Given : $\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{\circ}=-0.036 \mathrm{~V}, \quad…

Given : $\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{\circ}=-0.036 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.439 \mathrm{~V}$. The value of standard electrode potential for the change, $\mathrm{Fe}_{(\text {aq })}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})$ will be :
  1. $-0.072 \mathrm{~V}$
  2. $0.385 \mathrm{~V}$
  3. $0.770 \mathrm{~V}$
  4. $-0.270$

Solution

$ \begin{aligned} & \because \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; \mathrm{E}^{\circ}=-0.036 \mathrm{~V} \\ & \therefore \Delta \mathrm{G}_1^{\circ}=-\mathrm{nFE}^{\circ}=-3 \mathrm{~F}(-0.036) \\ & =+0.108 \mathrm{~F} \\ & \text { Also } \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; \mathrm{E}^{\circ}=-0.439 \mathrm{~V} \\ & \therefore \Delta \mathrm{G}_2^{\circ}=-\mathrm{nFE}^{\circ} \\ & =-2 \mathrm{~F}(-0.439) \\ & =0.878 \mathrm{~F} \\ & \text { To find } \mathrm{E}^{\circ} \text { for } \mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) \\ & \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}{ }^{\circ}=-1 \mathrm{FE} \circ \\ & \because \mathrm{G}^{\circ}=\mathrm{G}_1^{\circ}-\mathrm{G}_2^{\circ} \\ & \therefore \mathrm{G}^{\circ}=0.108 \mathrm{~F}-0.878 \mathrm{~F} \\ & \therefore-\mathrm{FE}^{\circ}=+0.108 \mathrm{~F}-0.878 \mathrm{~F} \\ & \therefore \mathrm{E}^{\circ}=0.878-0.108 \\ & =0.77 \mathrm{~V} \end{aligned} $

Asked in: JEE Main 2009

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