For the reaction,
`I^(-)+CIO_(3)^(-)+H_(2)SO_(4)rarrCI^(-)+HSO_(4)^(-)+I_(2)` the correct statement (s) in the balanced equation `is//are`

Options

A. stoichiometric coeffiecient of `HSO_(4)^(-)` is 6
B. iodide is oxidised
C. suphur is reduced
D. `H_(2)O` is one of the products

Answer

A, B, D

Explanation

The oxidation half reaction is:
`2I^(-)rarrI_(2)+2e^(-)`
The reduction half rection is :
`6H^(+)+CIO_(3)^(-)+6E^(-)rarrCI^(-)+3H_(2)O`
Multiply eqn.(i) by 3 and add to eqn.(i)
`6H^(+)+CIO_(3)^(-)+6e^(-)rarrCI^(-)+3H_(2)O`
`6I^(-)rarr3I_(2)+6e^(-)`
`(6I^(-)+6H^(+)+CIO_(3)^(-)rarr3I_(2)+3H_(2)O+CI^(-))/((a) "The stoichiometry of" HSO_(4)^(-) is 6`
The `H^(+) "ions are provided by" H_(2)SO_(4) "or" HsO_(4)^(-)` ion
(b) `I^(-)` ions are oxidisded to `I_(2)`
(d) `H_(2)O` is one of the products formed in the redox rection.