Given : $\mathrm{Co}^{3+} \mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+} ; \mathrm{E}^{\circ}=1.81 \mathrm{~V}$ $\mathrm{Pb}^{4}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}^{2+} ; \mathrm{E}^{\circ}=+1.67 \mathrm{~V}$ $\mathrm{Ce}^{4+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ce}^{3+} ; \mathrm{E}^{\circ}+1.61 \mathrm{~V}$ $\mathrm{Bi}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Bi} ; \mathrm{E}^{\circ}=+0.20 \mathrm{~V}$ Oxidizing power of the species will increase in the order

(C) $\mathrm{Bi}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Co}^{3+}$

(A) $\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Co}^{3+}$

(B) $\mathrm{Co}^{3+}<\mathrm{Pb}^{4+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}$

(D) $\mathrm{Co}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Pb}^{4+}$

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Redox Reactions and Electrochemistry

Given :

Co^{3 +} e^{-} ⟶ Co^{2 +}; E^{\circ} = 1.81 V

Pb^{4} + 2 e^{-} ⟶ Pb^{2 +}; E^{\circ} = + 1.67 V

Ce^{4 +} + e^{-} ⟶ Ce^{3 +}; E^{\circ} + 1.61 V

Bi^{3 +} + 3 e^{-} ⟶ Bi; E^{\circ} = + 0.20 V

Oxidizing power of the species will increase in the order

Answer: (C)

Oxidising power tracks the standard reduction potential: the higher the

E^{\circ}

, the greater the pull for electrons and the stronger the oxidant. The given values are

E^{\circ} (Co X^{3 +} / Co X^{2 +}) = 1.81 V

E^{\circ} (Pb X^{4 +} / Pb X^{2 +}) = 1.67 V

E^{\circ} (Ce X^{4 +} / Ce X^{3 +}) = 1.61 V

and

E^{\circ} (Bi X^{3 +} / Bi) = 0.20 V

. Ranking by increasing potential, the oxidising power increases in the order

Bi X^{3 +} < Ce X^{4 +} < Pb X^{4 +} < Co X^{3 +}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option