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