Question

Write a balanced oxidation reduction equation in acidic solution for: `C_2O_4^(2-)+Cr_2O_7^(2-)\toCO_2+Cr^(3+)`?

Answer 1

AS you have twigged this is a redox reaction, which we can solve by the method of half equations...

Explanation:

`"Oxalate ion"`, `C_2O_4^(2-)`, i.e. `C(+III)` is completely oxidized up to carbon dioxide `C(+IV)`;

`C_2O_4^(2-) rarr 2CO_2(g)+2e^(-)` `(i)`

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

And meanwhile dichromate, `Cr(+VI)` is reduced to `Cr(+III)`..

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

We take `3xx(i)+(ii)` to retire the electrons, virtual particles of convenience, to give....

`underbrace(Cr_2O_7^(2-))_"orange-red" +3C_2O_4^(2-)+14H^+ rarr underbrace(2Cr^(3+))_"green"+6CO_2(g) + 7H_2O`

….after cancellation...

And if this AIN'T balanced with respect to mass and charge, then it cannot be accepted as a representation of chemical reality .. Is it right?

We can usually assign oxidation numbers to organic compounds give the same rules that we employ in inorganic chemistry. Methane is `C(-IV)`...`CO-=C(+II)`...ethane is `C(-III)`...`CO_2-=C(+IV)`...and `"propane"` `-=` `H_3stackrel(-III)C-stackrel(-II)CH_2CH_3`

Answer 2

Please see the equation below

Explanation:

The half-reactions are

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

`C_2O_4^(2-)rarr 2CO_2+2e^-`

Therefore,

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

`3C_2O_4^(2-)rarr 6CO_2+6e^-`

Adding the last `2` equations

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