Q15 / 120 · VITEEE 2018 · Chemistry

Standard cell voltage for the cell $\mathrm{Pb}\left|\mathrm{Pb}^{2+} \| \mathrm{Sn}^{2+}\right| \mathrm{Sn}$ is $-0.01 \mathrm{~V}$. If the cell is to exhibit $E_{\text {cell }}=0$, the value of $\left[\mathrm{Sn}^{2+}\right] /\left[\mathrm{Pb}^{2+}\right]$ should be antilog of $-$

  • Correct answer
    A
    $+0.3$
  • B
    $0.5$
  • C
    $1.5$
  • D
    $-0.5$

Solution

Apply Nernst equation to the reaction $\mathrm{Pb}+\mathrm{Sn}^{2+} \rightarrow \mathrm{Pb}^{2+}+\mathrm{Sn}$ or $\quad E^{\circ}+\frac{0.059}{2} \log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=E_{\text {cell }}$ or $\log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=\frac{0.01 \times 2}{0.059}=0.3\left(\because E_{\text {cell }}=0\right)$ or $\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=$ antilog $(0.3)$

Hence, the correct answer is option: 'A'.

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