To solve this problem, we need to determine the quantity of electricity required to oxidize $$MnO$_4^{2-}$ to $$MnO$_4^{-}. \\$The oxidation reaction can be represented as:$\\$MnO$_4^{2-} \rightarrow$MnO$_4^{-} + e^- \\$This reaction shows that one mole of $$MnO$_4^{2-}$ releases one mole of electrons.$\\$Given:$\\$• Amount of $$MnO$_4^{2-} = 0.1$ mol$\\$Since 1 mole of $$MnO$_4^{2-}$ releases 1 mole of electrons,$\\$0.1 mol of $$MnO$_4^{2-}$ will release 0.1 mol of electrons.$\\$The charge of 1 mole of electrons (Faraday's constant) is $96,500$ C/mol.$\\$Therefore, the total charge required is:$\\$Charge$= 0.1$ mol$\times 96,500$ C/mol$\\$Charge$= 9,650$ C$\\$Thus, the quantity of electricity required is $9,650$ C, which corresponds to Option 3.$\\$