And so #"oxidation half-equation: C(+III)"rarr"C(+IV):"#
#C_2O_4^(2-) rarr 2CO_2(g)uarr+2e^(-)# #(i)#
Mass and charge are balanced as required.....
And so #"reduction half-equation: Mn(+VII)"rarr"Mn(+II):"#
#underbrace(MnO_4^(-))_"deep purple"+8H^+ + 5e^(-)rarr#
#underbrace(Mn^(2+))_"almost colourless" + 4H_2O(l)# #(ii)#
To produce the balanced equation, all we need do is to add (i) and (ii) such that electrons are removed in the final equation. We take #5xx(i)+2xx(ii)#....
#5C_2O_4^(2-) +2MnO_4^(-)+16H^+ + cancel(10e^(-))rarr 2Mn^(2+) + 8H_2O(l)+10CO_2(g)uarr+cancel(10e^(-))#
....to give.....
#5C_2O_4^(2-) +2MnO_4^(-)+16H^+ rarr 2Mn^(2+) +10CO_2(g)uarr+ 8H_2O(l)#
And note that as a bonus this reaction is self-indicating, given the fact that #Mn^(2+)#, as a #d^5# ion, is almost colourless, and permanganate ion is strongly coloured. You can get very precise endpoints with this titration. Of course, this is why it is used in an undergraduate lab....
