When $0.1 \mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }{\mathrm{M}\mathrm{n}\mathrm{O}}_4^{2-}$ is oxidized the quantity of electricity required to completely $\mathrm{M}\mathrm{n}{\mathrm{O}}_4^{2-}$ to $\mathrm{M}\mathrm{n}{\mathrm{O}}_4^{-}$ is

• A

  $96500\mathrm{ }\mathrm{C}$
• B

  $2\times 96500\mathrm{ }\mathrm{C}$
• Correct answer
  C

  $9650\mathrm{ }\mathrm{C}$
• D

  $96.50\mathrm{ }\mathrm{C}$

Hence, the correct answer is option: 'C'.