Given below are half cell reactions:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, E°(Mn²⁺/MnO₄⁻) = -1.510 V
½O₂ + 2H⁺ + 2e⁻ → H₂O, E°(O₂/H₂O) = +1.223 V
Will the permanganate ion, MnO₄⁻liberate O₂ from water in the presence of an acid?
Given below are half cell reactions:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, E°(Mn²⁺/MnO₄⁻) = -1.510 V
½O₂ + 2H⁺ + 2e⁻ → H₂O, E°(O₂/H₂O) = +1.223 V
Will the permanganate ion, MnO₄⁻liberate O₂ from water in the presence of an acid?
Option 1 -
Yes, because E°cell = +0.287 V
Option 2 -
No, because E°cell = -0.287 V
Option 3 -
Yes, because E°cell = +2.733 V
Option 4 -
No, because E°cell = -2.733 V
Asked by Shiksha User
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1 Answer
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Correct Option - 1
Detailed Solution:E°cell = E°cathode - E°anode = (+1.510) - (1.229) = +0.287 V. As E°cell is positive the cell will work.
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