Given: C o ^ 3 + + e ^ - → C o ^ 2 + ; E ° = + 1.81V Pb ^ 3 + + 2e ^ - → Pb ^ 2 + ; E ° = + 1.67V C e ^ 4 + + e ^ - → C e ^ 3 + ; E ° = + 1.61V Bi ^ 3 + + 3e ^ - → Bi; E ° = + 0.20V Oxidizing power of the species will

Correct answer: B. Bi ^ 3 + < C e ^ 4 + < Pb ^ 4 + < Co ^ 3 + Higher the value of E _ Red ^ 0 ; higher will be its oxidizing power. Hence correct order is B i ^ + 3

JEE Main 201912 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Redox Reactions

JEE Main 2019 · 12 April · Morning Shift · Chemistry · Redox Reactions

Given: C o ^ 3 + + e ^ - → C o ^ 2 + ; E ° = + 1.81V Pb ^ 3 + + 2e...

Given:

C{o}^{3 +} + {e}^{-} \to C{o}^{2 +}; E ° = + 1.81V

{Pb}^{3 +} + {2e}^{-} \to {Pb}^{2 +}; E ° = + 1.67V

C{e}^{4 +} + {e}^{-} \to C{e}^{3 +}; E ° = + 1.61V

{Bi}^{3 +} + {3e}^{-} \to Bi; E ° = + 0.20V

Oxidizing power of the species will increase in the order:

Options

A $C{o}^{3} + < C{e}^{4 +} < B{i}^{3 +} < P{b}^{4 +}$
B ${Bi}^{3} + < C{e}^{4 +} < {Pb}^{4 +} < {Co}^{3 +}$
C $C{o}^{3} + < {Pb}^{4 +} < {Ce}^{3 +} < {Bi}^{4 +}$
D $C{e}^{4} + < {Pb}^{4 +} < {Bi}^{3 +} < {Co}^{3 +}$

Correct answer

B. ${Bi}^{3} + < C{e}^{4 +} < {Pb}^{4 +} < {Co}^{3 +}$

Step-by-step solution

Higher the value of ${\text{E}}_{\text{Red}}^{\text{0}};$ higher will be its oxidizing power.

Hence correct order is $B{i}^{+ 3} < C{e}^{+ 4} < P{b}^{+ 4} < C{o}^{+ 3}$

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