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. - 0.70
• E. + 0.79

Answer 1

Answer: (B)

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