In the balanced chemical reaction
`IO_(3)^(ө)+al^(ө)+bH^(ө)rarrcH_(2)O+dI_(2)`
`a, b,c`, and `d`, respectively, correspond to

Options

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

Answer

A

Explanation

`overset(+5)(I)O_(3)^(-)+overset(-1)(I^(-))+H^(+) rarr H_(2)O+overset(0)(I_(2))`
Oxidation half-reaction:
`2I^(-) rarr I_(2)+2e^(-)`
Reduction half-reaction:
`2IO_(3)^(-)+10e^(-) rarr I_(2)`
Balance ionic charges by adding `H^(+)` ions:
`2I^(-) rarr I_(2) +2e^(-)`
`2IO_(3)^(-) +12H^(+) +10e^(-) rarr I_(2)`
Balance H and O b adding `H_(2)O` molecules:
`2I^(-) rarr I_(2)+2e^(-)`
`2IO_(3)^(-)+12H^(+) +10e^(-) rarr I_(2) +6H_(2)O`
To equalize electron transfer, we multiply oxidation half-reaction by 5 and add them, cancelling electrons on both sides:
`10I^(-)+2IO_(3)^(-) +12 H^(+)rarr 6I_(2)+6H_(2)O`
or `5I^(-)+IO_(3)^(-)+6H^(+) rarr 3I_(2)+3H_(2)O`
Thus, `a=5, b=6, c=3` and `d=3`