Question

Calculate the equilibrium constant for the reaction, $2Fe^{3+} +3I^{-} \rightleftharpoons 2Fe^{2+} +I^{-}_3$. The standard reduction potentials in acidic conditions are $0.77 \,V$ and $0.54\, V$ respectively for $Fe^{3+} / Fe^{2+}$ and $I^{-}_3 / I^{-}$ couples.

• A. $4.25 \times 10^{7}$
• B. $7.05 \times 10^{5}$
• C. $6.25 \times 10^{5}$
• D. $6.25 \times 10^{7}$

Answer 1

Answer: (D)

$2Fe^{3+} +3I^{-} \rightleftharpoons 2Fe^{2+} +I^{-}_3$
For the above change at equilibrium, $E = 0$.
Using the relation, $E=E^{\circ}-\frac{0.059}{2} log\,K_{c}$
or $0=E^{\circ}_{Fe^{3+} /Fe^{2+}}-E^{\circ}_{I^{-}_{3} /I^{-}}-\frac{0.059}{2} log\,K_{c}$
or $0=0.77-0.54-\frac{0.059}{2} log\,K_{c}$
or $0=0.23-\frac{0.059}{2} log\,K_{c}$
or $\frac{0.059}{2} log\,K_{c}=0.23$ or $log\,K_{c}=\frac{0.23\times2}{0.059}$
or $log\,K_{c}=7.796$ or $K_{c}=6.25\times10^{7}$