Q21 / 90 · KVPY 2011 (SB/SX) · Chemistry

If $E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.440 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.770 \mathrm{~V}$, then $\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{o}$ is-

  • A
    $0.330 \mathrm{~V}$
  • Correct answer
    B
    $-0.037 \mathrm{~V}$
  • C
    $-0.330 \mathrm{~V}$
  • D
    $-1.210 \mathrm{~V}$

Solution

$\mathrm{Fe}^{+3} / \mathrm{Fe}$ $\mathrm{Fe}^{+2}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathrm{E}^{0}=-0.44$ $\mathrm{Fe}^{+3}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{+2} \quad \mathrm{E}^{0}=0.770$ $\overline{\mathrm{Fe}^{+3}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}} \quad \mathrm{E}^{0}=?$ $\overline{\mathrm{E}^{0}=\frac{\mathrm{n}_{1} \mathrm{E}_{1}^{0}+\mathrm{n}_{2} \mathrm{E}_{2}^{0}}{\mathrm{n}_{3}}}=\frac{2(-0.44)+1 \times(0.770)}{3}=\frac{-0.88+0.770}{3}=-0.037 \mathrm{volt}$

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

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