Question

# For the reaction:${I^ - } + Cl{O_3}^ - + {H_2}S{O_4}\xrightarrow{{}}C{l^ - } + HS{O_4}^ - + {I_2}$ A.stoichiometric coefficient of $HS{O_4}^ -$ is $6$B.iodide is oxidizedC.sulphur is reducedD.${H_2}O$ is one of the products.

Hint: This is an example of redox reaction. Two half reactions are occurring in a reaction one is oxidation and the other reduction. In order to find the correct stoichiometric amount or the exact moles of reactants and products we need a balanced chemical reaction.

In order to find the exact stoichiometric balanced equation we have to identify the two individual half reactions. In one half oxidations occur and in the other reduction occurs.
To begin at first the reactants and products have to be evaluated. Here ${I^ - }$, $Cl{O_3}^ -$ and ${H_2}S{O_4}$ are reactants and $C{l^ - }$, $HS{O_4}^ -$ and ${I_2}$ are the products. The ${I^ - }$is converted to ${I_2}$, $Cl{O_3}^ -$ is converted to $C{l^ - }$ and ${H_2}S{O_4}$ is converted to $HS{O_4}^ -$.
In oxidation reaction addition of oxygen or release of electrons occurs. Therefore the oxidation half reaction is:
$2{I^ - }\xrightarrow{{}}{I_2} + 2{e^ - }$
Thus option B is correct i.e. iodide is oxidized.
In reduction reaction release of oxygen or addition of electrons occurs. Therefore the reduction half reaction is:
$6{H^ + } + Cl{O_3}^ - + 6{e^ - }\xrightarrow{{}}C{l^ - } + 3{H_2}O$
Thus option D is correct i.e. ${H_2}O$ is one of the products.
To get a balanced equation we have to multiply the oxidation half reaction by $3$ , so that the charges get cancelled on adding the oxidation and reduction half reactions.
$6{I^ - } + Cl{O_3}^ - + 6{H^ + }\xrightarrow{{}}C{l^ - } + 3{I_2} + 3{H_2}O$
Therefore the final balanced equation becomes
$6{I^ - } + Cl{O_3}^ - + 6{H_2}S{O_4}\xrightarrow{{}}C{l^ - } + 3{I_2} + 3{H_2}O + 6HS{O_4}^ -$.
Thus option A is correct i.e. stoichiometric coefficient of $HS{O_4}^ -$ is $6$.
And hence finally we can say that option A,B and D is correct.

Note: In the given reaction no change in oxidation state of sulphur is occurring. In both sides i.e. in ${H_2}S{O_4}$ the oxidation state of sulphur is $+ 6$ and in $HS{O_4}^ -$ the oxidation state is $+ 6$. Neither oxidation nor reduction is taking place with sulphur. The negative charge does not indicate gain of electrons by sulphur atom.