Question

E.M.F. diagram for some ions is given as
$FeO_{4}^{2 -}\overset{E^{o} = + 2.20 V}{ \rightarrow }Fe^{3 +}\overset{E^{o} = + 0.77 V}{ \rightarrow }Fe^{2 +}\overset{E^{o} = - 0.445 V}{ \rightarrow }Fe^{o}$
Determine the value of $E_{F e O_{4}^{2 -} / F e^{2 +}}^{o}$

• A. $1.84V$
• B. $1.42V$
• C. $1.3V$
• D. $2.0V$

Answer 1

Answer: (A)

$Fe^{6 +}+3e \rightarrow Fe^{3 +};$ $-\Delta G_{1}^{o}=3\times 2.20\times F$ …(i)
$Fe^{3 +}+e \rightarrow Fe^{2 +};$ $-\Delta G_{2}^{o}=1\times 0.77\times F$ …(ii)
$Fe^{2 +}+2e \rightarrow Fe^{o};$ $-\Delta G_{3}^{o}=2\times \left(\right.-0.445\left.\right)\times F$ …(iii)
Adding equation (i) and (ii)
$Fe^{6 +}+4e \rightarrow Fe^{2 +};$ $-\Delta G_{4}^{o}=-\Delta G_{1}^{o}-\Delta G_{2}^{o}$
Hence $4\times E_{4}^{o}\times F=3\times 2.20\times F+\left(1 \times 0.77 \times F\right)$
So $E_{4}^{o}=1.84V$
or $Fe^{6 +}+4e \rightarrow Fe^{2 +};$ $E^{o}=1.84V$