Question

Question #6b829

Answer 1

`Cr_2O_7^(-2) rarr Cr^(+3)+CrO_4^(-2) rarr Cr(OH)_3`

Let us take the first reaction
`Cr_2O_7^(-2) rarr Cr^(+3)+CrO_4^(-2)`
The reaction is reduction reaction. The oxidation state of Chromium in `Cr_2O_7^(-2)` is +6 which is reduced to +3 in `Cr^(+3)`

The second reaction also is a reduction reaction with same charge reduction on Chromium.

`Cr^(+3)+CrO_4^(-2) rarr Cr(OH)_3`

Calculating oxidation state is easy. The values of oxidation state of hydrogen and oxygen are +1 and -2 respectively. So, for `Cr_2O_7^(-2)`

`2x+7(-2) = -2`
`2x = 12`
`x=+6`

For `CrO_4^(-2)`
`x+4(-2)=-2`
`x=+6`

For `Cr(OH)_3`
`x+3*(-2+1) =0`
`x=+3`