1. Write the electrode reaction

At the cathode, zinc ions are reduced:

$\mathrm{Zn^{2+} + 2e^- \rightarrow Zn}$

So, 2 moles of electrons deposit 1 mole of Zn.

2. Calculate total charge passed

Given:

• Current, $I = 0.015\,\mathrm{A}$
• Time, $t = 15\,\mathrm{min} = 15 \times 60 = 900\,\mathrm{s}$

Using

$Q = It$

$Q = 0.015 \times 900 = 13.5\,\mathrm{C}$

3. Use Faraday's law

Mass deposited is given by:

$m = \frac{QM}{nF}$

where:

• $Q = 13.5\,\mathrm{C}$
• $M = 65.4\,\mathrm{g\,mol^{-1}}$
• $n = 2$
• $F \approx 96500\,\mathrm{C\,mol^{-1}}$

So,

$m = \frac{13.5 \times 65.4}{2 \times 96500}$

$m = \frac{882.9}{193000}$

$m \approx 4.57 \times 10^{-3}\,\mathrm{g}$

4. Express in the required form

We need:

$m = \underline{\hspace{1cm}} \times 10^{-4}\,\mathrm{g}$

Since

$4.57 \times 10^{-3} = 45.7 \times 10^{-4}$

Thus the required integer is approximately:

$46$

5. Comparison with stored answer

Derived answer = $46$

Stored correct answer = $46$

They agree.