.2) द्वारा परिभाषित द्विआधारी संक्रिया पर विचार कीजिए तथा
(i) (2 * 3) * 4 तथा 2 * (3 * 4) का परिकलन कीजिए।
(ii) क्या * क्रम विनिमेय है?
(iii) (2 * 3) * (4 * 5) का परिकलन कीजिए। (संकेत : निम्न सारणी का प्रयोग कीजिए।)

हल:
(i) दी गई सारणी से
(2 * 3) * 4 = 1 * 4 = 1
तथा 2 * (3 * 4) = 2 * 1 = 1

(ii) माना a, b ϵ {1, 2, 3, 4, 5}
∴ सारणी से, a * a = a (a ≠ b) तथा a, b विषम संख्या है
a * b = b * a = 1
2 * 4 = 4 * 2 = 2 जहाँ a तथा b सम संख्या तथा a ≠ b
अतः a * b = b * a
अतः द्विआधारी संक्रिया क्रम विनिमेय है।

(iii) सारणी से,
(2 * 3) * (4 * 5) = 1 *1
= 1

प्रश्न 5.
मान लीजिए कि समुच्चय {1, 2, 3, 4, 5} में एक द्विआधारी संक्रिया *’, a *’ b = a तथा b का HCF द्वारा परिभाषित है। क्या संक्रिया *’ उपर्युक्त प्रश्न 4 में परिभाषित संक्रिया * के समान है? अपने उत्तर का औचित्य भी बतलाइए।
हल:
यहाँ समुच्चय {1, 2, 3, 4, 5} संक्रिया a*’ b H.C.F. a, b द्वारा परिभाषित है।
इस संक्रिया की निम्न सारणी दी गयी है-

प्रश्न 4 में दी गई सारणी और यह सारणी समान है।
अतः संक्रिया *’ तथा * समान है।

प्रश्न 6.
मान लीजिए कि N में एक द्विआधारी संक्रिया, a * b = a तथा b का LCM द्वारा परिभाषित है। निम्नलिखित ज्ञात कीजिए:
(i) 5 * 7, 20 * 16
(ii) क्या संक्रिय * क्रम विनिमेय है?
(iii) क्या * साहचार्य है?
(iv) N में * का तत्समक अवयव ज्ञात कीजिए।
(v) N के कौन-से अवयव * संक्रिया के लिए व्युत्क्रमणीय है?
हल:
द्विआधारी संक्रिया (Binary Operations) * इस प्रकार परिभाषित है कि
a * b = a तथा b का L.C.M.
(i) 5 * 7 =5 तथा 7 का L.C.M.
= 35
तथा 20 * 16 = 20 तथा 16 का L.C.M.
= 80

(ii) a * b = a तथा b का L.C.M.
= b तथा a का L.C.M
a * b = b * a
अतः द्विआधारी संक्रिया क्रम विनिमेय है।

(iii) a * (b * c) = a * (b तथा c का L.C.M.)
= a तथा (b तथा c का L.C.M.) का L.C.M.
= a, b तथा c का L.C.M.
इसी प्रकार
(a * b) * c =(a तथा b का L.C.M.) * c
=a, b, c of L.C.M.
⇒ a *(b * c) =(a * b) * c
अतः द्विआधारी संक्रिया * साहचर्य है।

(iv) N में * संक्रिया की तत्समक अवयव 1 है।
∵ 1 * a = a * 1 = a
= 1 तथा a का L.C.M.

(v) माना * : N × N → N इस प्रकार परिभाषित है कि a * b = a तथा b का L.C.M.
∴ a = 1, b = 1 के लिए,
a * b = 1 = b * a
अत: 1a* संक्रिया के लिए व्युत्क्रमणीय है।

The step by step instructions on how to find the least common multiple of 12 and 16.

प्रश्न 7.
क्या समुच्चय {1, 2, 3, 4, 5} में a * b = a तथा b का LCM द्वारा परिभाषित * एक द्विआधारी संक्रिया है? अपने उत्तर का औचित्य भी बतलाइए।
हल:
दिया गया समुच्चय = {1, 2, 3, 4, 5} द्विआधारी संक्रिया द्वारा परिभाषित है कि a * b = a और b का LCM 2 * 6 = 6 जो कि समुच्चय {1, 2, 3, 4, 5} में नहीं है इसलिए * एक द्विआधारी संक्रिया है।

प्रश्न 8.
मान लीजिए कि N में a * b = a तथा b का HCF द्वारा परिभाषित एक द्विआधारी संक्रिया है। क्या * क्रमविनिमेय है? क्या * साहचर्य है? क्या N में इस द्विआधारी संक्रिया के तत्समक का अस्तित्व है?
हल:
यहाँ N, प्राकृत संख्याओं का समुच्चय है।
द्विआधारी संक्रिया a * b = a, b का H.C.F. द्वारा परिभाषित
(i) a, b का H.C.F. = b, a के H.C.F.
a * b = b * a
अतः संक्रिया क्रमविनिमेय है।

(ii) a * (b * c)= a * (b, c का H.C.F.)
=a व b, c का H.C.F.
= a, b, c का H.C.F.
(a * b) * c = (a, b का H.C.F.) * c
= a, b व c का H.C.F.
= a, b, c का H.C.F.
a * (b * c)= (a * b) * c (∵ संक्रिया साहचर्य है)

(iii) 1 * a = a * 1 = 1 ≠ a
अतः तत्समक अवयव का अस्तित्व नहीं है।

प्रश्न 9.
मान लीजिए कि परिमेय संख्याओं के समुच्चय में निम्नलिखित प्रकार से परिभाषित * एक द्विआधारी संक्रिया है:
(i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
(iv) a * b = (a – b)²
(v) a * b = \(\frac{a b}{4}\)
(vi) a * b = ab²
ज्ञात कीजिए कि इनमें से कौन-सी संक्रियाएँ क्रमविनिमेय हैं और कौन-सी साहचर्य हैं।
हल:
यहाँ परिमेय संख्याओं का समुच्चय Q दिया है।
(i) a * b = ab – b, द्विआधारी संक्रिया है।
(a) b * a = b – a
∴ a – b ≠ b – a ⇒ a * b ≠ b * a
अत: यह संक्रिया क्रमविनिमेय नहीं है।
(b) a * (b * c) = a * (b – c) = a – (b – c) = a – b + c
(a * b) * c = (a-b)* c = a – b -c
∴ a – b + c ≠ a – b – c = a * (b * c) * (a * b) * c
अतः यह संक्रिया साहचर्य नहीं है।

