Chemistry - Electrochemistry Question with Solution | TestHub

As oxidant ${\mathrm{H}}_2{\mathrm{O}}_2$ is reduced to water (1.77 V); as reductant it is oxidised to ${\mathrm{O}}_2$ (0.69 V).

(a) 0.15 V is below both. ${\mathrm{H}}_2{\mathrm{O}}_2$ oxidises ${\mathrm{Sn}}^{2+}$ ($1.77>0.15$). It cannot reduce ${\mathrm{Sn}}^{4+}$ since $0.15<0.69$. Oxidant only.

(b) 1.82 V is above 1.77 V, so ${\mathrm{Co}}^{3+}$ oxidises ${\mathrm{H}}_2{\mathrm{O}}_2$ and ${\mathrm{H}}_2{\mathrm{O}}_2$ cannot oxidise ${\mathrm{Co}}^{2+}$. Reductant only.

(c) 0.80 V lies between 0.69 and 1.77: ${\mathrm{H}}_2{\mathrm{O}}_2$ oxidises $\mathrm{Ag}$ ($1.77>0.80$) and reduces ${\mathrm{Ag}}^{+}$ ($0.80>0.69$). Both.

Answer: (C) - (a) oxidant, (b) reductant, (c) both.