JEE MAIN - Chemistry (2019 - 12th April Morning Slot - No. 2)
Given
CO3+ + e– $$ \to $$ CO2+ ; Eo = + 1.81 V
Pb4+ + 2e– $$ \to $$ Pb2+ ; Eo = + 1.67 V
Ce4+ + e– $$ \to $$ Ce3+ ; Eo = + 1.61 V
Bi3+ + 3e– $$ \to $$ Bi ; Eo = + 0.20 V
Oxidizing power of the species will increase in the order :
CO3+ + e– $$ \to $$ CO2+ ; Eo = + 1.81 V
Pb4+ + 2e– $$ \to $$ Pb2+ ; Eo = + 1.67 V
Ce4+ + e– $$ \to $$ Ce3+ ; Eo = + 1.61 V
Bi3+ + 3e– $$ \to $$ Bi ; Eo = + 0.20 V
Oxidizing power of the species will increase in the order :
Co3+ < Ce4+
< Bi3+ < Pb4+
Co3+ < Pb4+ < Ce4+
< Bi3+
Ce4+
< Pb4+ < Bi3+ < Co3+
Bi3+ < Ce4+
< Pb4+ < Co3+
Explanation
Higher the value of Standard reduction potential of metal ion electrode, the more strong oxidizing agent is that metal ion electrode and have greater oxidising power.
So, Bi3+ < Ce4+ < Pb4+ < Co3+
So, Bi3+ < Ce4+ < Pb4+ < Co3+
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