Oswal 36 Sample Question Papers ISC Class 12 Chemistry Solutions

Answer 21.

(i) (a) The cell can be represented as :

Mg|Mg²⁺ (0.001 M) || Cu²⁺ (0.01 M)|Cu

$$\text{E}_{\text{cell}} =\text{E}\degree_{\text{cell}} -\frac{0.059}{n}\text{log}k_{e}$$

Putting the given values,

$$\text{E}_{\text{cell}} =\text{E}\degree_{\text{cell}} -\frac{0.059}{2}\text{log}\frac{10^{\normalsize-3}}{10^{\normalsize-2}}$$

2.71 + 0.0295

E_cell = 2.7395V

(1) When an external opposite potential is applied less than 2.71 V the direction of flow of current would remain same.

(2) When an external opposite potential is applied more than 2.71 V the direction of flow of current would be reversed.

(b) Reactions which will occur in the solution are:

$$\text{CuCl}_{2}(s)\xrightarrow{}\text{Cu}^{2+} +\text{2 Cl}^{\normalsize-}\\\text{H}_{2}\text{O}\xrightarrow{}\text{H}^{\normalsize+} +\text{OH}^{\normalsize-}$$

Reaction at cathode :

$$\text{H}_{2}\text{O}\xrightarrow{}\text{H}^{\normalsize+} +\text{OH}^{\normalsize-}$$

$$\text{Cu}^{2+} + 2e^{\normalsize-}\xrightarrow{}\text{Cu(s)}$$

because E° Cu²⁺/Cu > E° H⁺/H₂

Reaction at anode :

$$\text{2Cl}^{-}\xrightarrow{}\text{Cl}_{2} + 2e^{\normalsize-}$$

This reaction occurs at the anode due to over-potential of O₂ oxidation of Cl° is preferred.

Hence, Cu will deposit at cathode and Cl₂ gas will be liberated at anode.

OR

(ii) (a) (1) Molar conductivity is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution.

Λ_m = K/C

(2) Fuel cell is an electrochemical cell that converts the chemical energy of a fuel and an oxidizing agent into electricity.

(b) (1) The cell can be represented as:

$$\text{AI}|\text{AI}^{3+}(0.01 \text{M})||\text{Ni}^{2+}\space(0.1)|\text{Ni}$$

Cell potential is given by the following equation in this case:

$$\text{E}_{\text{cell}} =\text{E}\degree_{\text{cell}}-\\\frac{0.059}{6}\text{log}\frac{[\text{AI}^{+3}]^{2}}{[\text{Ni}^{+2}]^{3}}$$

Given:

E˚cell = 1.41 V, concentration of Al³⁺ ions is 0.01 M and Ni²⁺ ions is 0.1 M

Putting the values in equation

$$\text{E}_{cell} = 1.41\text{V}-\\\frac{0.059}{6}\text{log}\frac{[0.01]^{2}}{[0.1]^{3}}$$

= 1.4198 V

= 1.42 V

Hence, the emf of the cell is 1.42 V.

(b) For strong electrolytes, the molar conductivity is increased only slightly on dilution. A strong electrolyte is completely dissociated in solution and thus, furnishes all ions for conductance.

However, at higher concentrations, the dissociated ions are close to each other and thus, the interionic attractions are greater. These forces retard the motion of the ions and thus, conductivity is low. With decrease in concentration (dilution), the ions move away from each other thereby feeling less attractive forces from the counter ions. This results in an increase in molar conductivity with dilution. The molar conductivity approaches a maximum limiting value at infinite dilution designated as Λ_m⁰.

In case of weak electrolytes, as the solution of a weak electrolyte is diluted, its ionisation is increased.

This results in more number of ions in solution and thus, there is an increase in molar conductivity, also there is a large increase in the value of molar conductivity with dilution, especially near infinite dilution. However, the conductance of a weak electrolyte never approaches a limiting value. In other words it is not possible to find conductance at infinite dilution (zero concentration).

So, limiting molar conductivity for weak electrolytes is obtained by using Kohlrausch law, from the limiting molar conductivities of individual ions (λ°).

Kohlrausch law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte.

Λ°_m = V⁺λ°⁺ + V^–λ°^–