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)

$0.770 \,V$

Answer 2

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