Write a balanced redox equation for the following in acidic solution?

#C_2O_4^(2-)+Cr_2O_7^(2-)\toCO_2+Cr^(3+)#

I understand that #C_2O_4^(2-)\to2CO_2+2e^(-)#
but I don't know how the #Cr_2O_7^(2-)# part works out.

My attempt (where do the electrons go?)
#Cr_2O_7^(2-)+14H^(+)...\to2Cr^(3+)+7H_2O#

1 Answer
Jul 17, 2018

Answer:

Well you gots a redox reaction here, which we address by the methods of half-equations that utilize electrons as stoichiometric particles...

Explanation:

Oxalate ion is oxidized to carbon dioxide...

#C_2O_4^(2-) rarr 2CO_2+2e^(-)# #(i)#

We gots here the transition, #C(+III)rarrC(+IV)#

And meanwhile dichromate ion is REDUCED to #Cr^(3+)#..the difference in oxidation numbers is as always the number of electrons we invoke....

#Cr_2O_7^(2-)+14H^+ +6e^(-) rarr 2Cr^(3+)+7H_2O(l)# #(ii)#

#underbrace(Cr(VI+))_"orange-red"# is REDUCED to

#underbrace(Cr(III+))_"green"#

And for the balanced redox equation we simply add the half-equations in such a way as to eliminate the electrons....i.e. #3xx(i)+(ii)#

#Cr_2O_7^(2-)+C_2O_4^(2-)+14H^+ +6e^(-) rarr 2Cr^(3+)+6CO_2+6e^(-)+7H_2O(l)#

....but we may cancel the electrons....

#Cr_2O_7^(2-)+C_2O_4^(2-)+14H^+ +cancel(6e^(-)) rarr 2Cr^(3+)+6CO_2+cancel(6e^(-))+7H_2O(l)#

...to give after cancellation...

#Cr_2O_7^(2-)+3C_2O_4^(2-)+14H^+ rarr 2Cr^(3+)+6CO_2+7H_2O(l)#

And so is this balanced with respect to MASS and charge...?

We could even further simplify this...by treating the #Cr(VI)# as chromic acid, #H_2Cr_2O_7#, and the oxalate ion as oxalic acid...#H(O=)C-C(=O)OH#...and deploying #H^+# as #HCl#.

#H_2Cr_2O_7+3C_2O_4H_2+6HCl rarr 2CrCl_3+6CO_2+7H_2O(l)#