Question

The electrode potential E$_{\left(\frac{Zn^{2+}}{Zn}\right)}$ of a zinc electrode at 25$^{\circ}$C with an aqueous solution of 0.1M $ZnSO_4$ is (E$^{\circ}$ $_{\left(\frac{Zn^{2+}}{Zn}\right)}$ = -0.76 V. Assume $\frac{2.303RT}{F} =$ 0.06 at 298 K)

• A. 0.73
• B. -0.79
• C. -0.82
• D. -0.7

Answer 1

Answer: (B)

-0.79

Answer 2

For $Zn^{+2} \longrightarrow Zn$ $E_{z_{n}}^{+} / z_{n} =E_{z_{n}^{+2} / z_{n}}^{\circ}-\frac{2.303 R T}{n F} \log \frac{\left[z_{n}\right]}{\left[z_{n}^{+2}\right]}$ $=-0.76-\frac{0.06}{2} \log \frac{1}{0.1}=-0.76-0.03$ $E_{z_{n}}^{+2} / z_{n} =-0.79\, V$