(ii) (a) a * b = a² + b²
∴ b * a = b² + a² = a² + b²
⇒ a * b = b * a
अत: यह संक्रिया क्रमविनिमेय है।
(b) a * (b * c) = a * (b² + c²) = a² + (b² + c²)²
(a + b) * c = (a² + b²) * c = (a² + b²)² + c²
⇒ a * (b * c) ≠ (a * b) * c.
अतः यह * संक्रिया साहचर्य नहीं है।

(iii) संक्रिया a * b = a + ab द्वारा परिभाषित है।
(a) a * b = a (1 + b), b * a = b + ba = b (1 + a)
∴ a * b + b * a
अतः यह * संक्रिया क्रमविनिमेय नहीं है।
(a) a * (b * c) = a + (b + bc)= a + a (b + bc)
= a + ab + abc
(a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c
= a + ab + ac + abc
∴ a * (b * c) (ab) * c
अतः यह * संक्रिया साहचर्य नहीं है।

(iv) दिया है : a * b = (a – b)²
(a) a * b = (a – b)², b * a = (b – a)² = (a – b)²
∴ a * b = b * a
अतः यह * संक्रिया क्रमविनिमेय है।
(b) a * (b * c) = a * (b – c) = [a – (b – c)²]²
(a * b) * c = (a – b)² * c = [(a – b)² – c]²
∴ a * (b * c) ≠ (a * b) * c
अतः यह * संक्रिया साहचर्य नहीं है।

(v) a * b = \(\frac{a b}{4}\)

अतः यह * संक्रिया साहचर्य है।

(vi) a * b = ab²
(a) a * b = ab², b * a = ba
∴ a * b ≠ b * a
अत: यह * संक्रिया क्रमविनिमेय नहीं है।
(b) a * (b * c) = a + bc² = a (bc²)² = ab²c⁴
(a * b) * c = ab² * c = ab²c² = ab²c²
∴ a + (b * c) ≠ (a * b) * c.
अतः यह * संक्रिया साहचर्य नहीं है।

प्रश्न 10.
सिद्ध कीजिए कि प्रश्न 9 में दी गई संक्रियाओं में किसी का तत्समक है, वह बतलाइए।
हल:
यहाँ (i) a * b = a – b
यदि e तत्समक अवयव हो तो
a * e = a – e, e * a = e – a
∴ a – e ≠ e – a ⇒ a * e ≠ e * a
अतः e का अस्तित्व नहीं है।

(ii) a * b = a² + b²
∴ a * e = a² + e², e * a = e² + a²
a * e = e * a ≠ a
अतः e का अस्तित्व नहीं है।

(iii) a * b = a + ab
a * e = a + ae, e * a = e + ea
∴ a * e # e * a # a
अत: e का अस्तित्व नहीं है।

(iv) a * b = (a – b)
a * e = (a – e)² # a, e * a = (e – a)² # a
a * e = e * a # a
अतः e का अस्तित्व नहीं है।

(v) a * b = \(\frac{a b}{4}\)
a * e = \(\frac{ae}{4}\) # a, e * a = \(\frac{ea}{4}\) # a
∴ a * e = e * a # a
अतः e का अस्तित्व नहीं है।

(vi) a * b = ab²
a * e = ae² # a, e * a = ea² # a
∴ a * e # e * a # a
अतः e का अस्तित्व नहीं है।

प्रश्न 11.
मान लीजिए कि A = N × N है तथा A में (a, b) * (c, d) = (a + c, b + d)द्वारा परिभाषित एक द्विआधारी संक्रिया है। सिद्ध कीजिए कि * क्रम विनिमेय तथा साहचर्य है। A में * का तत्समक अवयव, यदि कोई है, तो ज्ञात कीजिए।
हल:
माना A = N × N
द्विआधारी संक्रिया (Binary operation) * इस प्रकार परिभाषित है कि
(a, b) * (c, d) = (a + c, b + d)
इसलिए (c, d) * (a, b) = (c + a, d + b)
=(a + c, b + d)
=(a, b) * (c, d)
अतः द्विआधारी संक्रिया * क्रम विनिमेय है
पुनः (a, b)* [(c, d) * (e, f)]
= (a, b) * (c + e, d + f)
= (a + c + e, b + d + f)
तथा [(a, b) * (c, d)] * (e, f)
= (a + c, b + d) * (e, f)
= (a + c + e, b + d + f)
= (a, b) * [(c, d) * (e, f)]
= [(a, b) * (c, d)]* (e, f)
अतः दी गई संक्रिया * साहचर्य है।
A में तत्समक अवयव का अस्तित्व नहीं है।

प्रश्न 12.
बतलाइए कि क्या निम्नलिखित कथन सत्य हैं या असत्य हैं। औचित्य भी बतलाइए।
(i) समुच्चय N में किसी भी स्वेच्छ द्विआधारी संक्रिया* के लिए a * a = a, ∀a ϵ N
(ii) यदि N में * एक क्रमविनिमेय द्विआधारी संक्रिया है तो a * (b * c)=(c * b) * a
हल:
यहाँ द्विआधारी संक्रिया समुच्चय N पर इस प्रकार परिभाषित की गयी है कि
a * a = a ∀ a ϵ N
(i) यहाँ पर * संक्रिया में केवल एक ही अवयव का प्रयोग किया गया है।
अतः यह कथन असत्य है।
(ii) वास्तविक संख्याओं में समुच्चय पर संक्रिया क्रमविनिमेय है।
b * c = c * b
= (c * b) * a = (b * c) * a = a * (b * c)
∴ a * (b * c) = (c * b) * a
अतः यह कथन सत्य है।

प्रश्न 13.
a * b = a³ + b³ प्रकार से परिभाषित N में एक द्विआधारी संक्रिया * पर विचार कीजिए। अब निम्नलिखित में से सही उत्तर का चयन कीजिए।
(A) * साहचर्य तथा क्रमविनिमेय दोनों है
(B) * क्रमविनिमेय है किन्तु साहचर्य नहीं है
(C) * साहचर्य है किन्तु क्रमविनिमेय नहीं है
(D) * न तो क्रमविनिमेय है और न साहचर्य है
हल:
यहाँ द्विआधारी संक्रिया को समुच्चय पर इस प्रकार परिभाषित किया गया है कि
a * b = a³ + b³
(a) a * b = a³ + b³, b * a = b³ + a³ = a³ * b³
∴ a * b = b * a
अत: यह संक्रिया क्रमविनिमेय है।
(b) a * (b * c) = a * (b³ + c³) = a³ + (b³ + c³)³
(a * b) * c= (a³ + b³) * c = (a³ + b³) + c³
∴ a * (b * c) ≠ (a * b) * c
अतः यह * संक्रिया साहचर्य नहीं है।
∴ संक्रिया क्रमविनिमेय परन्तु साहचर्य नहीं है।
अतः विकल्प (B) सही है।

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 7 Integrals Ex 7.7 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.7

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.2

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 10 Vector Algebra Ex 10.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 10 Vector Algebra Ex 10.2

MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry

Electrochemistry NCERT Intext Exercises

Electrochemistry Class 12 NCERT Solutions help to score more marks.

