I would write two half equations as
C2O4^2- ==> 2CO2 + 2e
2H2O + MnO4^- + 3e ==> MnO2 + 4OH^-
and balance.
I would think the spectator ions (Al for eample) would end up as Al(OH)3(s) although that isn't a redox reaction.
what is the product of reaction dehydrated K3(Al(C2O4)3) with KMnO4 in the present of KOH?
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