Electrochemistry: Tricky Questions Solved (5)

Question

A current of $1.5\,\text{A}$ is passed through a $\text{CuSO}_4$ solution for $20\,\text{min}$. Calculate the mass of copper deposited at the cathode. Atomic mass of Cu $= 63.5\,\text{g/mol}$, $F = 96500\,\text{C/mol}$.

Solution — Step by Step

The mass of copper deposited is approximately $0.59\,\text{g}$.

Why This Works

Faraday’s first law says the mass deposited is proportional to the charge passed, with the proportionality constant being the equivalent weight divided by Faraday’s constant. The equivalent weight depends on the half-reaction — for $\text{Cu}^{2+}$, two electrons per atom, so divide atomic weight by $2$.

This is the workhorse formula for any electrolysis numerical. Identify the ion, count electrons, plug into $m = EIt/F$.

Alternative Method

Compute moles of electrons directly: $n_e = Q/F = 1800/96500 = 0.01865\,\text{mol}$. Then moles Cu $= n_e/2 = 0.00932\,\text{mol}$. Mass $= 0.00932 \times 63.5 = 0.592\,\text{g}$. Same answer.

Common Mistake

Using atomic mass instead of equivalent mass. Students compute $m = M \cdot Q/F$ and double the correct answer. The fix: always divide $M$ by the number of electrons in the half-reaction. NEET 2024 had this trap on silver vs copper electrolysis.