Question 1.
How would you determine the standard electrode potential of the system Mg²⁺ | Mg ?
Answer:
The standard electrode potential of the system Mg²⁺/ Mg can be determined as explained below. Prepare Mg²⁺/ Mg system by dipping Mg rod in a solution of Mg²⁺ ions and connect it to standard hydrogen electrode.
Mg | Mg²⁺ || H⁺_(aq) | H_2(g); Pt
When magnesium is connected with SHE, oxidation takes place at the Mg electrode. Hence, the potential of the magnesium electrode is taken as – ve The emf of the cell, determined potentiometrically, is equal to the potential of the magnesium electrode because the potential of SHE is taken as zero.

Question 2.
Can you store copper sulphate solutions in a zinc pot?
Answer:
No, it is not possible. The E° values of the copper and zinc electrodes are as follows :
Zn2+(aq) + 2e– → Zn(s) ; E° = – 0·76 V
Cu2+(aq) + 2e– → Cu(s) ; E° = + 0·34 V
This shows that zinc is a stronger reducing agent than copper. It will lose electrons to Cu2+ ions and a redox reaction will immediately set in.
Zn(s) + Cu2+ (aq) → Zn2+(aq) + Cu(s)
Thus, copper sulphate solution cannot be stored in zinc pot.

Question 3.
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Answer:
Substances having higher E^° values than Fe²⁺ ions can oxidise ferrous ions. Thus, Ag⁺, Hg²⁺, Br₂, Cl₂, F₂ etc. can oxidise Fe²⁺ ions to Fe³⁺ ions.

Question 4.
Calculate the potential of the hydrogen electrode in contact with a solution whose pH is 10.
Solution:
For hydrogen electrode, H+ + e– → 1/2H2
Applying Nernst equation,

Question 5.
Calculate the emf of the cell in which the following reaction takes place :
Ni(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag_(s) Given that E^∘_cell = 1.05 V.
Solution:

Question 6.
The cell in which the following reactions occurs :
2Fe³⁺(aq) + 2I^–(aq) → 2Fe²⁺(aq) + I₂(s) has E^∘_cell = 0-236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Solution:
The two half reactions are :
2Fe³⁺ + 2e^– → 2Fe²⁺ and 2I^– → I₂ + 2e^–
For the above reaction, n = 2

How to calculate emf of a cell PDF available in my website.

Question 7.
Why does the conductivity of a solution decrease with dilution?
Answer:
The number of ions per unit volume that carry the current in a solution decreases on dilution. Therefore the conductivity of a solution decreases with dilution.

Question 8.
Suggest a way to determine the \(\wedge_{m}^{0}\) value of water.
Answer:
Conductance of weak electrolytes can be determined by kohlraush’s law. Thus, molar conductance of water of infinite dilution can be determined by the molar conductances of NaOH, HCl and NaCl at infinite dilution.

Question 9.
The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1 Calculate its degree of dissociation and dissociation constant Given \(\lambda_{\left(\mathbf{H}^{+}\right)}^{\circ}\) = 349.6 S cm² mol^-1 and \(\lambda_{\left(\mathrm{H} \mathrm{COO}^{-}\right)}^{\circ}\) = 54.6 S cm² mol.
Solution:

Question 10.
If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire ?
Solution:
Q (coulomb) = 1 (ampere) × t (sec)
Q = 0.5 ampere × 2 × 60 × 60
= 3600 C
A flow of IF, i.e. 96500 C is equivalent to flow of 1 mole of electrons
i. e., = 6.023 × 10²³ electrons
3600 C is equivalent to flow of electrons

Question 11.
Suggest a list of metals that are extracted electrolytically.
Answer:
Li, Na, Mg, Ba, Ca, Al

Question 12.
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O^-2₇ ? Consider the reaction :

Answer:
1 mole Cr₂O^-2₇ requires 6 moles electrons for reduction.
∴ Required charge = 6F
= 6 × 96500 coulomb.
= 579000 coulomb.

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer:
Lead storage battery : It is a secondary cell i.e., a cell which is rechargeable because the products of cell reaction sticks to the electrode. It is also called as lead storage cells. It consists of six cells connected in series. Each cell consist of spongy lead anode and a grid of lead packed with lead dioxide (PbO₂) acts as cathode. An aqueous solution of H₂SO₄ (38% by mass) acts as electrolyte. The reactions which takes place at electrodes can be represented as:

Concentration of H₂SO₄ decreases as sulphate ions are consumed to form PbSO₄ during the working of the cell. As a result of this the density of solution also decreases.

Recharging the cell / battery: Lead storage battery can be recharged by connecting it to an external source of direct current. This reverses the flow of electron with the deposition of Pb on the anode and PbO₂ on the cathode. That is, during recharging operation the cell behaves as electrolytic cell. Following reaction occurs during recharging.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer:
Methanol (CH₃OH), propane (C₃H₈)

Question 15.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Answer:
Formation of carbonic acid takes place on the surface of iron

In presence of H⁺ ion, oxidation of iron takes place

The electrons are used at other spot where reduction takes place.

Overall reaction,

Electrochemistry NCERT Textbook Exercises

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
Answer:
A metal with lesser standard potential (more reactive) can displace the other metal from solution of its salts.

Question 2.
Given the standard electrode potentials, K⁺/K = -2.93V, Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = -0.74V. Arrange these metals in their increasing order of reducing power.
Answer:
The lower the reduction potential, the higher is the reducing power. Hence, the reducing power of the given metals increases in the following order.
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq)+ 2Ag_(s) takes place. Further show :
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Answer:

(i) Anode i.e., Zn electrode is negatively charged.
(ii) Electrons are current carriers.

Question 4.
Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

Calculate the ∆_rG°, and equilibrium constant of the reactions.
Solution:

Question 5.
Write the Nernst equation and emf of the following cells at 298 K :

Solution:

Question 6.
In the button cells widely used in watches and other devices the following reaction takes place:

Solution:

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:
Conductivity: The conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section.

Molar conductivity: Molar conductivity of a solution at a dilution (V) is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm³ of the solution when the electrodes are one cm apart and the area of cross-section of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by Λm.

