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 \[{{H}_{2}}O\] to \[{{O}_{2}}\]
D) \[M{{n}^{3+}}\] oxidises \[{{H}_{2}}O\] to \[{{O}_{2}}\]
Correct Answer: D
Solution :
[d] \[4M{{n}^{3+}}+2{{H}_{2}}o\to 4M{{n}^{2+}}+{{O}_{2}}+4{{H}^{+}}\] \[{{E}^{o}}_{M{{n}^{3+}}/M{{n}^{2+}}}+{{E}^{o}}_{{{H}_{2}}O/{{O}_{2}}}=1.50+\left( -1.23 \right)=0.27V\] Reaction is feasible. [\[\therefore \] \[{{E}^{o}}\] is positive]
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