Balance the following equation.
`Cr(OH)_(3)+IO_(3)^(-)overset(OH^(-))(to)I^(-)+CrO_(4)^(2-)`

Explanation

Oxidation half reaction `Cr(OH)_(3)to CrO_(4)^(2-)`
(The ox. No of Cr changes from +3 to +6 )
Reduction half reaction `IO_(3)^(-)to I^(-).` (The ox: no. Of .I. Changes from +5 to -1 )
(1) `Cr(OH)_(3)to Cr_(4)^(-2)`` IO_(3)^(-)to I^(-)`
(2) `Cr(OH)_(3) to Cr_(4)^(-2)``IO_(3)^(-)to I^(-)`
( Cr already balanced ) (I already balanced)
(3) `Cr(OH)_(3)+H_(2)O to CrO_(4)^(-2)``IO_(3)^(-)to I +3H_(2)O`
(O balanced) (O balanced)
`Cr(OH)_(3)+H_(2)O to Cr_(4)^(-2)+5H_(2)O+5OH.``IO_(3)^(-)+6H_(2)O to I + 6OH^(-)+3H_(2)O`
(H balanced in basic medium) (H balanced in basic medium)
(5) `Cr(OH)_(3)+H_(2)O+5OH^(-)to` `IO_(3)^(-)+6 H_(2)O^(-)+6e^(-)to I^(-)+6OH^(-)+3H_(2)O`
`CrO_(4)^(2-)+5H_(2)O+3e^(-)`("charges balanced")
(Charges balanced)
(6) `2Cr(OH)_(3)+H_(2)O+5OH^(-)``IO_(3)^(-)+6 H_(2)O+6e^(-)to I^(-)+6OH^(-)+3H_(2)O`
`to Cr(OH)_(4)^(-2)+5H_(2)O+3e^(-)`("electrons balanced") (electrons balanced)
(7) Adding
`2Cr(OH)_(3)+2H_(2)O+10 OH^(-)+IO_(3)^(-)+6H_(2)O+6e^(-)to 2CrO_(4)^(2-)+13H_(2)O+6e^(-)+I^(-)+6OH^(-)`
`2Cr(OH)_(3)+IO_(3)^(-)+4 OH to 2CrO_(4)^(-2)+5H_(2)O+ I^(-)`