Variation with concentration: The conductivity of a solution (Both for strong and weak electrolytes) decreases with a decrease in the concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increase in the decrease in the concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increases with decrease in the concentration of the electrolyte. This is because both the number of ions, as well as mobility of ions, increases with dilution. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

Question 8.
The conductivity of 0.20 M solution of KCI at 298 K is 0.0248 Scm^-1. Calculate its molar conductivity.
Solution:

Question 9.
The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500Ω What is the cell constant if the conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^-3 S cm^-1.
Solution:

Question 10.
The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below :
Concentration/M 0.001 0.010 0.020 0.050 0.100 10² × kSm^-1 1.237 11.85 23.15 55.53 106.74 for all concentrations and draw a plot between Λ_m and C^1/2 Find the value calculate Λ_m and C^1/2 Find the value calculate Λ_m of Λ⁰_m.
Answer:

Λ⁰_m = intercept on Λ_m axis = 124-0 S cm² mol^-1 (on extrapolation to zero concentration)

Question 11.
The conductivity of 0.00241 M acetic acid is 7.896 × 10^-5 S cm^-1. Calculate its molar conductivity and if Λ⁰_m for acetic acid is 390.5 S cm² mol^-1, what is its dissociation constant?
Solution:

Question 12.
How much charge is required for the following reductions :
(i) 1 mol of Al³⁺ to Al.
(ii) 1 mol of Cu²⁺ to Cu.
(iii) 1 mol of MnO^–₄ to Mn²⁺.
Solution:
(i) The electrode reaction is :

(ii) The electrode reaction is :

(iii) The electrode reaction is :

Question 13.
How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl₂.
(ii) 40.0 g of Al from molten Al₂O₃.
Solution:

Question 14.
How much electricity is required in coulomb for the oxidation of
1. 1 mole of H₂O to O₂
2. 1 mole of FeO to Fe₂O₃.
Answer:
1. 2H₂O → 4H⁺ + O₂ + 4e
2F of electricity is required for the oxidation of 1 mol of H₂O

2. Fe²⁺ → Fe³⁺ + e
1F of electricity is required for the oxidation of 1 mole of FeO

Question 15.
A solution of Ni(NO₃)₂ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Answer:
According to Faraday’s first law:
W = ZIr [z = \(\frac { M }{ nF } \)]

Question 16.
Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO_3, and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Solution:

Question 17.
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible :
(i) Fe³⁺_(aq) and I^–_(aq)
(ii) Ag⁺_(aq) and Cu_(s)
(iii) Fe³⁺_(aq) and Br^–_(aq)
(iv) Ag_(s) and Fe³⁺_(aq)
(v) Br_2(aq) and Fe²⁺_(aq)
Answer:
A reaction is feasible if EMF of the cell is +ve.
Cathode : At which reduction occurs.
Anode : At which oxidation occurs.

Question 18.
Predict the products of electrolysis in each of the following:
(a) An aqueous solution of AgNO₃ with silver electrodes.
(b) An aqueous solution of AgNO₃ with platinum electrodes.
(c) A dilute solution of H₂SO₄ with platinum electrodes.
(d) An aqueous solution of CuCl₂ with platinum electrodes.
Answer:
(i) At cathode:
The following reduction reactions compete to take place at the cathode
Ag⁺(aq) + e^– → Ag(s); E^θ = 0.80 V
H⁺ (aq) + e^– → \(\frac { 1 }{ 2 }\) H₂ (g); E^θ = 0.00V
The reaction with a higher value of E_θ takes place of the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:
The Ag anode is attacked by NO₃^– ions. Therefore, the silver electrode at the anode dissolves in the solution to from Ag⁺.

(ii) At cathode: Same as above
At anode: Anode is not attackable and hence OH^– ions have lower discharge potential than NO₃^– ions and OH^– ions react to give O₂
OH^– → OH + e^–
4OH → 2H₂O + O₂ (g)
(iii) H₂SO₄ → 2H⁺ + SO^2-₄
HO₂ ⇌ H⁺ + OH^–

At cathode:
2H⁺+ 2e^– → H₂
At anode: 4OH^– → 2H₂O + O₂ + 4e^–
i. e., H₂ will be liberated at cathode and O₂ at the anode.

(iv) CuCl₂ → Cu²⁺+2Cl^–

At Cathode: Cu²⁺ ions will be reduced in preference to H⁺ ions
Cu²⁺ + 2e → Cu
At anode: Cl’ ions will be oxidised in preference to OH^– ions.
2Cl^– → Cl₂ + 2e^–
i.e., Cu will be deposited on the cathode and Cl₂ will be liberated at the anode.

Electrochemistry Other Important Questions and Answers

Electrochemistry Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
If specific conductance = K, R = Resistance, l = distance between the electrodes A = Cross-sectional area of a conductor C_m = mol L^-1, C_eq = g.eq L^-1 then specific resistance of electrolyte will be equal to :

Question 2.
K is equal to :

Question 3.
Molar conductivity Λ_m is equal to :

Question 4.
Cell constant is :

Question 5.
For electrolyte (v₊ = v_– = 1) like NaCI:

Question 6.
Which among the following is not a conductor of electricity :
(a) NaCl_(aq)
(b) NaCl_(s)
(c) NaCl_(mol)
(d) Ag.

Question 7.
By which of the following the resistance of the solution is multiplied to obtain cell constant:
(a) Specific conductance (K^–)
(b) Molar conductance (Λ_m)
(c) Equivalent conductance (Λ_eq)
(d) None of these.

Question 8.
Increase in equivalent conductance of the solution of an electrolyte by dilution is due to:
(a) Increase in ionic attraction
(b) Increase in molecular attraction
(c) Increase in association of electrolyte
(d) Increase in ionization of electrolyte.

Question 9.
If the specific conductance and observed conductance of an electrolyte is same then its cell constant will be :
(a) 1
(b) 0
(c) 10
(d) 1000

Question 10.
Unit of cell constant is :
(a) ohm^-1 cm^-1
(b) cm
(c) ohm cm
(d) cm^-1.

Question 11.
Unit of specific conductance is :
(a) ohm^-1
(b) ohm^-1 cm^-1
(c) ohm^-2 cm^-1 equivalent^-1
(d) ohm^-1 cm^-2.

Question 12.
If the concentration of any solution is C gram equivalent/ litre and specific resistance is A, its equivalent conductance will be :

Question 13.
The coating of layer of zinc on iron to prevent it from corrosion is called :
(a) Galvanization
(b) Cathodic protection
(c) Electrolysis
(d) Photoelectrolysis.

Question 14.
A saturated solution of KNO₃ is used for salt bridge because :
(a) Speed of K⁺ is more than NO^–₃
(b) Speed of NO^–₃ is more than K⁺
(c) Speed of both is nearly same
(d) Solubility of KNO₃ is high in water.

