Question

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`

Answer 1

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)`