JEE Chemistry Electrochemistry MCQs Set 17

Answer: (c) \( E_{cell} \) is an intensive property while \( \Delta_r G \) of the cell reaction is an extensive property

Explanation:
1. \( \Delta_r G = -nFE_{cell} \), and n depends on the amount of reaction as written, so \( \Delta_r G \) is extensive.
2. \( E_{cell} \) does not change with the quantity of material, so it is intensive.

Teacher's Note:
1. Potentials are intensive. Gibbs energy, like the number of electrons n, is extensive.

Answer: (b) cell emf

Explanation:
1. The emf of a cell is the maximum potential difference measured when no current flows.
2. When current is drawn, the measured potential difference is lower than the emf.

Teacher's Note:
1. Emf is measured with a potentiometer or a high-resistance voltmeter, so that no current is drawn.

Answer: (b) Mn\(^{2+}\) < Cl\(^-\) < Cr\(^{3+}\) < Cr

Explanation:
1. Reducing power increases as the reduction potential of the couple decreases.
2. The couples have \( E^\circ \): Mn\(^{2+}\) (+1.51), Cl\(^-\) (+1.36), Cr\(^{3+}\) (+1.33) and Cr (\(-0.74\)).
3. So the order is Mn\(^{2+}\) < Cl\(^-\) < Cr\(^{3+}\) < Cr.

Teacher's Note:
1. Write each species with its couple, then rank by decreasing \( E^\circ \).

Answer: (a) Cr\(^{3+}\)

Explanation:
1. Among the oxidised forms in the data, the one in the couple with the lowest \( E^\circ \) is the hardest to reduce.
2. Cr\(^{3+}\) (couple \( Cr^{3+}/Cr \), \(-0.74\) V) is therefore the most stable.

Teacher's Note:
1. Most negative \( E^\circ \) means the oxidised form is the hardest to reduce and so the most stable.

Answer: (d) Cl\(^-\) \(\rightarrow\) 1/2 Cl\(_2\)(g) + e\(^-\); \( E^\circ = -1.36 \) V

Explanation:
1. Oxidation occurs at the anode, so options A and C (reductions) are ruled out.
2. On potentials alone water (1.23 V) would be oxidised more easily, but oxygen evolution has a very high overpotential.
3. In a reasonably concentrated NaCl solution chloride is oxidised, giving Cl\(_2\).

Teacher's Note:
1. Overpotential explains why chlorine, not oxygen, is obtained in the chlor-alkali process.

Answer: (c) Cu\(^{2+}\)/Cu is a stronger oxidising agent than H\(^+\)/H\(_2\)

Explanation:
1. \( E^\circ(Cu^{2+}/Cu) = +0.34 \) V is higher than \( E^\circ(H^+/H_2) = 0 \).
2. So Cu\(^{2+}\) is the stronger oxidising agent, and Cu cannot reduce H\(^+\), so it cannot displace hydrogen from acid.

Teacher's Note:
1. A metal displaces hydrogen from acids only if its \( E^\circ \) is negative.

Answer: (c) A and C

Explanation:
1. In dilute acid, H\(^+\) is reduced at the cathode (A is true).
2. At the anode water is oxidised, because its oxidation needs only 1.23 V against 1.96 V for sulphate (C is true).
3. In concentrated acid sulphate is oxidised to peroxodisulphate, so B is false.
4. The product is peroxodisulphate, not tetrathionate, so D is false.

Teacher's Note:
1. The species with the less negative oxidation potential (easier to oxidise) reacts at the anode.

Answer: (a) A and B only

Explanation:
1. Conductivity \( \kappa \) is the conductance of a unit cube of solution, so it depends on the nature and concentration of the electrolyte (and the temperature and solvent).
2. It is independent of the cell geometry (electrode distance) and of the power of the source.

Teacher's Note:
1. Conductivity is a property of the solution; conductance also depends on the cell geometry.

Answer: (c) A and B

Explanation:
1. Cu\(^{2+}\) is reduced and deposited at the cathode.
2. Copper is more easily oxidised than water (\( E^\circ_{ox} = -0.34 \) V against \(-1.23\) V), so the copper anode dissolves instead of oxygen being released.
3. This is the principle of electrolytic refining of copper.

Teacher's Note:
1. With an active electrode of the same metal, the anode dissolves and the cathode grows.

Answer: (a) \( 5 \times 10^{-4} \)

Explanation:
1. Cell constant \( = \kappa R = 1.4 \times 50 = 70 \) m\(^{-1}\).
2. For 0.5 M: \( \kappa = \frac{70}{280} = 0.25 \) S m\(^{-1}\).
3. \( \Lambda_m = \frac{\kappa}{C} = \frac{0.25}{500\ \text{mol m}^{-3}} = 5 \times 10^{-4} \) S m\(^2\) mol\(^{-1}\).

Teacher's Note:
1. In SI units, concentration must be in mol per cubic metre (1 M = 1000 mol m\(^{-3}\)).

