In basic medium balance the half-reactions following below:
`Cr(OH)_(3)toCrO_(4)^(2)`

Explanation

In order to equalise the number of O- atomic one `H_(2)O` molecule is added to the right side (because it has an excess O- atom) and for this I molecule of `H_(2)O. two OH^(-)` ions are added to the left side .Now ,we get `Cr(OH)_(3)+2OH^(-)toCrO_(4)^(2-)+H_(2)O`
In this equation five H -atoms on left side and two H- atoms are present on t he right side. To equation the number of H -atoms on both sides, three `OH^(-) (aq)` and three `H_(2)O` molecules are added to the left side and right side respectively.Finally ,we get -
`Cr(OH)_(3)+2OH^(-)+3OH^(-)toCrO_(4)^(2-)+H_(2)O+3H_(2)O`.
`Cr(OH)_(3)+5OH^(-)toCrO_(4)^(2-)+4H_(2)O`
For balancing the charge 3. electrons are added to the right side. Therefore ,the final half -raction becomes,
`Cr(OH)_(3)+5OH^(-)toCrO_(4)^(2-)+4H_(2)O`