Q25 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Chemistry

$\mathrm{Mg}^{2+}$ displaces hydrogen from acids but copper does not. A galvanic cell prepared by combining $\mathrm{Cu} / \mathrm{Cu}^{2+}$ and $\mathrm{Mg} / \mathrm{Mg}^{2+}$ has an EMF of $2.71 \mathrm{~V}$ at $298 \mathrm{~K}$. If the potential of copper electrode is $0.34 \mathrm{~V}$, what is the reduction potential of $\mathrm{Mg}$ electrode?

  • A
    + 3 05. V
  • Correct answer
    B
    -2 37. V
  • C
    +2 37. V
  • D
    2 V

Solution

The galvanic cell can be represented as : $\begin{gathered} \mathrm{Mg}\left|\mathrm{Mg}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu} \\ E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}-E_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ} \\ \Rightarrow E_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}-E_{\mathrm{Cell}}^{\circ}(\text { Reduction potential }) \\ =0.34-2.71 \mathrm{~V}=-2.37 \mathrm{~V} \\ =(0.34-2.7 \mathrm{l}) \mathrm{V}=-2.37 \mathrm{~V} \end{gathered}$

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

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