Answer: (a) \(-2.69\) V, the reaction will not occur

Explanation:
1. Mn\(^{2+}\) is reduced to Mn at the cathode: \( E^\circ = -1.18 \) V.
2. Mn\(^{2+}\) is oxidised to Mn\(^{3+}\) at the anode: \( E^\circ_{ox} = -1.51 \) V.
3. \( E^\circ_{cell} = -1.18 + (-1.51) = -2.69 \) V. Being negative, the reaction does not occur spontaneously.

Teacher's Note:
1. Electrode potentials are intensive and are not multiplied by the number of moles in a balanced half-reaction.

Answer: (d) X = Zn, Y = Ni

Explanation:
1. \( E^\circ_{cell} = E^\circ(Y^{2+}/Y) - E^\circ(X^{2+}/X) \) must be positive.
2. For X = Zn, Y = Ni: \(-0.23 - (-0.76) = +0.53\) V.
3. The other combinations give \(-0.21\), \(-0.53\) and \(-0.32\) V.

Teacher's Note:
1. The metal with the more negative potential is the better reductant and displaces the other.

Answer: (d) 1.57 V

Explanation:
1. \( Q = \frac{[Fe^{2+}]^2}{p_{O_2}[H^+]^4} = \frac{(10^{-3})^2}{0.1 \times (10^{-3})^4} = 10^7 \).
2. n = 4, so \( E = 1.67 - \frac{0.0591}{4}\log 10^7 = 1.67 - 0.103 = 1.57 \) V.

Teacher's Note:
1. Put the concentrations of products over reactants (raised to their coefficients) into Q, and use n = 4 here.

Answer: (c) 2.5 V

Explanation:
1. The reaction releases one O\(_2\), which needs 4 electrons: \( 2O^{2-} \rightarrow O_2 + 4e^- \), so n = 4.
2. \( E = \frac{\Delta_r G}{nF} = \frac{966 \times 10^3}{4 \times 96500} = 2.5 \) V.

Teacher's Note:
1. Count n from the balanced equation as written, here four electrons per O\(_2\).

Answer: (c) 0.770 V

Explanation:
1. \( \Delta G^\circ_1 = -3F(-0.036) = +0.108F \) for Fe\(^{3+}\) + 3e\(^-\) \(\rightarrow\) Fe, and \( \Delta G^\circ_2 = -2F(-0.439) = +0.878F \) for Fe\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Fe.
2. The required reaction is (1) \(-\) (2): \( \Delta G^\circ_3 = 0.108F - 0.878F = -0.770F = -1 \times F \times E^\circ \).
3. So \( E^\circ = +0.770 \) V.

Teacher's Note:
1. Potentials cannot be subtracted directly when the number of electrons differs; combine free energies instead.

Answer: (a) \( \Lambda^\circ \) of NaCl

Explanation:
1. \( \Lambda^\circ(CH_3COOH) = \lambda(CH_3COO^-) + \lambda(H^+) \).
2. \( \Lambda^\circ(CH_3COONa) + \Lambda^\circ(HCl) - \Lambda^\circ(NaCl) = \lambda(CH_3COO^-) + \lambda(H^+) \).
3. So only \( \Lambda^\circ(NaCl) \) is needed.

Teacher's Note:
1. Choose the extra salt that removes the unwanted ions Na\(^+\) and Cl\(^-\).

Answer: (d) \( \Lambda^\circ(NaCl) \)

Explanation:
1. \( \Lambda^\circ(HOAc) = \Lambda^\circ(NaOAc) + \Lambda^\circ(HCl) - \Lambda^\circ(NaCl) \) by Kohlrausch's law.
2. So the value of NaCl is needed.

Teacher's Note:
1. Add and subtract salts so that Na\(^+\) and Cl\(^-\) cancel.

Answer: (d) \(-16.13\)

Explanation:
1. Reverse the first reaction: AgI + e\(^-\) \(\rightarrow\) Ag + I\(^-\), \( E^\circ = -0.152 \) V.
2. Add Ag \(\rightarrow\) Ag\(^+\) + e\(^-\) (\(-0.800\) V): AgI \(\rightarrow\) Ag\(^+\) + I\(^-\), \( E^\circ = -0.152 - 0.800 = -0.952 \) V.
3. \( \log K_{sp} = \frac{nE^\circ}{0.059} = \frac{-0.952}{0.059} = -16.13 \).

Teacher's Note:
1. The overall reaction is the dissolution of AgI, so its equilibrium constant is \( K_{sp} \).

Answer: (c) C, A

Explanation:
1. CN\(^-\) is a strong ligand and forms complexes such as [Ni(CN)\(_4\)]\(^{2-}\), so it shows the complexing property.
2. It also acts as a reducing agent, for example reducing Cu(II) to Cu(I) (while being oxidised to cyanogen).
3. It is not an oxidising agent.

Teacher's Note:
1. Cyanide is both a strong-field ligand and a reducing agent.

Answer: (b) create a potential difference between the two electrodes

Explanation:
1. In a fuel cell hydrogen is oxidised at the anode and oxygen is reduced at the cathode, in separate compartments.
2. The aim is to convert chemical energy directly into electrical energy through the potential difference created between the electrodes, not to produce heat.

Teacher's Note:
1. Fuel cells give electricity directly, with higher efficiency than burning fuel to make heat.