The reduction potential (E^{circ},in V) of Mn | vedclass

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(B) The overall reaction is $MnO_4^{-} + 7e^{-} \rightarrow Mn$. The total change in Gibbs free energy is the sum of the changes for the individual st

Vedclass · Accepted Answer

$0.77$

Vedclass · Answer

(B) The overall reaction is $MnO_4^{-} + 7e^{-} ightarrow Mn$. The total change in Gibbs free energy is the sum of the changes for the individual steps:
$\Delta G^{\circ}_{total} = \Delta G^{\circ}_{1} + \Delta G^{\circ}_{2} + \Delta G^{\circ}_{3}$
Since $\Delta G^{\circ} = -nFE^{\circ}$,we have:
$-7 	imes F 	imes E^{\circ} = -(3 	imes F 	imes 1.68) + (-2 	imes F 	imes 1.21) + (-2 	imes F 	imes (-1.03))$
$7 	imes E^{\circ} = (3 	imes 1.68) + (2 	imes 1.21) + (2 	imes (-1.03))$
$7 	imes E^{\circ} = 5.04 + 2.42 - 2.06$
$7 	imes E^{\circ} = 5.40$
$E^{\circ} = \frac{5.40}{7} \approx 0.7714 \ V$
Thus,the reduction potential is approximately $0.77 \ V$.

The reduction potential ($E^{\circ}$,in $V$) of $MnO_4^-{_{\text{(aq)}}} / Mn_{\text{(s)}}$ is. . . . . $[Given : E^{\circ}_{(MnO_4^{-(aq)} / MnO_{2(s)})} = 1.68 \ V ; E^{\circ}_{(MnO_{2(s)} / Mn^{2+}_{(aq)})} = 1.21 \ V ; E^{\circ}_{(Mn^{2+(aq)} / Mn_{(s)})} = -1.03 \ V]$

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