Q22 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Chemistry

The emf of the following cell $\mathrm{Mg}\left|\mathrm{Mg}^{+2}(0.01 \mathrm{M})\right|\left|\mathrm{Sn}^{+2}(0.1 \mathrm{M})\right| \mathrm{Sn}$ at $298 \mathrm{~K}$ in ' $\mathrm{V}$ ' is $\left(\right.$ Given, $\left.E_{\mathrm{Mg}^{+2} \mid \mathrm{Mg}}^{\circ}=-2.34 \mathrm{~V}, E_{\mathrm{Sn}^{+2} \mid \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}\right)$

  • A
    2.17
  • Correct answer
    B
    2.23
  • C
    2.51
  • D
    2.45

Solution

At anode, $\quad \mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}+2 e^{-}$ At cathode, $\mathrm{Sn}^{2+}+2 e^{-} \longrightarrow \mathrm{Sn}$ From Nernst equation, $$ \begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{\mathrm{Mg}^{+2}}{\mathrm{Sn}^{+2}} \\ E_{\text {cell }} & =-0.14-(-2.34)-\frac{0.0591}{2} \log 10^{-1} \quad\left(\because n=2 e^{-}\right) \\ & =2.2+\frac{0.0591}{2} \\ & =2.2+0.0295 \approx 2.23 \end{aligned} $$

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

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