Question

Balance the chemical equation by using oxidation number method (not by ion electron method) [Cr(OH)4]^- + H2O2 --->CrO4^2- + H2O(Basic medium)?

Answer 1

Well `Cr(III+)` is oxidized to `Cr(VI+)`....

`2Cr(OH)_4^(-) +3H_2O_2 +2HO^(-) rarr 2CrO_4^(2-)+8H_2O`

Explanation:

And so....

`stackrel(III)Cr(OH)_4^(-) rarr stackrel(VI)CrO_4^(2-)+4H^+ + 3e^(-)`

...and we use electrons as virtual particles, to account for the difference in oxidation numbers, but basic conditions were specified so we add `4xxHO^-` to each side to give....

`stackrel(III)Cr(OH)_4^(-) +4HO^(-) rarr stackrel(VI)CrO_4^(2-)+underbrace(4H^+ +4HO^(-))_(4H_2O)+ 3e^(-)`

and thus....

`Cr(OH)_4^(-) +4HO^(-) rarr CrO_4^(2-)+4H_2O+ 3e^(-)` `(i)`

And for every oxidation, there must be a corresponding reduction, and here peroxide, `H-stackrel(-I)O-stackrel(-I)O-H`, is REDUCED to hydroxide.....

`H_2O_2 +2HO^(-)+ 2e^(-) rarr 4HO^-` `(ii)`

And we take `2xx(i)+3xx(ii)` to balance the electrons.....

`2Cr(OH)_4^(-) +cancel(14)2HO^(-) +3H_2O_2 + cancel(6e^(-))rarr 2CrO_4^(2-)+8H_2O+ cancel(6e^(-))+cancel(12HO^-)`

To give (finally)....

`2Cr(OH)_4^(-) +3H_2O_2 +2HO^(-) rarr 2CrO_4^(2-)+8H_2O`

And is this balanced with respect to mass and charge? All care taken but no responsibility admitted.

Now this might seem a lot of work (well it might if you did it yourself!), but it is tedious rather than hard....ALL I have done is to balance mass and balance charge.....

Answer 2

`2[Cr(OH)_4]^(-) +3H_2O_2 =2 CrO_4^(2-) + 8 H_2O`

Explanation:

Cr has, at first, N° of oxydation +3 and becomes +6. For make this passage Cr gives 3 electrons
`[Cr(OH)_4]^(-) = CrO_4^(2-) + 3e^-`for balance the charge, i put on the left 4 negative charges as `OH^-` and i obtain on the right 4 mol of water
`[Cr(OH)_4]^(- ) +4OH^(-) = CrO_4^(2-) + 3e^(-) + 4 H_2O`.

The Oxygen, at first,has oxydation number -1 and becomes -2 gaining 1 electrons for each atom
`H_2O_2 + 2e^(-) =2 H_2O`
for balance the charge i put on the right 2 negative charges as `OH^-` but now to balance oxigen and hydrogen i must put on the left two molecules of water.
`2 H_2O+H_2O_2 + 2e^(- ) =2 H_2O +4OH^-` and semplifiing water in both the parts:
`H_2O_2 + 2e^(-) =+4OH^-`

Since Cr gives 3 electrons and oxygen needs two electrons, i multiply the first reaction for two and the second for three
`2[Cr(OH)_4]^(-) +8OH^- =2 CrO_4^(2-) + 6e^(-) + 8 H_2O`
`3H_2O_2 + 6e^(-) =+8OH^-`
joining the two reaction and semplifying the common species, you have
`2[Cr(OH)_4]^(-) +3H_2O_2 =2 CrO_4^(2-) + 8 H_2O`