Question

Calculate the reduction potential for the following half cell at $25{}^{o}C$.

$Cu\left(s\right)|C{u}^{2+}(aq,0.2M);E_{C{u}^{2+}/Cu}^0=0.34V$

Take :
[$lo{g}_{10}5=0.7$]

• A. $E_{C{u}^{2+}/Cu}=0.39V$
• B. $E_{C{u}^{2+}/Cu}=0.50V$
• C. $E_{C{u}^{2+}/Cu}=-0.2V$
• D. $E_{C{u}^{2+}/Cu}=0.32V$

Answer 1

The correct option is D $E_{C{u}^{2+}/Cu}=0.32V$

$Cu\left(s\right)|C{u}^{2+}(aq,0.2M);E_{C{u}^{2+}/Cu}^0=0.34V$

$E_{C{u}^{2+}/Cu}=E_{C{u}^{2+}/Cu}^0-\frac{2.303RT}{nF}log\frac{\left[M\right(s\left)\right]}{\left[M^{n+}\right(aq\left)\right]}$

$n=2$

$E_{C{u}^{2+}/Cu}=0.34-\frac{0.059}{2}log\frac{1}{\left[C{u}^{2+}\right]}$

$E_{C{u}^{2+}/Cu}=0.34-\frac{0.059}{2}log\frac{1}{\left[0.2\right]}$

$E_{C{u}^{2+}/Cu}=0.34-\frac{0.059}{2}\times log5$

$E_{C{u}^{2+}/Cu}=0.34-\frac{0.059}{2}\times 0.7\approx 0.32V$