Chemistry - Electrochemistry Question with Solution | TestHub

$M{n}^{2+}\to M{n}^{3+}{E}^{\circ }=1.5V$

$M{n}^{3+}\to Mn{O}_2{E}^{\circ }=1.0V$

Reaction:

$2M{n}^{3+}+2H_{2}O\to M{n}^{2+}+Mn{O}_2+4H^{+}$

disproportionation reaction of $M{n}^{3+}$.

Cell Potential

$E_{cell}^{\circ }=E_{reduction}^{\circ }-E_{oxidation}^{\circ }$

$E_{cell}^{\circ }=1.5-1.0$

$E_{cell}^{\circ }=0.5V$

Use Gibbs Energy Formula

$\Delta G^{\circ }=-nF{E}^{\circ }$

Where

$n=1$

$F=96500C$

$\Delta G^{\circ }=-(1)(96500)(0.5)$

$\Delta G^{\circ }=-48250J$

Convert to kJ

$\Delta G^{\circ }=-48.25kJ$