Question

Given $: E^{o}_{Fe^{3+}/Fe}=-0.036V, E^{o}_{Fe^{2+}/Fe}=-0.439V$. The value of standard electrode potential for the change, $Fe^{3+}_{\left(aq\right)}+e^{-} \rightarrow Fe^{2+}\left(aq\right)$ will be :

• A. $-0.072\,V$
• B. $0.385 \,V$
• C. $0.770 \,V$
• D. $-0.270 \,V$

Answer 1

Answer: (C)

$\because Fe^{3+}+3e^{-} \rightarrow Fe; E^{o}=-0.036V$
$\therefore \Delta G^{o}_{1}=-nFE^{o}=-3F\left(-0.036\right)$
$=+0.108\,F$
Also $Fe^{2+}+2e^{-} \rightarrow Fe; E^{o}=-0.439\,V$
$\therefore \Delta G^{o}_{2}=-nFE^{o}$
$=-2\,F\left(-0.439\right)$
$=0.878\,F$
To find $E^{o}$ for $Fe^{3+}_{\left(aq\right)}+e^{-} \rightarrow Fe^{2+}\left(aq\right)$
$\Delta G^{o}=-nFE^{o}=-1FE^{o}$
$\because G^{o}=G^{o}_{1}-G^{o}_{2}$
$\because G^{o}=G^{o}_{1}-G^{o}_{2}$
$\therefore G^{o}=0.108F-0.878F$
$\therefore -FE^{o}=+0.108F-0.878F$
$\therefore E^{o}=0.878-0.108$
$=0.77v$