NEETChemistry-Electrochemistry
NEET Chemistry Redox Reactions MCQ Question
Type: MCQ-numerical-Medium-Class 12
Given below are half-cell reactions: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O; E° = +1.510 V 1/2 Ω₂ + 2H⁺ + 2e⁻ → H₂O₂; E° = +1.223 V Will the permanganate ion, MnO₄⁻ liberate O₂ from water in the presence of an acid?
A
Yes, because E°cell = +0.287 V
B
No, because E°cell = -0.287 V
C
Yes, because E°cell = +2.733 V
D
No, because E°cell = -2.733 V
Correct Answer
Option A
Detailed Explanation
The cell potential for the reaction is calculated by subtracting the reduction potential of O₂ from that of MnO₄⁻. Since the calculated E°cell is positive (+0.287 V), the reaction is spontaneous, and MnO₄⁻ can liberate O₂.
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