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]
