How can I balance this redox reaction?

Question

#KMnO_4 +H_2C_2O_4 +H_2SO_4 -> K_2SO_4 + MnSO_4 + CO_2 + H_2O #

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Answer 1 · 2018-03-18T04:02:24

By the method of half equations...

Permanganate is reduced to COLOURLESS #Mn^(2+)#...and the colour change makes the reaction self-indicating...

#MnO_4^(-) +8H^+ +5e^(-) rarrMn^(2+) + 4H_2O(l)#

Oxalate ion, #C(+III)# is oxidized to #CO_2#...

#C_2O_4^(2-) rarr2CO_2+2e^(-)#...

and we take two of the former and five of the latter....

#2MnO_4^(-)+5C_2O_4^(2-) +16H^+ +10e^(-) rarr2Mn^(2+) + 8H_2O(l)+10CO_2+10e^(-)#
...to give after cancellation...

#2MnO_4^(-)+5C_2O_4H_2 +6H^+ rarr2Mn^(2+) + 8H_2O(l)+10CO_2#

And if you like we could add in some sulfate counterions...

#2HMnO_4(aq)+5C_2O_2H_2(aq) +2H_2SO_4(aq)rarr2Mn(SO_4)_2(aq) + 8H_2O(l)+10CO_2(g)uarr#