A copper plate of area $10{\text{cm}}^2$ is to be electroplated with nickel to a thickness of $20\mu \text{m}$. Density of nickel $=8.90{\text{g cm}}^{-3}$, molar mass $=58.69{\text{g mol}}^{-1}$. (a) Calculate the total charge required. (b) Find the time needed if a constant current of $2.0\text{A}$ is passed. Assume 100% current efficiency. Reaction: $\mathrm{N}{\mathrm{i}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Ni}$. Use $F=96485{\text{C mol}}^{-1}$.

Step-by-step solution

1. Convert thickness to centimetres: $20\mu \text{m}=20\times 10^{-4}\text{cm}=0.002\text{cm}$.
2. Calculate the volume of nickel deposited: $V=\text{area}\times \text{thickness}=10{\text{cm}}^2\times 0.002\text{cm}=0.02{\text{cm}}^3$.
3. Find the mass: $m=V\times \text{density}=0.02{\text{cm}}^3\times 8.90{\text{g cm}}^{-3}=0.178\text{g}$.
4. Determine moles of Ni: $n_{\text{Ni}}=\frac{m}{M}=\frac{0.178}{58.69}=3.03\times 10^{-3}\text{mol}$.
5. For the reduction $\mathrm{N}{\mathrm{i}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Ni}$, $z=2$ electrons per Ni atom. Total charge $Q=n_{\text{Ni}}\times z\times F=3.03\times 10^{-3}\times 2\times 96485=5.85\times 10^2\text{C}$.
6. Time required $t=\frac{Q}{I}=\frac{5.85\times 10^2\text{C}}{2.0\text{A}}=2.93\times 10^2\text{s}$, which is about $4.9\text{min}$.

Concept tested: Application of Faraday’s laws of electrolysis

Published 5 September 2026 · How this solution was prepared and checked