Question #d2fbd

1 Answer
Jun 5, 2016

Answer:

Manganate, #Mn(VI)#, is reduced to #Mn(III)#. Oxalic acid is oxidized to #CO_2#.

Explanation:

Reduction half equation:

#2MnO_4^(2-) + 10H^(+) + 6e^(-) rarr Mn_2O_3 +5H_2O# #(i)#

Oxidation half equation:

#HO(O=)C-C(=O)OH rarr 2CO_2uarr +2H^(+) +2e^(-)# #(ii)#

So #(i)+3xx(ii)=#

#2MnO_4^(2-) + 3{HO(O=)C}_2 +4H^(+)rarr Mn_2O_3 +5H_2O +6CO_2#

Which (I think) is balanced with respect to mass and charge as required.

You've got potassium oxide in your equation, for which I have not accounted. The redox couple is manganese and carbon.