Given $E_{{\text{Fe}}^{3+}/{\text{Fe}}^{2+}}^{\circ }=0.77\text{ V}$ and $E_{{\text{Fe}}^{2+}/\text{Fe}}^{\circ }=-0.44\text{ V}$. Calculate $E_{{\text{Fe}}^{3+}/\text{Fe}}^{\circ }$.

Solution

1. The reaction ${\text{Fe}}^{3+}+3e^{-}\to \text{Fe}$ is the sum of ${\text{Fe}}^{3+}+e^{-}\to {\text{Fe}}^{2+}$ and ${\text{Fe}}^{2+}+2e^{-}\to \text{Fe}$.
2. The Gibbs free energy change is $\Delta G_3^{\circ }=\Delta G_1^{\circ }+\Delta G_2^{\circ }$.
3. Using $\Delta G^{\circ }=-nF{E}^{\circ }$, we get $3E_3^{\circ }=1(0.77)+2(-0.44)$.
4. $$\begin{array}{rl}3E_3^{\circ }& =0.77-0.88\\ & =-0.11\end{array}$$.
5. $$\begin{array}{rl}{E}_3^{\circ }& =-0.11/3\\ & =-0.036\text{ V}\end{array}$$.
6. Hence the answer is (A).