Faraday's first and second law of electrolysis — numerical problems

Question

A current of 5 A is passed through an aqueous solution of CuSO₄ for 32 minutes 10 seconds. Calculate the mass of copper deposited at the cathode. (Atomic mass of Cu = 63.5, 1 Faraday = 96500 C)

(NEET 2022, similar pattern)

Solution — Step by Step

The mass of substance deposited at an electrode is directly proportional to the quantity of charge passed:

m = \frac{M \times I \times t}{n \times F}

where $M$ = molar mass, $I$ = current, $t$ = time in seconds, $n$ = number of electrons transferred per ion, $F$ = Faraday constant (96500 C/mol).

At the cathode (reduction):

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}

So $n = 2$ (each Cu²⁺ ion needs 2 electrons).

$I = 5$ A, $t = 32 \text{ min } 10 \text{ s} = 32 \times 60 + 10 = 1930$ s

Q = I \times t = 5 \times 1930 = 9650 \text{ C}

Notice: $Q = 9650$ C = $\frac{9650}{96500} = 0.1$ Faraday. The question was designed to give a clean number.

m = \frac{M \times Q}{n \times F} = \frac{63.5 \times 9650}{2 \times 96500}

m = \frac{63.5 \times 9650}{193000} = \frac{63.5}{20} = \boxed{3.175 \text{ g}}

Why This Works

Electrolysis is essentially a stoichiometric process. One Faraday of charge (96500 C) deposits one gram equivalent of a substance. The gram equivalent of Cu (valency 2) is $63.5/2 = 31.75$ g. Since we passed 0.1 Faraday, we deposited $0.1 \times 31.75 = 3.175$ g.

Faraday’s second law says: when the same charge passes through different electrolytes in series, the masses deposited are proportional to their equivalent weights. So if CuSO₄ and AgNO₃ solutions were in series, the same 9650 C would deposit 3.175 g of Cu and 10.8 g of Ag.

Alternative Method — Using equivalents

Equivalents deposited = $Q/F = 9650/96500 = 0.1$

Equivalent weight of Cu = $63.5/2 = 31.75$ g/eq

Mass = equivalents $\times$ equivalent weight = $0.1 \times 31.75 = 3.175$ g

NEET often gives time in odd formats (like “32 minutes 10 seconds”) to test if you convert properly. Always convert to seconds first. Also, check if $Q/F$ gives a clean fraction — NEET question setters design the numbers to come out neatly (0.1, 0.5, etc.).

Common Mistake

The most frequent error: using $n = 1$ for copper instead of $n = 2$ . Copper is Cu²⁺ in CuSO₄, so it needs 2 electrons for reduction. If the question involved Cu⁺ (from CuCl), then $n = 1$ . Always write the electrode reaction first to determine $n$ — don’t assume.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using equivalents

Common Mistake

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