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. `3, 5, 6`
B. `5, 6, 3`
C. `3, 6, 5`
D. `6, 3, 5`

Answer

C

Explanation

`{:(overset(0)I_(2)rarroverset(+5)(IO_(3))," ..................(i) Oxidation"),(overset(0)I_(1)rarroverset(-1)I^(-)," ..................(ii) Reduction"):}`
Balancing atoms of iodine on both sides, we get,
`" "I_(2)rarr2IO_(3)^(-)`
`or "I_(2)+12OH^(-)rarr2IO_(3)^(-)+6H_(2)O`
Blancing charge,
`" "I_(2)+12OH^(-)rarr2IO_(3)^(-)+6H_(2)O+10e^(-)`
`" "I_(2)rarr2I^(-)`
`"and "(I_(2)+2e^(-)rarr2I^(-))xx5`
`"adding "6I_(2)+12OH^(-)rarr2IO_(3)^(-)+10I^(-)+6H_(2)O`
`"or, "3I_(2)+6OH^(-)rarrIO_(3)^(-)+5I^(-)+3H_(2)O`