The electrode potential $\mathrm{E}_{\left(\mathrm{Zn}^{2+} / \mathrm{Zn}ight)}$ of a zinc electrode at…

The electrode potential $\mathrm{E}_{\left(\mathrm{Zn}^{2+} / \mathrm{Zn}ight)}$ of a zinc electrode at $25^{\circ} \mathrm{C}$ with an aqueous solution of $0.1 \mathrm{M} \mathrm{ZnSO}_{4}$ is
$\left[E_{\left(\mathrm{Zn}^{2+} / \mathrm{Zn}ight)}^{\circ}=-0.76 \mathrm{~V} .ight.$ Assume $\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{at} 298 \mathrm{~K}ight]$
  1. $+0.73$
  2. $-0.79$
  3. $-0.82$
  4. $-0.70$

Solution

For $\begin{aligned} \mathrm{Zn}^{2+} & ightarrow \mathrm{Zn} \\ \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}} &=\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{[\mathrm{Zn}]}{\left[\mathrm{Zn}^{2+}ight]} \\ &=-0.76-\frac{0.06}{2} \log \frac{1}{[0.1]}=-0.76-0.03 \\ \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}} &=-0.79 \mathrm{~V} \end{aligned}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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