A) 10
B) 2
C) 1
D) 5
Correct Answer: C
Solution :
\[2\,Mn{{O}^{-}}_{4}+5{{C}_{2}}{{O}^{2-}}_{4}+16\,{{H}^{+}}\to \,\,2\,M{{n}^{2+}}+\,10C{{O}_{2}}\,+\,8\,{{H}_{2}}O\] The number of \[{{e}^{-}}\] involved in producing 10 mole of \[C{{O}_{2}}\] is 10, so number of \[{{e}^{-}}\] involved in producing 1 mole of \[C{{O}_{2}}\] will be 1.
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