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## Q. 16.3

Controlled-Potential Electrolysis

What cathode potential is required to reduce 99.99% of 0.10 M $Cu^{2+}$ to Cu(s)? Is it possible to remove this $Cu^{2+}$ without reducing 0.10 M $Sn^{2+}$ in the same solution?

$Cu^{2+} + 2e^{−} \rightleftharpoons Cu(s)$             E° = 0.339 V         (16-7)

$Sn^{2+} + 2e^{−} \rightleftharpoons Sn(s)$              E° = −0.141 V        (16-8)

## Verified Solution

If 99.99% of $Cu^{2+}$ were reduced, the concentration of remaining $Cu^{2+}$ would be 1.0 × $10^{−5}$ M, and the required cathode potential would be

$E(cathode) = 0.339 − \frac{0.059 16}{2} \log (\frac{1}{\begin{matrix} 1.0 × 10^{−5} \\ \quad \quad \quad \quad \quad \quad \nwarrow [Cu^{2+}] \end{matrix} }) = 0.19$ V

The cathode potential required to reduce $Sn^{2+}$ is

$E(cathode, for reduction of Sn^{2+} ) = −0.141 − \frac{0.059 16}{2} \log(\frac{1}{\begin{matrix} 0.10 \\ \quad \quad \quad \quad \quad \quad \nwarrow [Sn^{2+}] \end{matrix} }) = −0.17$ V

We do not expect reduction of $Sn^{2+}$ at a cathode potential more positive than −0.17 V. The reduction of 99.99% of $Cu^{2+}$ without reducing $Sn^{2+}$ appears feasible.

Test Yourself    Will E(cathode) = 0.19 V reduce 0.10 M $SbO^{+}$ at pH 2 by the reaction $SbO^{+} + 2H^{+} + 3e^{−} \rightleftharpoons Sb(s) + H_{2}O$, E° = 0.208 V? (Answer: E(cathode) for $SbO^{+}$ = 0.11 V, so reduction should not occur at 0.19 V)