Question

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

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

Answer 1

Answer: (B)

For the reactions
$Fe^{2+}+2e^{-}\rightarrow Fe;E^{\circ}_{1}=-0.440\, V .....\left(i\right)$
$Fe^{3+}+e^{-}\rightarrow Fe^{2+};E^{\circ}_{2}=0.770\,V ....\left(ii\right)$
To obtain reaction,
$Fe^{3+}+3e^{-}\rightarrow Fe; E^{\circ}_{3}= ?$
We add $Eq. (i) + (ii)$
$\therefore \Delta G^{\circ}_{3}=\Delta G^{\circ}_{1}+\Delta G^{\circ}_{2} $
$=-n_{3}FE^{\circ}_{3}=-n_{1}FE^{\circ}_{1}+\left(-n_{2}FE^{\circ}_{2}\right)$
$\Rightarrow E^{\circ}_{3}=\frac{-n_{1}E^{\circ}_{1}-n_{2}E^{\circ}_{2}}{n_{3}E^{\circ}_{3}}$
$=\frac{2\left(-0.44\right)+1\times0.770}{3}$
$=\frac{-88+0.770}{3}=-0.037\,V$