Given:E^o_{Fe^{3+} /Fe} = -0.036 V, E^o_{Fe^ | vedclass

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(B) The given half-cell reactions are:$(i) Fe^{3+} + 3e^- \rightarrow Fe, E^{\circ}_1 = -0.036 \ V, n_1 = 3$$(ii) Fe^{2+} + 2e^- \rightarrow Fe, E^{\c

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$0.770$

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(B) The given half-cell reactions are:$(i) Fe^{3+} + 3e^- ightarrow Fe, E^{\circ}_1 = -0.036 \ V, n_1 = 3$$(ii) Fe^{2+} + 2e^- ightarrow Fe, E^{\circ}_2 = -0.439 \ V, n_2 = 2$We need to find $E^{\circ}$ for the reaction:$(iii) Fe^{3+} + e^- ightarrow Fe^{2+}, n_3 = 1$This reaction can be obtained by $(i) - (ii)$.Using the relation $\Delta G^{\circ} = -nFE^{\circ}$,we have:$n_3 E^{\circ}_3 = n_1 E^{\circ}_1 - n_2 E^{\circ}_2$$1 	imes E^{\circ} = 3 	imes (-0.036) - 2 	imes (-0.439)$$E^{\circ} = -0.108 + 0.878$$E^{\circ} = 0.770 \ V$

Given:
$E^o_{Fe^{3+} /Fe} = -0.036 \ V, E^o_{Fe^{2+} /Fe} = -0.439 \ V$
The value of standard electrode potential for the change,
$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$ will be ........ $V$.

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Given:$E^o_{Fe^{3+} /Fe} = -0.036 \ V, E^o_{Fe^{2+} /Fe} = -0.439 \ V$The value of standard electrode potential for the change,$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$ will be ........ $V$.