Question

How do we represent the oxidation of `Cr^(3+)` ion to `CrO_4^(2-)` by hydrogen peroxide, using the method of half-equations?

Answer 1

By the method of half-equations............

`2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+)`

Explanation:

`"Chromic chloride"`, `CrCl_3`, is oxidized from `Cr(+III)` to `Cr(+VI)` in chromate:

`Cr^(3+) + 4H_2O rarr stackrel(VI+)(Cr)O_4^(2-) +8H^(+)+ 3e^(-)` `(i)`

Is this balanced with respect to mass and charge? It is your problem not mine.

Hydrogen peroxide, `H_2O_2`, is reduced from `O^(-I)` to `O^(-II)`.

`H_2O_2 +2H^(+) + 2e^(-) rarr 2H_2O` `(ii)`

Again is the thing balanced?

To give the overall redox equation, we cross-multiply the individual redox equations: `2xx(i) + 3xx(ii)`:

`2Cr^(3+) + cancel(8)2H_2O +3H_2O_2 +cancel(6H^(+)) + cancel(6e^(-)) rarr cancel(6H_2O)+2CrO_4^(2-) +cancel(16)10H^(+)+ cancel(6e^(-))`

And then we SUBTRACT the superfluous reagents from each side:

`2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+)`

But you wanted the reaction done under basic conditions. And so we add `10xxHO^(-)` to BOTH SIDES OF THE REACTION:

`2Cr^(3+) +3H_2O_2 +10HO^(-) rarr 2CrO_4^(2-) +8H_2O`

Is this balanced with respect to mass and charge?