1. Find the volume of nickel to be deposited

Given:

• Area of metal surface, $A = 100\,\text{cm}^2$
• Thickness of nickel layer, $d = 0.001\,\text{mm}$

Convert thickness into cm: $0.001\,\text{mm} = 10^{-3}\,\text{mm} = 10^{-4}\,\text{cm}$ since $1\,\text{mm} = 0.1\,\text{cm}$.

So volume deposited is: $V = A \times d = 100 \times 10^{-4} = 10^{-2}\,\text{cm}^3 = 0.01\,\text{mL}$

2. Find the mass of nickel deposited

Given density of Ni: $\rho = 10\,\text{g mL}^{-1}$

Thus mass: $m = \rho V = 10 \times 0.01 = 0.1\,\text{g}$

3. Calculate moles of nickel deposited

Molar mass of Ni = $60\,\text{g mol}^{-1}$ $n_{\text{Ni}} = \frac{0.1}{60} = \frac{1}{600}\,\text{mol}$

4. Use the electrode reaction

Nickel is deposited as: $\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$

So, 1 mole of Ni requires 2 moles of electrons.

Hence moles of electrons needed: $n_{e^-} = 2 \times \frac{1}{600} = \frac{1}{300}\,\text{mol}$

5. Calculate total charge required

Using Faraday constant $F = 96500\,\text{C mol}^{-1}$: $Q = n_{e^-}F = \frac{1}{300} \times 96500 = 321.67\,\text{C}$

6. Find time using current

Given current $I = 2\,\text{A}$ and $Q = It$ so $t = \frac{Q}{I} = \frac{321.67}{2} = 160.83\,\text{s}$

Nearest integer: $x = 161$

7. Comparison with stored answer

Stored correct answer = $161$

This matches our derived answer.