How do we represent the disproportionation of oxalic acid, #HO_2C-CO_2H# to carbon dioxide and carbon monoxide, in an acidic medium?

Question

287 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-10T19:04:51

Oxalic acid undergoes disproportionation in a strongly acidic medium....

We got #HO_2C-CO_2H#, #C(+III)#, which is oxidized to #C(IV+)# in #CO_2#:

#HO_2C-CO_2H(aq) rarr 2CO_2 + 2H^+ +2e^(-)# #(i)#

And #"oxalic acid"# is reduced to #"carbon monoxide"#, #C(+II)#:

#HO_2C-CO_2H(aq) +2H^+ + 2e^(-) rarr 2CO + 2H_2O(l)# #(ii)#

And we add #(i) + (ii)# to eliminate the electrons....

#2HO_2C-CO_2H(aq) +cancel(2H^+ + 2e^(-)) rarr 2CO_2+2CO + 2H_2O(l) + cancel(2H^+ +2e^(-))#

#2HO_2C-CO_2H(aq) rarr 2CO_2+2CO + 2H_2O(l)#

Charge is balanced, and mass is balanced, so this is kosher...