Question

The equilibrium constant of the following redox reaction at $298\, K$ is $1 \times 10^{8}$
$2 Fe ^{3+}$ (aq. $)+2 I ^{-}$(aq. $) \rightleftharpoons 2 Fe ^{2+}$ (aq. $)+ I _{2}$ (s)
If the standard reduction potential of iodine becoming iodides is $+0.54 V$. What is the standard reduction potential of $Fe ^{3+} / Fe ^{2+}$ ?

• A. + 1.006 V
• B. - 1.006 V
• C. + 0.77 V
• D. - 0.77 V

Answer 1

Answer: (C)

$E^\circ _{cell} = \frac{2.303 RT}{nF} \log \, K_{eq}$
$= \frac{2.303 \times 8.314 \times 298}{2 \times 96500} \times \log [10^8] = 0.236 \, V$
Now,
$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$ (At cathode); Reduction $2I^- \rightarrow I_2 + 2e^-$ (At anode) ; Oxidation and we know that
$E^\circ _{\text{cell}} = E^\circ _{\text{cathode}} - E^\circ _{\text{anode}}$
given, $E _{\text{anode}} = \text{0.54 \, V}$
$\therefore \:\:\:\: E _{\text{cathode}} = E^\circ _{\text{cell}} + E _{\text{anode}} = 0.236 + 0.54$
$= +0.776\, V$