Question 15.
Acts as dipolar in dry cells :
(a) NH₄Cl
(b) Na₂CO₃
(c) pbSO₄
(d) MnO₂

Question 16.
Reduction is known as :
(a) Electronation
(b) De-electronation
(c) Protonation
(d) De-protonation.

Question 17.
What is the path of electric current in a Daniel cell when Zn and Cu electrodes are connected:
(a) From Cu to Zn inside the cell
(b) From Cu to Zn outside the cell
(c) From Zn to Cu inside the cell
(d) From Zn to Cu outside the cell.

Question 18.
In a cell containing Zn electrode and normal hydrogen electrode (NHE), Zn acts like:
(a) Anode
(b) Cathode
(c) Neither cathode nor anode
(d) Both anode and cathode.

Question 19.
If salt bridge is removed from half cells then voltage :
(a) Reduces and becomes zero
(b) Increases
(c) Immediately increases
(d) Does not change.

Question 20.
When lead storage battery is discharged :
(a) SO₂ is released
(b) Pb is manufactured
(c) PbSO₄ is used
(d) H₂SO₄ is used.

Question 21.
Process of Rusting of iron is :
(a) Oxidation
(b) Reduction
(c) Corrosion
(d) Polymerisation.

Question 22.
Value of standard potential of hydrogen electrode is :
(a) Positive
(b) Negative
(c) Zero
(d) No definite value.

Answers:
1. (b), 2. (c), 3. (d), 4. (b), 5. (a), 6 (b), 7. (a), 8. (d), 9. (a), 10. (d), 11. (b), 12. (a), 13. (a), 14. (c), 15. (d), 16. (a), 17. (d), 18. (a), 19. (a), 20. (d), 21. (c), 22. (c).

Question 2.
Fill in the blanks :

1. Acetic acid is a ………………… electrolyte.
2. Conductance of electrolyte ………………… with the increase in temperature.
3. On increasing dilution the value of specific conductance of a solution …………………
4. ………………… decreases with increase in size of ion.
5. Unit of specific resistance is …………………
6. Primary cells cannot be ………………… again.
7. Device which converts chemical energy into electrical energy is known as ………………… cell.
8. Amount of electric current which produces one gram equivalent of a substance is known as …………………
9. In metallic conduction ………………… property remains unchanged.
10. Reciprocal of resistance is known as …………………
11. Conductance of 1 cm cube of a conductor is called …………………
12. 1 Faraday is equal to ………………… coulomb.
13. Rusting of iron is an example of …………………
14. Potential of standard hydrogen is assumed to be …………………

Answers:

1. Weak
2. Increases
3. Decreases
4. Conductance
5. ohm cm
6. Charged
7. Electrochemical
8. Faraday
9. Chemical property
10. Conductance
11. Specific conductance
12. 96500 coulomb
13. Corrosion process
14. 0.0 volt
15. Electrochemical cell.

Question 3.
Match the following :

I.

Answers:

1. (e)
2. (c)
3. (d)
4. (b)
5. (a).

II.


Answer:

1. (c)
2. (d)
3. (b)
4. (a)
5. (e)
6. (f).

Question 4.
Answer in one word / sentence :

1. Give two examples of strong electrolyte.
2. Give two examples of weak electrolytes.
3. What is the effect of temperature on electrolytic conductivity ?
4. Write the formula of Kohlrausch’s law.
5. Write the formula of equivalent conductance.
6. Write the formula of molar conductance.
7. Device which converts electrical energy into chemical energy.
8. What is the potential of both the electrodes of the cell due to which electric current flows in the cell ?
9. What is the potential produced due to redox reaction between a metal electrode and its ions called ?
10. What is the unit of potential difference ?
11. Write the formula of cell constant.
12. Cell which can be recharged are known as.
13. State the unit of Equivalent conductance.
14. What is the chemical composition of rust ?
15. Write the relation between Electromotive force and Equilibrium constant of a cell.
16. What is the name of the reaction in which oxidation and reduction occur simultaneously ?
Answers:
1. Strong electrolyte : HCl, NaOH, NaCl
2. Weak electrolyte : CH₃COOH, H₂CO₃,
3. Conductivity increases
4.

5. Equivalent conductance

6.

7. Electrolytic cell
8. Electromotive force
9. Electrode potential
10. Volt
11. Cell constant = \(\frac { l }{ A } \)
12. Secondary cell
13. Ohm cm² gm eq^-1
14. Fe₂O₃ . xH₂O
15.

16. Redox reaction.

Electrochemistry Very Short Answer Type Questions

Question 1.
Write the definition of Electrochemical cell.
Answer:
System in which chemical energy is converted to electrical energy by oxidation reduction is known as electrochemical cell or voltaic cell.

Question 2.
What is an Electrolytic cell ?
Ans.
The container or system in which electrical energy is passed by which chemical reaction takes place thus, electrical energy is converted to chemical energy is known as electrolytic cell.

Question 3.
What is Electrode potential ?
Answer:
The potential difference developed between the electrodes and electrolyte of an Electrolytic cell is known as Electrode potential.

Question 4.
What is a strong electrolyte ? Write two examples.
Answer:
Electrolyte which completely dissociate in aqueous solution are known as strong electrolyte.
Example : NaCl, KCl, NH₄Cl etc.

Question 5.
What is a weak Electrolyte ? Write two examples.
Answer:
Electrolyte which dissociate partially in aqueous solution are known as weak electrolyte.
Example : NH₄OH, CH₃COOH, HCN etc.

Question 6.
What is meant by standard electrode potential ?
Answer:
Standard electrode potential (E°) of a half cell is the potential difference when one electrode is dipped in molar solution of its ion at 298 K. If electrode is gaseous the pressure of gas must be one atmosphere.

In IUPAC system, reduction potential are known as standard electrode potential.

Question 7.
Write Ohm’s law.
Answer:
According to Ohm’s law “It states that potential difference across the conductor is directly proportional to the current (I) flowing through it” i.e.,
Mathematically, it can be written as :
I α V
V = IR (R = Resistance, unit = ohm, Ω)

Question 8.
What is cell constant ?
Answer:
For a conductivity cell, the ratio of distance between two electrodes (l) and area of cross-section of electrode (A) is called as cell constant.
∴ Cell constant = \(\frac { l }{ A } \) or x = \(\frac { l }{ A } \)
Unit of cell constant = cm^-1.

Question 9.
What is galvanization ? Explain.
Answer:
Iron is coated with the layer of zinc to protect it from rusting. This process is known as galvanization. The galvanized iron articles keep their lustre due to the coating of invisible protective layer of basic zinc carbonate, ZnCO₃.Zn(OH)₂.

