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Given CO⁢3++e⁢− ->CO⁢2+; E° = +1.81 V Pb⁢4++2e⁢− ->Pb⁢2+; E° = +1.67 V Ce⁢4++e⁢−->Ce⁢3+; E° = +1.61 V Bi⁢3++3e⁢−->Bi; E° = +0.20 V Oxidizing power of the species will increase in the order: - Chemistry (Theory)

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Question

Given

\[\ce{CO^{3+} + e- -> CO^{2+}}\]; E° = +1.81 V

\[\ce{Pb^{4+} + 2e- -> Pb^{2+}}\]; E° = +1.67 V

\[\ce{Ce^{4+} + e- -> Ce^{3+}}\]; E° = +1.61 V

\[\ce{Bi^{3+} + 3e- -> Bi^}\]; E° = +0.20 V

Oxidizing power of the species will increase in the order:

Options

  • Co3+ < Ce4+ < Bi3+ < Pb4+

  • Co3+ < Pb4+ < Ce4+ < Bi3+

  • Ce4+ < Pb4+ < Bi3+ < Co3+

  • Bi3+ < Ce4+ < Pb4+ < Co3+

MCQ
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Solution

Bi3+ < Ce4+ < Pb4+ < Co3+

Explanation:

The higher E° means stronger oxidising ability. So, Co3+ is the strongest oxidising agent among the given ions.

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Chapter 8: d-and ƒ-Block Elements - OBJECTIVE (MULTIPLE CHOICE) TYPE QUESTIONS [Page 502]

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Nootan Chemistry Part 1 and 2 [English] Class 12 ISC
Chapter 8 d-and ƒ-Block Elements
OBJECTIVE (MULTIPLE CHOICE) TYPE QUESTIONS | Q 89. | Page 502
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