Verified Solution • 2025–26 Session moderate

QJEEM19Chae6a: JEE Main 2019 Chemistry - Class 12 JEEM19Chae6a Solution: Given CO 3+ + e – CO 2 + ; E o = + 1.81 V Pb 4+ + 2e – Pb 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 :

Multiple Choice Options

  • A Co3+ < Ce4+ < Bi3+ < Pb4+
  • B Co3+ < Pb4+ < Ce4+ < Bi3+
  • C Ce4+ < Pb4+ < Bi3+ < Co3+
  • D Bi3+ < Ce4+ < Pb4+ < Co3+

Step-by-Step Verified Solution

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+

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring