A) 4 to 2
B) 6 to 4
C) +7 to +2
D) 7 to 4
Correct Answer: C
Solution :
\[\overset{+7}{\mathop{2KMn{{O}_{4}}}}\,+3{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+\overset{+2}{\mathop{2MnS{{O}_{4}}}}\,\] \[+3{{H}_{2}}O+5[O]\] \[5{{H}_{2}}{{C}_{2}}{{O}_{4}}+5[O]\xrightarrow[{}]{{}}10C{{O}_{2}}+5{{H}_{2}}O\] Here,\[M{{n}^{+7}}\]in\[KMn{{O}_{4}}\]has been reduced to \[M{{n}^{2+}}\]in\[MnS{{O}_{4}}\].
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