In the reaction of oxalate with permaganate in acidic medium, the number of electrons involved in producing one molecule of CO₂ is

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answer is C.

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Detailed Solution

$\large MnO_4^ - \; + \;{C_2}O_4^{ - 2}\;\xrightarrow{{\;\;{H^ + }\;\;}}\;M{n^{ + 2}}\; + \;C{O_2}$

$\large \begin{array}{*{20}{c}} {{RHR}} \\ {{MnO_4^ - \; \to \;M{n^{ + 2}}}} \\ {Balancing\;'O'\;atom} \\ {MnO_4^ - \; \to \;M{n^{ + 2}}\; + \;4{H_2}O} \\ {Balancing\;H - atom\;in\;acidic\;medium} \\ {MnO_4^ - \; + \;8{H^ + }\; \to \;M{n^{ + 2}}\; + \;4{H_2}O} \\ {Balancing\;charges} \\ {MnO_4^ - \; + \;8{H^ + }\; + \;5\bar e\; \to \;M{n^{ + 2}}\; + \;4{H_2}O} \end{array}\left| {\begin{array}{*{20}{c}} {{OHR}} \\ {{{C_2}O_4^{ - 2}\; \to \;C{O_2}}} \\ {{Balancing\;'C'\;atoms}} \\ {{C_2}O_4^{ - 2}\; \to \;2C{O_2}} \\ {Balancing\;charge} \\ {{C_2}O_4^{ - 2}\; \to \;2C{O_2}\; + \;2\bar e} \end{array}} \right.$
$\large \begin{array}{*{20}{c}} {(RHR)\; \times 2\;\;\;2MnO_4^ - \; + \;16{H^ + }\; + \;10\bar e\; \to \;2M{n^{ + 2}}\; + \;8{H_2}O} \\ {(OHR)\; \times 5\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;5{C_2}O_4^{ - 2}\; \to \;10C{O_2}\; + \;10\bar e\;\;\;\;\;\;\;\;\;\;\;\;} \\ {\overline {\underline {Net\;reaction\;\;2MnO_4^ - + \;5{C_2}O_4^{ - 2} + 16{H^ + }\;\xrightarrow{{\;10\bar e\;}}\;2M{n^{ + 2}} + 10C{O_2} + 8{H_2}O\;\;} } } \end{array}$