Question 10.
What is Electrochemical Equivalent ?
Answer:
Electrochemical equivalent of a substance is that mass of substance released or deposited on an electrode when a current of one ampere is passed for one second.

Electrochemistry Short Answer Type Questions

Question 1.
What is salt bridge ? Write its two functions.
Answer:
‘U’ shaped tube filled with KCl or KNO₃ in Agar-Agar solution or gelatin, is known as salt bridge. It connects the two half cell.
Functions : (i) It allows the flow of current by completing the circuit.
(ii) It maintains the electrical neutrality.

Question 2.
Derive relation between standard electromotive force and equilibrium constant
Answer:
Relation between standard electromotive force and equilibrium constant can be derived using van’t Hoff isochore. For any given reaction equilibrium constant IQ is equal to the ratio of rate constant of forward reaction and rate constant of backward reaction.

Value of equilibrium constant Kc can be calculated using standard free energy change (∆G°) because

Question 3.
What do you understand by oxidation-reduction reactions ?
Answer:
Oxidation-Reduction reactions: Chemical reactions in which valency of elements changes are known as oxidation-reduction reactions. In this process both oxidation and reduction reactions occur simultaneously, in which one of the substance is oxidized and the other substance is reduced. Like

In this reaction FeCl₃ is reduced to FeCl₂ and SnCl₂ is oxidized to SnCl₄. In the reaction valency of Fe decreases and valency of Sn increases.

Question 4.
Write difference between Metallic conduction and Electrolytic conduction.
Answer:
Differences between Metallic conduction and Electrolytic conduction :

Metallic conduction:

1. Metallic conduction takes place by movement of electrons.
2. There is no chemical change.
3. There is no transfer of matter.
4. In metallic conduction conductivity decreases with increase in temperature.

Electrolytic conduction

1. Electrolytic conduction takes place by movement of ions.
2. Due to chemical change decomposition of electrolyte takes place.
3. Transfer of matter takes place as ions.
4. In electrolytic conduction conductivity increases with increase in temperature.

Question 5.
What are the difference between emf (Cell potential) and potential difference
Answer:
Difference between EMF and Potential difference :

EMF / Cell potential:

1. It is the potential difference between the two terminals of the cell when no current is flowing in the circuit, i.e., in an open circuit.
2. It is the maximum voltage which can be obtained from a cell.
3. It can be measured by potentiaometrie method.
4. Work performed by electromotive force is the maximum work done by a cell.
5. It is responsible for continuous flow of current in electric circuit.

Potential difference:

1. It is the difference of the electrodes potentials of the two electrodes when the cell is sending current through the circuit.
2. It is the less than the maximum voltage as it is the difference of electrode potential.
3. It can be measured by simple voltmeter also.
4. Work performed by potential difference is less than the maximum work done by a cell.
5. It is not responsible for the continuous flow of current in circuit.

Question 6.
What is specific conductance ? Give its unit.
Answer:
Specific conductivity: The reciprocal of resistivity is called specific conductiv¬ity. It is defined as the conductance between the opposite faces of one centimeter cube of a conductor. It is denoted by K (kappa).


The specific conductivity of a solution at a given dilution is the conductance of one cm cube of the solution. It is represented by K (kappa).
Note : The specific conductivity of a solu¬tion of electrolyte depends upon the dilution or molar concentration of the solution.

Question 7.
What is resistivity of any solution ?
Answer:
Resistivity : When current flow in the solution through two electrodes the resistance is proportional to length and inversely proportional to cross-sectional area A.

The constant p (rho) is called resistivity or specific resistance.
Unit: If l is expressed in cm, A in cm2 and R in ohm, the unit of resistivity will be

or Resistivity of any solution is the resistance of 1 cm cube.

Question 8.
Differentiate between Electrochemical cell (Galvanic cell) and Electrolytic cell.
Answer:
Differences between Electrochemical and Electrolytic cells :

Electrochemical cell or Galvanic cell:

1. It is a device to convert chemical energy into electrical energy.
2. It consists of two electrodes in different compartments joined by a salt bridge.
3. Redox reactions occurring in the cell are spontaneous.
4. Free energy decreases with operation of cell, i.e., ∆G < 0.
5. Useful work is obtained from the cell.
6. Anode works as negative and cathode as positive electrodes.
7. Electrons released by oxidation process at anode go into external circuit and pass to cathode.
8. To set-up this cell, a salt bridge/porous pot is used.

Electrolytic cell:

1. It is a device to convert electrical energy into chemical energy.
2. Both the electrodes are in same solution.
3. Redox reactions occurring in the cell are non-spontaneous.
4. Free energy increases with operation of cell, i.e., ∆G > 0.
5. Work is done on the system.
6. Anode is positive and cathode is negative.
7. Electrons enter into cathode electrode from external source and leave the cell at anode.
8. No salt bridge is used in this cell.

Question 9.
What is equivalent conductance ?
Answer:
Equivalent conductance:
“Conductance of total ion produced by one gram equivalent of electrolyte in the solution is called equivalent conductance.” It is denoted by Λ_eq.

Question 10.
What is molar conductance ?
Answer:
Molar Conductivity : The molar conductivity of a solution at definite concentration (or dilution) and temperature is the conductivity of that volume which contains one mole of the solute and is placed between two parallel electrodes 1 cm apart and having sufficient area to hold whole of the solution. It is denoted by Λ_m.
Mathematically,
Λ_m = K × V …(1)
Where V is the volume in ml in which one gram mole of substance is dissolved.
If M is molarity or m moles are dissolved in 1000 ml.

Question 11.
Define cell constant Develop a relation between specific conductance and cell constant.
Answer:
Cell constant: In any conductive cell, the distance between two electrodes and surface area of electrode A are constant. The ratio of l and a is called cell constant i.e.

Unit of cell constant is cm^–1 and it is generally expressed by x.
Relation between specific conductance and cell constant : For a conductor, the resistance R is directly proportional to length R and inversely proportional to area of cross – section of electrolyte.

Question 12.
What are the factors which influence the electrical conductance of electrolytes ?
Answer:
Factor which influence electrical conductivity of electrolytes :

The main factor which influence the electrical conductivity are following:

1. Temperature: It influence following interactions.
(a) Interionic attractions: It depends upon the solute-solute interactions. Which is found between the ions of solute.
(b) Solvation of ions: It depends upon solute-solvent interactions. It is relation between ions of solute and solvent molecules.
(c) Viscosity of solvent: It depends upon solvent-solvent interactions. Solvent molecules are related with each other.
With increase in temperature all these three effects decrease and average kinetic energy of ions increases. Thus, with increase of temperature, resistance of solution decreases and hence conductance increases.

