Question #d44a5

Apr 29, 2015

${E}_{r e a c t} = - 2.18 \text{V}$

The proposed reaction is:

$2 A {u}_{\left(s\right)} + 3 Z {n}_{\left(a q\right)}^{2 +} \rightarrow 2 A {u}_{\left(a q\right)}^{3 +} + 3 Z {n}_{\left(s\right)}$

So the 1/2 equations are :

$Z {n}_{\left(a q\right)}^{2 +} + 2 e \rightarrow Z {n}_{\left(s\right)}$ ${E}_{1} = - 0.76 \text{V}$

$A {u}_{\left(s\right)} \rightarrow A {u}_{\left(a q\right)}^{3 +} + 3 e$ ${E}_{2} = - 1.42 \text{V}$

So total potential = ${E}_{1} + {E}_{2} = - 1.42 + \left(- 0.76\right) = - 2.18 \text{V}$

$\Delta G = - n F {E}_{r e a c t} = - n F \left(- 2.18\right)$

This means $\Delta G$ is +ve so the reaction is not feasible.

A preferred approach is to use standard electrode potentials which should be listed negative to positive:

$Z {n}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s Z n$ ${E}^{0} = - 0.76 \text{V}$

$A {u}^{3 +} + 3 e r i g h t \le f t h a r p \infty n s A u$ ${E}^{0} = + 1.42 \text{V}$

Use the rule "Bottom left oxidises top right" or "top right reduces bottom left".

From this we can see that $A u$ is not able to reduce $Z {n}^{2 +}$ to $Z n$ as the Au3+/Au half cell ${E}^{0}$ value needs to be < -0.76V.