Q. Using the following data, for the electrode potentials calculate $\mathrm{\Delta}G{^\circ},$ in kJ, for the indicated reaction $5{Ce}^{4 +}(aq) + {Mn}^{2 +}(aq) + 4H_{2}O(l) \rightarrow 5{Ce}^{3 +}(aq) + {MNO}_{4}^{-}(aq) + 8H^{+}(aq)$ ${MnO}_{4}^{-}(aq) + 8H^{+}(aq) + 5e^{-} \rightarrow {Mn}^{2 +}(aq) + 4H_{2}O(l)\ E{^\circ} = + 1.51\ V$ ${Ce}^{4 +}(aq) + e^{-} \rightarrow {Ce}^{3 +}(aq)\ E{^\circ} = + 1.61\ V$
A) $- 36.24$
B) $- 48.25$
C) $- 31.54$
D) $- 19.65$
Explanation
(b)
$E_{cell}^{{^\circ}} = E_{(RHS)}^{{^\circ}} - E_{(LHS)}^{{^\circ}} = 1.61 - 1.51 = 0.10\ V$
$\ \mathrm{\Delta}G^{{^\circ}} = - nFE^{{^\circ}}$
$= - 5 \times 96500 \times 0.10\ J$
$\ \mathrm{\Delta}G^{{^\circ}} = - 48.25\ kJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.