2. Nature of electrolyte : The conductance of solution depends upon the nature of electrolyte. On the basis of conductance measurement electrolytes are classified as strong electrolyte and weak electrolyte. Strong electrolytes have high value of conductance even at higher concentration also.

3. Dilution or concentration : It is main factor which influence electrical conductance. Effect of dilution or concentration can be studied indivisually in equivalent conductance, specific conductance and molar conductance. But for a general concept of electrical conductance of solution as the concentration is lowered or dilution increases, electrical conductance of whole solution increases.

Question 13.
On what factors does the various conductivities of an electrolytic solution depend ?
Answer:
Conductivities of electrolytic solution depend on the following factors :

1. Dilution : On increasing dilution, value of specific conductance of a solution decreases, value of equivalent conductance and molar conductance increases.
2. Nature of solvent: A solvent with high dielectric constant has high conductivity and with low dielectric constant has low conductivity.
3. Number of ions present in solution: Conductivity of strong electrolytes is higher than the conductivity of weak electrolytes.
4. Size of ion : In aqueous solution, small ions are heavily hydrated due to which their conductivity decreases.
5. Effect of Temperature: With the increase in temperature conductivity increases.

Question 14.
With the increase in dilution how do specific conductance, Equivalent conductance and molar conductance change ?
Answer:
With the increase in dilution, specific conductance decreases. This is because by the increase in dilution number of ions present in 1 cm cube of solution decreases.
But, Equivalent conductance Λ_eq = K × V
and Molar conductance Λ_m = K × V
Equivalent conductance and molar conductance are the product of specific conductance and dilution. By the increase in dilution (or decrease in concentration) magnitude of K decreases but that of V increases.
Increase in magnitude of V is comparatively much more than the decrease in magnitude of K. Thus, by the combined effect of both, by the increase in dilution Λ_eq and Λ_m increases.

Question 15.
What is an Electrolytic cell and how does it work ?
Answer:
Electrolytic cells : In these cells electric current is supplied through an external source, as a result of which chemical reactions take place which is called electrolysis like : Electrolysis of water, NaCl, Al₂O₃ etc. For example in Solvay trough cell electrode is immersed in sodium chloride solution and electric current is passed due to which NaCl electrolyses.

At mercury cathode sodium is released and at anode chlorine is released. Sodium forms amalgam with mercury and is taken out of the cell.

Question 16.
What is meant by electromotive force of an electrochemical cell ?
Answer:
The difference in electrode potentials of the two electrodes of an electro- chemical cell is known as electromotive force or cell potential. It is expressed in volt.

Due to difference in potential electric current flows from an electrode of lower potential to an electrode of higher potential. EMF of the cell can be expressed in terms of reduction potential as :

Cell potential = Standard electrode potential – Standard electrode potential
of R.H.S. electrode of L.H.S electrode

EMF of a cell is measured by connecting the voltmeter between the two electrodes of a cell. EMF of a cell depend on the concentration of solutions of both half cells and nature of the two electrodes. For example, In Daniel cell, concentration of CuSO₄ and ZnSO₄ solutions in the two half cells is 1M and at 298 K EMF of the cell is 1.10 volt.

Electrochemistry Long Answer Type Questions

Question 1.
What is standard hydrogen electrode ? How is it prepared ?
Answer:
Standard hydrogen electrode : This consists of gas at 1 atmospheric pressure bubbling over a platinum electrode immersed in 1 M HC1 at 25°C (298 K) as shown in figure. The platinum electrode is coated with platinum black to increase its surface. The hydrogen electrode thus con¬structed forms a half cell which on coupling with any other half cell begins to work on the principle of oxidation or reduction. Electrode depending upon the circumstances works both as anode or cathode.

Standard hydrogen electrode (SHE) is arbitrarily assigned a potential of zero.

Question 2.
Derive Nernst Equation for single electrode potential.
Answer:
Value of standard electrode potential given in electrochemical series is applicable only when the concentration of electrolyte is 1M and temperature is 298 K. But in electrochemical cells the concentration of electrolyte is not definite and electrode potential depends on concentration and temperature. In such condition single electrode potential can be expressed by Nernst equation.
For a reduction half reaction, Nernst equation can be expressed as follows :

Where E = Reduction electrode potential
E° = Standard electrode potential (Mⁿ⁺ concentration 1M and at 298 K)
R = Gas constant = 8.31 JK^-1 mol^-1 = Temperature (in kelvin) = 298 K
n = Valency of metal ion, F = 1 Faraday (96,500 coulomb)

Equation (2) is Nernst equation for single electrode potential.

Question 3.
Write the Faraday’s laws of electrolysis.
Answer:
Faraday’s first law of electrolysis : The law states that, “The mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed.”
Thus, if W gm of the substance is deposited on passing Q coulomb of electricity, then
W α Q or W = ZQ
Where, Z is a constant of proportionality and is called electrochemical equivalent of the substance deposited. If a current of I ampere is passed for t second, then Q = I × t. So that,
W = Z × Q = Z × I × t
Thus, if Q = 1 coulomb, I = 1 ampere and t = 1 second, then W = Z. Hence, electrochemical equivalent of a substance may be defined as, “The mass of the substance deposited when a current of one ampere is passed for one second.”

As one faraday (96500 C) deposits one gram equivalent of the substance, hence electrochemical equivalent can be calculated from the equivalent mass.
Faraday’s second law of electrolysis : It states that, “When the same quantity of electricity is passed through solutions of different electrolytes connected in series, the weight of the substances produced at the electrodes are directly proportional to their equivalent mass.”

For example, for CuSO₄ solution and AgNO₃ solution connected in series, if the same quantity of electricity is passed, then

Question 4.
What is rusting of iron ? Describe Electrochemical theory of rusting.
Answer:
Corrosion: Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric condition are called corrosion. Rusting of iron is an example of corrosion, chemically it is Fe₂O₃xH₂O.

Electrochemical theory of rusting :
Anode reaction : On one spot of iron sheet, oxidation takes place and this spot behaves as an anode.

The electrons which are released at this spot travel through the metal and reach another spot on the metal which acts as cathode. These electrons cause the reduction of oxygen in the presence of hydrogen ions (H⁺). H⁺ ions are formed due to decomposition of carbonic acid formed by dissolution of CO₂ in H₂O.

Question 5.
What is corrosion ? Write three factors affecting it and any three methods to prevent it.
Answer:
Corrosion : Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric conditions are called corrosion.

