A) 7 to 5
B) 2 to 7
C) 5 to 7
D) 7 to 2
Correct Answer: D
Solution :
\[2KMn{{O}_{4}}+5{{H}_{2}}{{C}_{2}}{{O}_{4}}+6HCl\xrightarrow{{}}2MnC{{l}_{2}}\] \[+10C{{O}_{2}}+8{{H}_{2}}O\] \[\therefore \] \[MnO_{4}^{-}\xrightarrow[{}]{{}}M{{n}^{2+}}\] Oxidation state of Mn in\[MnO_{4}^{-}\] \[x+4{{x}^{-2}}=-1\] \[x+(-8)=-1\] \[x=7\] Oxidation state of Mn in\[M{{n}^{2+}}=2\] \[\therefore \]Change in oxidation state = 7 to 2You need to login to perform this action.
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