Question

The emf of the cell, $Pt|C{e}^{4+}(90$ normal calomel electrode is 1.464 V at 25?C. Find the value of equilibrium constant of the reaction: $2C{e}^{3+}+2H^{+}\to 2C{e}^{4+}+H_2$ The electrode potential of the normal calomel electrode is 0.28 V.

• A. $2.38\times {10}^{38}$
• B. $1.08\times {10}^{42}$
• C. $1.67\times {10}^{39}$
• D. $3.24\times {10}^{44}$

Answer 1

Answer: (A)

$E_{cell}=E_c-E_a=1.464V$ $E_{C{e}^{4+}/C{e}^{2+}}=E_{C{e}^{4+}/C{e}^{3+}}^o-\frac{0.059}{1}\log \frac{[C{e}^{3+}]}{[C{e}^{4+}]}$ $=E_{C{e}^{4+}/C{e}^{3+}}^o-\frac{0.059}{1}\log \frac{10}{90}$ $E_{cell}=E_{calo}-E^{o}C{e}^{4+}/C{e}^{3+}$ $1464=E_{calo}-\left[\left({E^o}_{C{e}^{4+}/C{e}^{3+}}-0.059\log \frac{1}{8}\right)\right]$ $1464=0.28-\left({E^o}_{C{e}^{4+}/C{e}^{3+}}-0.059\log \frac{1}{8}\right)$ ${E^o}_{C{e}^{4+}/C{e}^{3+}}=-1.24V$ Now, $2C{e}^{3+}+2H^{+}C{e}^{4+}+H_2$ At equilibrium, $E_{cell}=0$ $\therefore$ $E^o=\frac{0.059}{2}\log k$ At anode $2C{e}^{3+}\to 2C{e}^{4+}+2e^{-}$ (oxidation) At cathode $2H^{+}+2e^{-}\to H_2$ (reduction) Cell reaction $2C{e}^{3+}+2H^{+}\to 2C{e}^{4+}+H_2$ $E_{cell}^o=E^o$ red(cathode) $-E^o$ red(anode) $E_{red(H_2)}^o-E_{red}^o(C{e}^{4+}/C{e}^{3+})$ $=0-(-124)=124$ $E^o=\frac{0.059}{2}\log k$ $1.24=\frac{0.059}{2}\log k$ $\log k=42.03$ $k=1.08\times {10}^{42}$