Question

Balance the following equation by the method of oxidation-reduction showing all electron tranfers. Please show work.

___ H₂C₂O₄ + ____ KMnO₄ ---> ____ CO₂ + ____ Mn₂O₃ + ____ H₂O

Identify (a) the atom oxidized, (b) the atom reduced, (c) the reducing agent, (d) the oxidation half reaction, and (e) the reduction half reaction.

Answer 1

(a) Since OS of H is +1, O is -2 and K is +1, Oxidation state of C in H2C2O4 can be calculated as

2 + 2x +(-2*4) = 0 => x = +3

Hence Oxidation state of C in H2C2O4 is +3

Oxidation state of C in CO2 can be calculated as

x+(-2*2) = 0 => x = +4

Hence C atom is oxidised from +3 to +4.

(b) Oxidation state of Mn in KMnO4 can be calculated as

(+1)+ x+(-2*4) = 0 => x = +7

Oxidation state of Mn in Mn2O3 can be calculated as

2x+(-2*3) = 0 => x = +3

Hence Mn atom is reduced from +7 to +4.

(c) A reducing agent helps in reducing and itself gets oxidised. Since here H2C2O4 is oxidised to CO2, hence H2C2O4 acts as a reducing agent.

(d) Oxidation half reaction:

C³⁺  ---------------> C⁴⁺  + 1e-

(e) Reduction half reaction:

Mn⁺⁷ + 4e- ---------> Mn³⁺

Here in the balanced reaction, there would be one extra prdcut which is K2O. and the balanced reaction is

4H2C2O4 + 2KMnO4 -------------> 8CO2 + K2O + Mn2O3 + 4H2O