Question

If the standard electrode potential of $Cu^{2+}/ Cu$ electrode is $0.34 \,V$, what is the electrode potential of $0.01 \,M $ concentration of $Cu^{2+}?(T = 298 \,K)$

• A. 0.399 V
• B. 0.281 V
• C. 0.222 V
• D. 0176 V

Answer 1

Answer: (B)

$Cu^{2+} + 2e^- \rightarrow Cu$
At $298\,K$
$E = E^0 - \frac{0.0591}{n} log \frac{1}{[M^{n+}]}$
$E_{{Cu}^{2+}/Cu} - \frac{0.0591}{2} log \frac{1}{10^2}$
$ = (0.34 - 0.0591) V $
$ = 0.2809\,V$