Ion electron method MnO43- + C2O42-. …

First step:- The skeletal equation is to be written:- (MnO4)3- + (C2O4)2- ------> (Mn)+2 + CO2 Second step:- We've to seperately the balanced oxidation half and reduction half The oxidation half:- (C2O4)2- ---> 2CO2 The reduction half:- (MnO4)3- ---> (Mn)2+ Step three:- To balance oxygen, add required no. of water molecules in both halves Reduction half:- (MnO4)3- + (4H)+ ---> (Mn)2+ + 4H2O (Add H+ because it is in Acidic medium) Step four:- To balance the charge, add required no. of electrons Reduction half:- (MnO4)3- + 4H+ ---> Mn2+ + 4H2O + e- Oxidation Half:- (C2O4)2- ---> 2CO2 + 2e- Step five:- To equal the no. of electrons on both sides , multiply suitable constant Here, -2 is to be multiplied to Reduction Half, -2(MnO4)3- -8H+ ---> -2Mn2+ -8H2O - 2e- Now adding both final equations of Reduction and Oxidation We get, (C2O4) 2- - (MnO4)3- -8H+ ---> 2CO2 - 2Mn2+ -8H2O Done !!!

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