Redox Reactions And Electrochemistry Ques 745

Question: While $ F{{e}^{3+}} $ is stable, $ M{{n}^{3+}} $ is not stable in acid solution because

Options:

A) $ O_2 $ oxidizes $ M{{n}^{2+}} $ to $ M{{n}^{3+}} $

B) $ O_2 $ oxidizes both $ M{{n}^{2+}} $ to $ M{{n}^{3+}} $ and $ F{{e}^{2+}} $ to $ F{{e}^{3+}} $

C) $ F{{e}^{3+}} $ oxidizes $ H_2O $ to $ O_2 $

D) $ M{{n}^{3+}} $ oxidises $ H_2O $ to $ O_2 $

Show Answer

Answer:

Correct Answer: D

Solution:

[d] $ 4M{{n}^{3+}}+2H_2o\to 4M{{n}^{2+}}+O_2+4{{H}^{+}} $

$ E^{o} _{M{{n}^{3+}}/M{{n}^{2+}}}+E^{o} _{H_2O/O_2}=1.50+( -1.23 )=0.27V $

Reaction is feasible. [ $ \therefore $ $ E^{o} $ is positive]



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