Question

The equilibrium constant of the following redox reaction at $298K$ is $1\times 10^8$
$2F{e}^{3+}$ (aq. $)+2I^{-}$(aq. $)\rightleftharpoons 2F{e}^{2+}$ (aq. $)+I_2$ (s)
If the standard reduction potential of iodine becoming iodides is $+0.54V$. What is the standard reduction potential of $F{e}^{3+}/F{e}^{2+}$ ?

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

Answer 1

Answer: (C)

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