Electrochemistry: Numerical Problems Set (1)

easy • 3 min read
Tags Electrochemistry

Question

A current of 5 A5 \text{ A} is passed through a CuSO4\text{CuSO}_4 solution for 3030 minutes. Calculate the mass of copper deposited at the cathode. Atomic mass of Cu = 63.5 g/mol63.5 \text{ g/mol}, Faraday constant F=96500 C/molF = 96500 \text{ C/mol}.

Solution — Step by Step

Q=I⋅t=5×(30×60)=5×1800=9000 CQ = I \cdot t = 5 \times (30 \times 60) = 5 \times 1800 = 9000 \text{ C}

ne=QF=900096500≈0.0933 moln_e = \frac{Q}{F} = \frac{9000}{96500} \approx 0.0933 \text{ mol}

For copper deposition: Cu2++2e−→Cu\text{Cu}^{2+} + 2e^- \to \text{Cu}. Each mole of Cu requires 22 mol of electrons.

nCu=ne2=0.09332≈0.0466 moln_{\text{Cu}} = \frac{n_e}{2} = \frac{0.0933}{2} \approx 0.0466 \text{ mol}

m=nCu×MCu=0.0466×63.5≈2.96 gm = n_{\text{Cu}} \times M_{\text{Cu}} = 0.0466 \times 63.5 \approx 2.96 \text{ g}

Why This Works

This is Faraday’s first law of electrolysis in action: mass deposited is proportional to the charge passed. The proportionality constant — the “electrochemical equivalent” zz — equals M/(nF)M/(nF) where nn is the number of electrons per ion (here 22 for Cu²⁺).

Compact formula:

m=M⋅I⋅tn⋅Fm = \frac{M \cdot I \cdot t}{n \cdot F}

Plug in: m=(63.5×5×1800)/(2×96500)=571500/193000≈2.96 gm = (63.5 \times 5 \times 1800)/(2 \times 96500) = 571500/193000 \approx 2.96 \text{ g}. ✓

Alternative Method

Equivalent mass approach: the equivalent mass of Cu is M/n=63.5/2=31.75 g/equivM/n = 63.5/2 = 31.75 \text{ g/equiv}. Charge of one equivalent is F=96500 CF = 96500 \text{ C}. Equivalents deposited: 9000/96500=0.09339000/96500 = 0.0933. Mass: 0.0933×31.75≈2.96 g0.0933 \times 31.75 \approx 2.96 \text{ g}. Same answer.

For NEET-level shortcut: m=(I×t×equivalent mass)/96500m = (I \times t \times \text{equivalent mass})/96500. Memorise the equivalent masses of common metals: Cu = 31.7531.75, Ag = 108108, Al = 99, Zn = 32.532.5. These come up in PYQs every year.

Common Mistake

Three classic slips:

  1. Forgetting nn in the half-reaction. Cu²⁺ needs 2e−2e^-, Ag⁺ needs 1e−1e^-, Al³⁺ needs 3e−3e^-. Plugging the wrong nn scales your answer by a factor.

  2. Time in minutes instead of seconds. Current is in amperes (coulombs/second). Always convert minutes to seconds before computing QQ.

  3. Using molar mass instead of equivalent mass. If you use the formula m=M⋅Q/Fm = M \cdot Q / F (without dividing by nn), you get twice the right answer for Cu²⁺. The nn in the denominator is non-negotiable.

Final answer: Mass of Cu deposited ≈2.96 g\approx 2.96 \text{ g}.

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