Faraday's laws — how much copper deposits when 2A flows for 1 hour

Question

How much copper is deposited at the cathode when a current of 2 A flows through a copper sulphate solution for 1 hour? (Atomic mass of Cu = 63.5 g/mol, Faraday constant $F = 96500$ C/mol)

Solution — Step by Step

We need Faraday’s first law of electrolysis: the mass of substance deposited is proportional to the charge passed.

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

Where:

$m$ = mass deposited (g)
$M$ = molar mass of the element (g/mol)
$I$ = current (A)
$t$ = time (s)
$n$ = number of electrons transferred per ion
$F$ = Faraday constant = 96500 C/mol

The cathode reaction for copper is: $\text{Cu}^{2+} + 2e^- \to \text{Cu}$ . So $n = 2$ .

The current flows for 1 hour. Convert to SI units:

t = 1 \text{ hour} = 60 \times 60 = 3600 \text{ s}

Always convert time to seconds before substituting — a classic unit error trap.

Total charge = current × time:

Q = I \times t = 2 \times 3600 = 7200 \text{ C}

Substituting into the formula:

m = \frac{M \times Q}{n \times F} = \frac{63.5 \times 7200}{2 \times 96500}

m = \frac{457200}{193000} = 2.369 \text{ g}

\boxed{m \approx 2.37 \text{ g of copper is deposited}}

Why This Works

The electrode reaction $\text{Cu}^{2+} + 2e^- \to \text{Cu}$ tells us that 2 moles of electrons deposit 1 mole of copper (63.5 g). One mole of electrons carries a charge of 1 Faraday = 96500 C. So 2 moles of electrons carry $2 \times 96500 = 193000$ C.

We passed 7200 C, which is $\frac{7200}{193000}$ moles of “2-electron packages,” each depositing one Cu atom. This is exactly what the formula computes.

The formula works because electrical charge is quantised — electrons carry a fixed charge, and stoichiometry is fixed. Every $\frac{2 \times 96500}{6.022 \times 10^{23}}$ coulombs deposits exactly one Cu atom.

Alternative Method

Using equivalents (older but still valid in some textbooks):

Equivalent weight of Cu = $\frac{M}{n} = \frac{63.5}{2} = 31.75$ g/equivalent.

One Faraday deposits one equivalent = 31.75 g of Cu.

Mass deposited = $\frac{Q}{F} \times$ (equivalent weight) = $\frac{7200}{96500} \times 31.75 = 0.07461 \times 31.75 \approx 2.37$ g.

Same answer, different route.

Common Mistake

The most common mistake is using $n = 1$ instead of $n = 2$ for copper. Copper ions in CuSO₄ are Cu²⁺, so two electrons are needed per copper atom deposited. Using $n = 1$ doubles the calculated mass — this is a very common error in CBSE board papers and costs 1–2 marks.

Always write the electrode equation first ( $\text{Cu}^{2+} + 2e^- \to \text{Cu}$ ) and read off $n$ from there.

For JEE, also know Faraday’s second law: when the same charge passes through different electrolytes in series, the masses deposited are in the ratio of their equivalent weights. This is tested frequently when two electrolytic cells are connected in series.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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