NTA Abhyas JEE Main2020ChemistryElectrochemistryPractice
Given that : E O 2 / H 2 O ° = + 1.23 V , E S 2 O 8 2 - / S O 4 2 - ° = + 2.05 V E B r 2 / B r - ° = + 1.09 V E A u 3 + / Au ° = + 1.4 V The strongest oxidizing agent is
Options
- AO 2
- BB r 2
- CS 2 O 8 2 -
- DA u 3 +
Correct answer
C. S 2 O 8 2 -
Step-by-step solution
Oxidizing power ∝ Tendency to undergo reduction ∝ reduction potential value. S 2 O 8 2 - > A u 3 + > O 2 > B r 2