Consider the half-cell reduction reactions:Mn | vedclass

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(D) The given half-cell reactions are:$(I) \ Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ}_{1} = -1.18 \ V$$(II) \ Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,

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$-2.69 \ V$ and no

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(D) The given half-cell reactions are:$(I) \ Mn^{2+} + 2e^{-} ightarrow Mn$,$E^{\circ}_{1} = -1.18 \ V$$(II) \ Mn^{3+} + e^{-} ightarrow Mn^{2+}$,$E^{\circ}_{2} = +1.51 \ V$To obtain the reaction $3Mn^{2+} ightarrow Mn + 2Mn^{3+}$,we perform the operation: $(I) - 2 	imes (II)$.The standard Gibbs free energy change is given by $\Delta G^{\circ} = -nFE^{\circ}$.For reaction $(I)$,$\Delta G^{\circ}_{1} = -2 	imes F 	imes (-1.18) = +2.36F$.For reaction $(II)$,$\Delta G^{\circ}_{2} = -1 	imes F 	imes (+1.51) = -1.51F$.For the target reaction,$\Delta G^{\circ}_{net} = \Delta G^{\circ}_{1} - 2 	imes \Delta G^{\circ}_{2} = +2.36F - 2 	imes (-1.51F) = +2.36F + 3.02F = +5.38F$.Since $\Delta G^{\circ}_{net} = -nFE^{\circ}_{cell}$,where $n = 2$ electrons are transferred in the balanced equation:$5.38F = -2 	imes F 	imes E^{\circ}_{cell}$$E^{\circ}_{cell} = -5.38 / 2 = -2.69 \ V$.Since $E^{\circ}_{cell}$ is negative,the reaction is non-spontaneous.

Consider the half-cell reduction reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ} = -1.18 \ V$
$Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ} = +1.51 \ V$
The $E^{\circ}$ for the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,and the possibility of the forward reaction are respectively:

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Consider the half-cell reduction reactions:$Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ} = -1.18 \ V$$Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ} = +1.51 \ V$The $E^{\circ}$ for the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,and the possibility of the forward reaction are respectively: