The $\mathrm{E}^{\circ}$ of $\mathrm{M} \mid \mathrm{M}^{2+} \| \mathrm{Cu}^{2+} / \mathrm{Cu}$ is 0.3 V ,…

The $\mathrm{E}^{\circ}$ of $\mathrm{M} \mid \mathrm{M}^{2+} \| \mathrm{Cu}^{2+} / \mathrm{Cu}$ is 0.3 V , At what concentration of $\mathrm{Cu}^{2+}$ (in $\mathrm{mol} \mathrm{L}^{-1}$ ), the $\mathrm{E}_{\text {cell }}$ value becomes zero? $\left(\frac{2.303 R T}{F}=0.06\right)\left(\text { Conc. of } \mathrm{M}^{2+}=0.1 \mathrm{M}\right)$
  1. $10^{-9}$
  2. $10^{-8}$
  3. $10^{-11}$
  4. $10^{-10}$

Solution

According to Nernst equation, $\mathrm{E}=\mathrm{E}^{\circ}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{[\text { product }]}{[\text { reactant }]}$
$\mathrm{E}=\mathrm{E}^{\circ}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{\left[\mathrm{M}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$ Given, $\begin{aligned} & \mathrm{E}^{\circ}=0.3 \mathrm{~V} \\ & \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\end{aligned}$ $\begin{aligned} & \mathrm{n}=2 \\ & \mathrm{M}^{2+}=0.1 \mathrm{M}\end{aligned}$ We have to take $\mathrm{E}_{\text {cell }}=0$. According to question. $0=0.3 \mathrm{~V}-\frac{0.06}{3} \log _{10} \frac{\left[10^{-1}\right]}{\left[\mathrm{Cu}^{2+}\right]}$ $-0.3=-0.03 \log _{10} \frac{\left[10^{-1}\right]}{\left[\mathrm{Cu}^{2+}\right]}$ $10=\log _{10} \frac{\left[10^{-1}\right]}{\left[\mathrm{Cu}^{2+}\right]}$ $\therefore \frac{\left[10^{-1}\right]}{\left[\mathrm{Cu}^{2+}\right]}=10^{10}$ or, $\left[\mathrm{Cu}^{2+}\right]=\frac{10^{-1}}{10^{10}}=10^{-1-10}=10^{-11}$ $\mathrm{Cu}^{2+}=10^{-11}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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