A) Reduction of\[Mn\]
B) Reduction of\[{{C}_{2}}O_{4}^{2-}\]
C) Oxidation of\[Mn\]
D) None of these
Correct Answer: A
Solution :
Key Idea: (i) Loss of electron and increase in oxidation number is oxidation. (ii) Gain of electron and decrease in oxidation number is reduction. \[MnO_{4}^{-}+{{H}^{+}}+{{C}_{2}}O_{4}^{2-}\xrightarrow{{}}\] \[M{{n}^{2+}}+{{H}_{2}}O+C{{O}_{2}}\] Oxidation number of \[Mn\] in\[MnO_{4}^{-}=+7\] Oxidation number of \[Mn\] in\[M{{n}^{2+}}=+2\] Oxidation number of \[C\] in\[{{C}_{2}}O_{4}^{2-}=+3\] Oxidation number of \[C\] in\[C{{O}_{2}}=+4\] For\[Mn\], oxidation number is decreasing from\[+7\]to\[+2\] \[\therefore \]\[Mn\]is getting reduced during reaction. For\[C\], oxidation number is increasing from\[+3\] to\[+4\] \[\therefore \]\[C\]is oxidised during reaction.
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