Electrochemistry

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

The conductivity in an ionic solution is due to the ions in the solution. It depends on the following factors;

(a)Temperature: On an increase in temperature, the molar conductivity of ionic solutions increases.

(b)Concentration of Electrolytes: An increase in the concentration of electrolytes decreases the molar conductivity as the number of ions per unit volume decreases.

Therefore, options (i, iii) are correct

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, ii)

Conductance is the reciprocal of resistance and conductivity is the reciprocal of resistivity. It can be written as follows

k = $\frac{1}{\mathbf{?}}$

R= $\mathbf{?}\mathbf{}\frac{1}{\mathbf{A}}$

$\mathbf{?}$ = $\frac{\mathbf{R}\mathbf{A}}{\mathbf{l}}$

k  = $\frac{1}{\frac{\mathbf{R}\mathbf{A}}{\mathbf{l}}}$

k = $\frac{1}{\mathbf{R}}\frac{\mathbf{l}}{\mathbf{A}}$

k = $\frac{1}{\mathbf{R}}$ G*

k = $\frac{{\mathbf{G}}^{\mathbf{*}}}{\mathbf{R}}$

G* is cell constant.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, ii)

In the electrolysis process of CuSO₄ the following reactions occur at the half cell which is as follows:

At the cathode, the reaction goes this way,

Cu²⁺+2e⁻→Cu (s)

At the anode, the reaction goes this way,

Cu (s)→Cu²⁺+2e−

Here copper will be deposited at the cathode and copper will dissolve at the anode.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

Electrolysis of copper sulfate solution is as follows

CuSO₄ ? Cu²⁺ + SO₄²⁻

H₂O? H⁺ + OH⁻

At the cathode, the reaction goes this way,

Cu²⁺ + 2e− → Cu; E? _Cell = 0.34V

H₂O⁻ → $\frac{1}{2}$ H₂ E? _Cell = 0.00V

At the anode, the reaction goes this way,

2SO₄^2- + 2e−→S₂O₈²⁻ E? _Cell =1.96V

2H₂O→O₂ + 4H⁺ + 4e⁻ E? _Cell =1.23V

The reaction will lower the value of E? _Cell is preferred at anode so the second reaction is feasible.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

 Kohlrausch law of states that limiting molar conductivity of any salt species is equal to the sum of the limiting molar conductivity of cations and anions of the electrolyte

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, ii)

Conductivity is due to the movement of ions in the solution. The conductivity of ions depends on the following factors:

(i) nature of electrolyte added

(ii) size of ion produced

(iii) concentration of electrolyte

(iv) nature of the solvent

(v) temperature

Distance between electrodes does not affect the conductivity of an electrolytic solution.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (ii, iii)

At equilibrium, ?G $°$  = -2.303 RT ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

-nFE $°$  = -2.303 RT ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

E $°$ = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{\mathrm{n}\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

For Daniel cell, n=2

E $°$ = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

At equilibrium, E $°$ = 1.1V

1.1V = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$

(i) Since ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$  this option is incorrect

(ii) As derived 1.1V = $\frac{2.303\mathrm{}\mathrm{R}\mathrm{T}}{2\mathrm{F}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}$

(iii) As derived _{${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$} so this option is correct

(iv) As ${\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{K}}_{\mathrm{c}}=\mathrm{}\frac{2.2}{0.059}$  _,so this option is incorrect

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: Option (i, iii)

In the electrolysis of sulphuric acid, the following reactions occur

2SO₄²⁻ (aq)→S₂O₈²⁻ (aq) + 2e⁻ E? _Cell =1.96V

2H₂O (l)→O₂ (g)+4H⁺ (aq) + 4e⁻ E? _Cell =1.23V

The reaction will lower the value of E? _Cell is preferred at anode so the second reaction is feasible.

H⁺ + e⁻ → $\frac{1}{2}$ H₂ E? _Cell = 0.00V

At the cathode, reduction of water occurs. Therefore, in dilute sulphuric acid solution, hydrogen will be reduced at the cathode.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (ii, iv)

The lesser the E? value of the redox couple higher is the reducing power

For Cu²⁺ + 2e⁻→Cu; E? =0.34V

For 2H⁺ + 2e⁻→H₂ E? =0.00V

Since the second redox couple has less standard reduction potential than the first so it can be concluded that the redox couple is a stronger oxidizing agent than H⁺/H₂ and copper cannot displace H₂ from acid.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct Option: (iv)

The electrolysis of aqueous sodium chloride is as follows

NaCl→Na⁺ + Cl⁻

H₂O→H⁺ + OH⁻

At the cathode, the reaction goes this way,

H₂O + e⁻ → $\frac{1}{2}$ H₂ + OH⁻

At the anode, the reaction goes this way,

Cl⁻ (aq)→ $\mathrm{}\frac{1}{2}$ Cl₂ (g)+ e⁻ E? _Cell =1.36V

2H₂O (g)→O₂ (g) + 4H⁺ (aq) + 4e⁻ E? _Cell =1.23V

At the anode, the reaction with a lower E? value will be preferred and oxidation of oxygen is a slow process and requires high voltage so the first reaction will take place.