If E_{Au^{+}/Au}^o is 1.69 text{ V} and E_{Au | vedclass

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(C) The given half-reactions are:$(1) Au^{+} + e^- ightarrow Au$,$E_1^o = 1.69 	ext{ V}$,$\Delta G_1^o = -1 	imes F 	imes 1.69 = -1.69F$$(2) Au^{

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$-1.255$

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(C) The given half-reactions are:$(1) Au^{+} + e^- ightarrow Au$,$E_1^o = 1.69 	ext{ V}$,$\Delta G_1^o = -1 	imes F 	imes 1.69 = -1.69F$$(2) Au^{3+} + 3e^- ightarrow Au$,$E_2^o = 1.40 	ext{ V}$,$\Delta G_2^o = -3 	imes F 	imes 1.40 = -4.20F$For the target reaction $Au^{+} ightarrow Au^{3+} + 2e^-$,we subtract reaction $(2)$ from reaction $(1)$:$\Delta G_3^o = \Delta G_1^o - \Delta G_2^o$$-2FE_3^o = (-1.69F) - (-4.20F)$$-2FE_3^o = 2.51F$$E_3^o = -1.255 	ext{ V}$

If $E_{Au^{+}/Au}^o$ is $1.69 \text{ V}$ and $E_{Au^{3+}/Au}^o$ is $1.40 \text{ V}$,then $E_{Au^{+}/Au^{3+}}^o$ will be ... $\text{V}$.

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