Which will be the value of x, y and z in the following equaton.
`xI_(2)+yOH^(-)rarrIO_(3)^(-)+zI+3H_(2)O`

Options

A. `{:(x,y,z),(6,3,5):}`
B. `{:(x,y,z),(3,2,3):}`
C. `{:(x,y,z),(3,6,5):}`
D. `{:(x,y,z),(3,3,3):}`

Answer

C

Explanation

`I_(2)+OH^(-)rarr IO_(3)^(-)+I^(-)+H_(2)O`
Oxidation half : `I_(2)+OH^(-)rarr IO_(3)^(-)+H_(2)O`
Reduction half : `I_(2)rarr I^(-)`
Oxidation half-
Adding `OH^(-), I_(2)+12OH^(-)rarr 2IO_(3)^(-)+6H_(2)O`
Adding electrons :
`I_(2)+12OH^(-)rarr 2I_(3)^(-)+6H_(2)O+10e^(-)`
Reduction half -
`I_(2)+2e^(-)rarr 2I^(-)`
Balancing `e^(-), I_(2)+12OH^(-)rarr 2IO_(3)^(-)+6H_(2)O+10e^(-)`
`I_(2)+2e rarr 2I^(-)] xx 5`
Adding both reactions
`6I_(2)+12OH^(-)rarr 2IO_(3)^(-)+10I^(-)+6H_(2)O`
Dividing by 2,
`3I_(2)+6OH^(-)rarr IO_(3)^(-)+5I^(-)+3H_(2)O`