MnO_4^- (aq) + 8H^+ (aq) + 5e^- to Mn^{2+} (a | vedclass

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(A) The given half-reactions are:$(1) MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$; $\Delta G_1^o = -n_1 F E_1^o = -5 	imes F 	imes 1.51 = -7.55 F$$(2

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$1.71$

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(A) The given half-reactions are:$(1) MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$; $\Delta G_1^o = -n_1 F E_1^o = -5 	imes F 	imes 1.51 = -7.55 F$$(2) MnO_2 + 4H^+ + 2e^- 	o Mn^{2+} + 2H_2O$; $\Delta G_2^o = -n_2 F E_2^o = -2 	imes F 	imes 1.21 = -2.42 F$We need the reaction: $MnO_4^- + 4H^+ + 3e^- 	o MnO_2 + 2H_2O$This can be obtained by subtracting reaction $(2)$ from reaction $(1)$:$(1) - (2): (MnO_4^- + 8H^+ + 5e^-) - (MnO_2 + 4H^+ + 2e^-) 	o (Mn^{2+} + 4H_2O) - (Mn^{2+} + 2H_2O)$$MnO_4^- + 4H^+ + 3e^- 	o MnO_2 + 2H_2O$The change in Gibbs free energy is $\Delta G_3^o = \Delta G_1^o - \Delta G_2^o = -7.55 F - (-2.42 F) = -5.13 F$Since $\Delta G_3^o = -n_3 F E_3^o$ and $n_3 = 3$,we have:$-3 F E_3^o = -5.13 F$$E_3^o = \frac{5.13}{3} = 1.71 \ V$

$MnO_4^- (aq) + 8H^+ (aq) + 5e^- \to Mn^{2+} (aq) + 4H_2O (l)$; $E_1^o = 1.51 \ V$
$MnO_2 (s) + 4H^+ (aq) + 2e^- \to Mn^{2+} (aq) + 2H_2O (l)$; $E_2^o = 1.21 \ V$
$MnO_4^- (aq) + 4H^+ (aq) + 3e^- \to MnO_2 (s) + 2H_2O (l)$; $E_3^o = ?$
Value of $E_3^o$ will be ............ $V$

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$MnO_4^- (aq) + 8H^+ (aq) + 5e^- \to Mn^{2+} (aq) + 4H_2O (l)$; $E_1^o = 1.51 \ V$$MnO_2 (s) + 4H^+ (aq) + 2e^- \to Mn^{2+} (aq) + 2H_2O (l)$; $E_2^o = 1.21 \ V$$MnO_4^- (aq) + 4H^+ (aq) + 3e^- \to MnO_2 (s) + 2H_2O (l)$; $E_3^o = ?$Value of $E_3^o$ will be ............ $V$