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

For the reaction, I^(-)+CIO_(3)^(-)+H_(2)SO_(4)rarrCI^(-)+HSO_(4)^(-)

1.	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
Answer» stoichiometric coeffiecient of `HSO_(4)^(-)` is 6 iodide is oxidised suphur is reduced `H_(2)O` is one of the products Solution :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.