Question

If $E{°}_{F{e}^{2+}/Fe}=-0.440V$ and $E{°}_{F{e}^{3+}/F{e}^{2+}}=0.770V$, then $E{°}_{F{e}^{3+}/Fe}$ is

• A. 0.330 V
• B. -0.37 V
• C. -0.330 V
• D. -1.210 V

Answer 1

Answer: (B)

For the reactions
$F{e}^{2+}+2e^{-}\to Fe;E_1^{\circ }=-0.440V.....\left(i\right)$
$F{e}^{3+}+e^{-}\to F{e}^{2+};E_2^{\circ }=0.770V....\left(ii\right)$
To obtain reaction,
$F{e}^{3+}+3e^{-}\to Fe;E_3^{\circ }=?$
We add $Eq.(i)+(ii)$
$\therefore \Delta G_3^{\circ }=\Delta G_1^{\circ }+\Delta G_2^{\circ }$
$=-n_{3}F{E}_3^{\circ }=-n_{1}F{E}_1^{\circ }+\left(-n_{2}F{E}_2^{\circ }\right)$
$\Rightarrow E_3^{\circ }=\frac{-n_1{E}_1^{\circ }-n_2{E}_2^{\circ }}{n_3{E}_3^{\circ }}$
$=\frac{2\left(-0.44\right)+1\times 0.770}{3}$
$=\frac{-88+0.770}{3}=-0.037V$