Question

While $F{e}^{3+}$ is stable, $M{n}^{3+}$ is not stabe in acid solution because:
$(M{n}^{3+}+e^{-}\to M{n}^{2+},E^{\circ }=1.5V;O_2+4H^{+}+4e^{-}\to 2H_{2}O,E^{\circ }=+1.23V;F{e}^{3+}+e^{-}\to F{e}^{2+},E^{\circ }=0.77V)$

• A. $O_{2}oxidisesM{n}^{2+}toM{n}^{3+}$
• B. $O_{2}oxidisesM{n}^{2+}andF{e}^{2+}toF{e}^{3+}$
• C. $F{e}_{3+}oxidises{H}_{2}Oto{O}_2$
• D. $M{n}^{3+}oxidises{H}_{2}OtoO$

Answer 1

The correct option is D

$M{n}^{3+}oxidises{H}_{2}OtoO$

$M{n}^{3+}$ oxides $H_{2}O$ to $O_2$ because the standard reduction potential of $(M{n}^{3+}+e^{-}\to M{n}^{2+};E^{\circ }=+1.5V)$ is greater than that $(O_2+4H^{+}+4e^{-}\to 2H_{2}O;E^{\circ }=+1.23V)$