Question

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

• A. +0.3
• B. 0.5
• C. 1.5
• D. -0.5

Answer 1

Answer: (A)

Apply Nernst equation to the reaction
$Pb + Sn^{2+} \to Pb^{2+} + Sn$
or $ E^{\circ} + \frac{0.059}{2}\log \frac{\left[Sn^{2+}\right]}{\left[Pb^{2+}\right]} = E_{\text{cell}}$
or$ \log \frac{\left[Sn^{2+}\right]}{\left[Pb^{2+}\right]} =\frac{0.01\times2}{0.059} = 0.3$
$ \left(\because E_{\text{cell}} = 0\right)$
or $ \frac{\left[Sn^{2+}\right]}{\left[Pb^{2+}\right]} =$ antilog $\left(0.3\right)$