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MCQMediumJEE 2024Faraday's Laws of Electrolysis

JEE Chemistry 2024 Question with Solution

A constant current was passed through a solution of AuCl4−\text{AuCl}_4^- ions between gold electrodes. After 10 minutes10 \, \text{minutes}, the increase in the cathode’s mass was 1.314 g1.314 \, \text{g}. The total charge passed through the solution is:

  • A

    2 F2 \, \text{F}

  • B

    4 F4 \, \text{F}

  • C

    6 F6 \, \text{F}

  • D

    8 F8 \, \text{F}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Mass of gold deposited at the cathode is 1.314 g1.314 \, \text{g}. The ion reduced is AuCl4−\text{AuCl}_4^-.

Find: Total charge passed through the solution in Faradays.

At the cathode, the reduction reaction is

AuCl4−+3e−→Au+4Cl−\text{AuCl}_4^- + 3e^- \rightarrow \text{Au} + 4\text{Cl}^-

So, deposition of 11 mole of Au requires 33 moles of electrons.

Moles of gold deposited:

moles of Au=1.314197=0.00667\text{moles of Au} = \frac{1.314}{197} = 0.00667

Moles of electrons transferred:

n=3×0.00667=0.02001n = 3 \times 0.00667 = 0.02001

Since 11 Faraday corresponds to 11 mole of electrons, the charge passed is

Q=0.02001 F≈2×10−2 FQ = 0.02001 \, \text{F} \approx 2 \times 10^{-2} \, \text{F}

Therefore, the total charge passed is 2×10−2 F2 \times 10^{-2} \, \text{F}. The solution concludes this value, but the listed options are 2 F2 \, \text{F}, 4 F4 \, \text{F}, 6 F6 \, \text{F}, and 8 F8 \, \text{F}, which do not contain the computed answer. Using the solution's’s marked correct option, the most defensible mapped answer is A.

Equivalent Weight Shortcut

Given: Deposited mass of gold is 1.314 g1.314 \, \text{g}.

Find: Charge in Faradays.

Using Faraday’s law directly,

WE=Q1 F\frac{W}{E} = \frac{Q}{1 \, \text{F}}

where the equivalent weight of gold for Au3+\text{Au}^{3+} deposition is

E=1973E = \frac{197}{3}

Substitute the values:

Q=1.314197/3 FQ = \frac{1.314}{197/3} \, \text{F} Q=2×10−2 FQ = 2 \times 10^{-2} \, \text{F}

This shortcut works because equivalent weight already accounts for the 33 electrons needed per atom of gold deposited. Hence the charge passed is 2×10−2 F2 \times 10^{-2} \, \text{F}.

Common mistakes

  • A common mistake is to treat AuCl4−\text{AuCl}_4^- as requiring 11 electron instead of 33 electrons. This is wrong because gold changes to elemental Au from oxidation state +3+3. Use the cathode half-reaction first, then count electrons correctly.

  • Students often compare the computed value 2×10−2 F2 \times 10^{-2} \, \text{F} with the options and ignore the exponent. This is wrong because 2×10−2 F2 \times 10^{-2} \, \text{F} is not equal to 2 F2 \, \text{F}. Keep powers of ten throughout the calculation and note the source-page discrepancy explicitly.

  • Another mistake is to use molar mass directly in Faraday’s law without converting to equivalent weight. That misses the electron factor. If using WE=Q1 F\frac{W}{E} = \frac{Q}{1 \, \text{F}}, first compute E=1973E = \frac{197}{3}, not 197197.

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