A) + 4 to + 2
B) + 6 to + 4
C) + 7 to +2
D) + 7 to + 4
Correct Answer: C
Solution :
\[[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\xrightarrow{{}}M{{n}^{2+}}+4{{H}_{2}}O]\times 2\] \[\begin{align} & \underline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[{{C}_{2}}o_{4}^{2-}\xrightarrow{{}}2C{{O}_{2}}+2{{e}^{-}}]\times 5} \\ & \underline{\begin{align} & 2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\xrightarrow{{}}2M{{n}^{2+}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+10C{{O}_{2}}+8{{H}_{2}}O \\ \end{align}} \\ \end{align}\] So, oxidation number of Mn changes in the above reaction from + 7 to + 2.
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