Balance the following redox equaiton by both methods.
`[Cr(OH)_(4)]^(ө)+H_(2)O_(2)rarrCrO_(4)^(2-)+H_(2)O`(basic medium)

Explanation

(i) `overset(+3-2+1)([Cr(OH)_(4)]^(-))+overset(+1-2)(H_(2)O_(2))rarroverset(+6-2)((CrO_(4))^(2-))+overset(+1-2)(H_(2)O)` Skeleton equation
Note. Charge on the ion is indicated on the right of the numeral digit whereas O.N. is indicated on the left side
(ii) Atoms undergoing change in O.N
(iii) `2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)rarrCrO_(4)^(2-)+H_(2)O` Equating increase and decrease in O.N
(iv) `2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)rarr2CrO_(4)^(2-)+H_(2)O` Balancing all other atoms except H and O
(v) `2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)rarr2CrO_(4)^(2-)+6H_(2)O` Balance the no. of oxygen atoms by `6H_(2)O`
(vi) `2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)+2OH^(-)rarr2CrO_(4)^(2-)+6H_(2)O+2H_(2)O`
Balance the no. of H atoms by adding `2OH^(-)` to left and `2H_(2)O` to right
Note : R.H.S. of the equation has two H-atoms less than the number of H-atoms on L.H.S.
or `2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)+2OH^(-)rarr2CrO_(4)^(2-)+8H_(2)O`.