Factors affecting corrosion :

1. Nature of metal: Reactive metal corrodes readily.
2. Impurities in metal: Impure metal corrodes quickly to a greater extent.
3. Environment: If environment around metal contains oxygens, carbon dioxide, moisture, salts and acidic gases like CO₂, SO₂, SO₃ etc. then corrosion occurs quickly.

Methods to prevent corrosion :

1. Barrier protection : In this method, the surface of the metal is coated with paint, oil or grease. Due to this the surface of the metal remain unexposed to atmospheric conditions and hence corrosion is prevented. Surface of the metal can also be protected by :
(i) Coating metal surface with non-corroding metals like nickel and chromium is called electroplating.
(ii) Dipping iron article in molten metal like zinc. This process is called galvanization.

2. Sacrificial protection : The rusting of iron can be prevented by covering the iron with more electropositive metals like zinc. Zinc metal has more tendency to get oxidized as compared to iron. So, iron articles will not be harmed till the layer of zinc present on its surface hence zinc metal is called the sacrificial metal.

3. Antirust solution : The alkaline antirust solution are employed to prevent rusting, alkaline solution prevent the availability of H⁺ ions.

In this method, iron articles are dipped in alkaline sodium phosphate or chromate solution. Due to this an insoluble sticking film of iron phosphate is formed on the surface which prevents rusting.

Question 6.
Describe dry cell with labelled diagram.
Answer:
Dry cell: It is a primary cell based on Leclanche cell invented by G. Leclanche in 1868. In a primary cell, the electrode reactions cannot be reversed by an external source of electrical energy. In this cell, the cell reaction takes place only once i.e., this cell is not rechargeable.

It is generally used in torches, transistors, radios, calculators, tape recorders, etc. It consists of a hollow zinc cylinder which is filled with a paste of NH₄Cl and a little ZnCl₂. This paste is made with the help of water. The zinc cylinder acts as anode while cathode is a graphite rod (Carbon). The carbon rod is surrounded by a black paste of MnO₂ and carbon powder. The zinc cylinder has an outer insulation of cardboard case.

Dry cells are sealed with wax or other material to protect the moisture from evaporation. When the electrodes are connected, the cell operates.

The electrode reactions are complex. Metallic Zn is oxidized to Zn²⁺ and the electrons liberated are left on the container. The reactions which take place at electrodes can be represented as :

This reaction prevents polarization due to formation of ammonia. It also prevents the substantiaL increase of concentration of Zn²⁺ ions which would decrease the cell potential. This potential of dry cell is approximately 1.5 V.

Defect: Due to acidic nature of NH₄Cl zinc container corrodes due to which holes develop through which the chemicals come out.

Nowadays, the cells are made leakage resistant. In it KOH is used in place of NH₄Cl by which zinc does not corrode.

Question 7.
What is Kohlrausch law ? Give its two applications.
Answer:
Kohlrausch in 1875 gave a generalisation known as Kohlrausch’s law, “At infinite dilution when the dissociation of the electrolyte is complete, each ion makes a definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion with which it is associated.”
Or
“The value of molar conductance at infinite dilution is given by the sum of the contributions of ions (cation and anion).”

Where, λ^∞₊ and λ^∞_– are ionic contributions or ionic conductances of cation and anion while v₊ and v_– are the number of cations and anions in the formula unit of electrolyte.

Applications of Kohlrausch’s law :

(i) Calculation of molar conductance at infinite dilution for weak electrolytes :

Molar conductance or equivalent conductance of weak electrolytes cannot be obtained graphically by extrapolation method, since these are feebly ionized. Kohlrausch’s law enables indirect evaluation in such cases. For example, molar conductances of acetic acid can be obtained from the knowledge of molar conductances at infinite dilution of HCl, CH₃COONa and NaCl which are strong electrolytes.

From Kohlrausch’s law, it is clear that

(ii) Determination of degree of dissociation :

Question 8.
Draw a labelled diagram of Daniel cell and explain cell reaction.
Or,
Draw a labelled diagram of electrochemical cell and write cell reaction.
Answer:
Electrochemical cell: In the redox reactions, the transfer of electrons between oxidizing and reducing agents occurs through wire and thus chemical energy changes into electrical energy. The device on which chemical energy changes into electrical energy is called electro chemical cell. These are also known as galvanic or voltaic cells. Working of these cells can be understood with the example of Daniel cell.

Daniel cell : In this cell, Zn rod is dipped in ZnSO4 solution and Cu rod in copper sulphate solution. Both solutions are connected through KC1 salt bridge. When Zn and Cu electrodes are connected by wire and galvanometer, flow of electrons from Zn to Cu occurs. Zinc atoms change into Zn2+ and electrons reach at Cu electrode, where Cu2+ changes into Cu metal and this copper deposits on electrode.

Electrochemistry Numerical Questions

Question 1.
On passing 5 ampere electric current for 30 minute through a container filled with AgNO₃ 10.07 gram silver is deposited, then determine the chemical equivalent of silver. If electrochemical equivalent of hydrogen is 0-00001036 then calculate the equivalent mass of Silver.
Solution:
Given : W = 10-07 gm, i = 5 ampere, t = 30 × 60 second
According to Faraday’s first law W = Zit
∴ 10.7 = Z × 30 × 60

Question 2.
What are weak electrolytes ? Give one example. Find out molar conductivity of LiBr aqueous solution infinite dilution when joint conductance of Li⁺ ion and Br^– ion are 38.7 Scm² mol^-1 and 78.40 Scm² mol^-1 respectively.
Solution:
Weak electrolytes : These are the substances which dissociate only to a small extent.
Examples: CH₃COOH,NH₄OH

Question 3.
What are strong electrolytes ? Find out the molar conductivity of aqueous solution of BaCl₂ at infinite dilution when ionic conductance of Ba⁺² ion and Cl^– ion are 127.30 Scm² mol^-1 and 76.34 Scm² mol^-1 respectively.
Solution:
Strong electrolytes : These are substances which dissociate almost completely into ions under all dilutions.
Examples: NaCl, HCl,CH₃COONa

Question 4.
Calculate the molar conductance of Al₂(SO₄)₃ if


Solution:
By the ionisation of Al₂(SO₄)₃ solution

Question 5.
Molar conductivity of \(\frac { M }{ 30 } \) CH₃COOH is 9.625 mho and molar conductance of CH₃COOH at infinite dilution (Λ^∞_m) is 385 mho. Calculate the percentage of dissociation of \(\frac { M }{ 30 } \) CH₃COOH.
Solution:
Degree of dissociation

Question 6.
Calculate molar conductance of acetic acid at infinite dilution from following values:

Solution:
From Kohlrausch’s law,

MP Board Class 12th Chemistry Solutions