Q3 / 120 · VITEEE 2014 · Chemistry

$\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{E}^{\circ}=1.51 \mathrm{~V}$ $\mathrm{MnO}_{2}+4 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{H}_{2} \mathrm{O}$ $E^{\circ}=1.23 \mathrm{~V} E_{\mathrm{MnO}_{4}^{-} \mid \mathrm{MnO}_{2}}^{\circ}$ is

  • Correct answer
    A
    $1.70 \mathrm{~V}$
  • B
    $0.91 \mathrm{~V}$
  • C
    $1.37 \mathrm{~V}$
  • D
    $0.548 \mathrm{~V}$

Solution

On subtracting eqn. (ii) from (i) we get $\begin{aligned} & \mathrm{MnO}_{4}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \\ \therefore \quad &-\mathrm{E}_{3}=\frac{-1.51 \times 5+2 \times 1.23}{3} \\ \therefore \quad & \mathrm{E}_{3}=1.70 \mathrm{~V} \end{aligned}$

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

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