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Chapter 16

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)

Step-by-Step

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)

FIGURE 16-7