Question

Calculate the equilibrium constant for the reaction, $2F{e}^{3+}+3I^{-}\rightleftharpoons 2F{e}^{2+}+I_3^{-}$. The standard reduction potentials in acidic conditions are $0.77V$ and $0.54V$ respectively for $F{e}^{3+}/F{e}^{2+}$ 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)

$2F{e}^{3+}+3I^{-}\rightleftharpoons 2F{e}^{2+}+I_3^{-}$
For the above change at equilibrium, $E=0$.
Using the relation, $E=E^{\circ }-\frac{0.059}{2}log{K}_c$
or $0=E_{F{e}^{3+}/F{e}^{2+}}^{\circ }-E_{I_3^{-}/I^{-}}^{\circ }-\frac{0.059}{2}log{K}_c$
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 $log{K}_c=\frac{0.23\times 2}{0.059}$
or $log{K}_c=7.796$ or $K_c=6.25\times 10^7$