Question:
Suppose that gold is being plated on to another metal in an electrolytic cell. The half-cell reaction producing Au(s) is
\[
\text{AuCl}_4^- \rightarrow \text{Au}(s) + 4\text{Cl}^- - 3e^-
\]
If a 0.30 A current runs for 15.00 min, what mass of Au(s) will be plated? (Faraday constant = 96485 C/mol, molar mass of Au = 197)
Suppose that gold is being plated on to another metal in an electrolytic cell. The half-cell reaction producing Au(s) is
\[
\text{AuCl}_4^- \rightarrow \text{Au}(s) + 4\text{Cl}^- - 3e^-
\]
If a 0.30 A current runs for 15.00 min, what mass of Au(s) will be plated? (Faraday constant = 96485 C/mol, molar mass of Au = 197)
Show Hint
Always divide by number of electrons in half-reaction.
Updated On: Apr 23, 2026
- 0.184 g Au
- 0.551 g Au
- 1.84 g Au
- 0.613 g Au
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
Faraday’s law: mass $\propto$ charge passed.
Step 1: Charge passed $Q = It = \textcolor{red}{0.30 \times (15 \times 60)} = 270$ C.
Step 2: Moles of electrons = $Q/F = 270/96485 \approx 0.0028$ mol.
Step 3: From reaction, 3 mol e$^-$ $\Rightarrow$ 1 mol Au.
\[ \text{Moles of Au} = \frac{0.0028}{3} \approx 0.00093 \]
Step 4: Mass of Au = $0.00093 \times 197 \approx 0.184$ g.
Conclusion:
Mass deposited = 0.184 g
Step 1: Charge passed $Q = It = \textcolor{red}{0.30 \times (15 \times 60)} = 270$ C.
Step 2: Moles of electrons = $Q/F = 270/96485 \approx 0.0028$ mol.
Step 3: From reaction, 3 mol e$^-$ $\Rightarrow$ 1 mol Au.
\[ \text{Moles of Au} = \frac{0.0028}{3} \approx 0.00093 \]
Step 4: Mass of Au = $0.00093 \times 197 \approx 0.184$ g.
Conclusion:
Mass deposited = 0.184 g
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