# Question 18173

Jan 6, 2018

$\textsf{{E}^{\circ} = + 1.29 \textcolor{w h i t e}{x} V}$

#### Explanation:

You haven't given the relevant $\textsf{{E}^{\circ}}$ values so I looked them up:

$\textsf{A {u}^{3 +} + 3 e r i g h t \le f t h a r p \infty n s A u \textcolor{w h i t e}{\times \times \times \times} {E}^{\circ} = + 1.42 \textcolor{w h i t e}{x} V}$

$\textsf{A {u}^{+} + e r i g h t \le f t h a r p \infty n s A u \textcolor{w h i t e}{\times \times \times \times \times} {E}^{\circ} = + 1.68 \textcolor{w h i t e}{x} V}$

sf(Au^(3+)+2erightleftharpoonsAu^(+)color(white)(xxxxxxx)sf(E^@# is unknown

We need to convert these to free energy values and then apply the conservation of energy.

$\textsf{\Delta {G}^{\circ} = - n F {E}^{\circ}}$

$\textsf{F}$ is the Faraday Constant

$\textsf{n}$ is the no. of moles of electrons transferred.

Let $\textsf{{E}^{\circ}}$ be the value we want to find.

We can say that:

$\textsf{\Delta {G}^{\circ} \left[A {u}^{3 +} \rightarrow A u\right] = \Delta {G}^{\circ} \left[A {u}^{3 +} \rightarrow A {u}^{+}\right] + \Delta {G}^{\circ} \left[A {u}^{+} \rightarrow A u\right]}$

$\therefore$$\textsf{- 3 \cancel{F} \times 1.42 = - 2 \cancel{F} {E}^{\circ} - \cancel{F} \times 1.68}$

$\therefore$$\textsf{- 2 {E}^{\circ} = - 4.26 + 1.68 = - 2.58 \textcolor{w h i t e}{x} V}$

$\textsf{{E}^{\circ} = + \frac{2.58}{2} = + 1.29 \textcolor{w h i t e}{x} V}$