Question

Standard electrode potentials of
$F{e}^{2+}+2e^{–}\to Fe$ and $F{e}^{3+}+3e^{–}\to Fe$
are $–0.440\text{V}$ and $–0.036\text{V}$ respectively.
The standard electrode potential ($E^0$) for
$F{e}^{3+}+e^{–}\to F{e}^{2+}$ is:

• A. $–0.476\text{V}$
• B. $–0.404\text{V}$
  
• C. $+0.404\text{V}$
• D. $+0.772\text{V}$

Answer 1

The correct option is D $+0.772\text{V}$

$F{e}^{2+}+2e^{-}\to Fe;E_1^0=-0.440\text{V}......\left(1\right)$

$F{e}^{3+}+3e^{-}\to Fe;E_2^0=-0.036\text{V}......\left(2\right)$

$\Delta G^0=-nF{E}_{cell}^0$​

It is given in the question that
$F{e}^{3+}+e^{-}\to F{e}^{2+}......\left(3\right)$
To obtain the equation $3$, we must subtract the equation $\left(1\right)$ from $\left(2\right)$
$\Delta G_{cell}^0=\Delta G_2-\Delta G_1\Rightarrow -nF{E}_{cell}^0=-n_{2}F{E}_2^0-(-n_{1}F{E}_1^0)$

$n_2=3$ and $n_1=2$

For the equation $\left(3\right)$, $n=1$
$-E_{cell}^0=-3\times E_2^0+2\times E_1^0\Rightarrow -E_{cell}^0=-3\times (-0.036)+2\times (-0.44)\Rightarrow -E_{cell}^0=0.108-0.88\Rightarrow E_{cell}^0=-0.108+0.88=+0.772$