A) Reduction of Mn
B) Reduction of \[{{\text{C}}_{\text{2}}}\text{O}_{4}^{2-}\]
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 \[\text{MnO}_{4}^{-}=+7\] Oxidation number of \[\text{Mn}\]in \[\text{M}{{\text{n}}^{\text{2+}}}\text{=}\,\text{+}\,\text{2}\] 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 \[+\,4\] \[\therefore \]is oxidised during reaction.
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