For the reaction
$I^{⊖} + C l O_{3}^{⊖} + H_{2} S O_{4} \to C l^{⊖} + H S O_{4}^{-} + I_{2}$
The correct statement(s) in the balanced equation is/ are:
(IIT-JEE-Advanced 2014)

Stoichiometric coefficient of

H S O_{4}^{-}

is 6.

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Iodide is oxidized.

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Sulphur is reduced.

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H_{2} O

is one of the products.

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Solution

The correct option is D $H_{2} O$ is one of the products.
$6 I^{⊖} + C l O_{3}^{⊖} + 6 H_{2} S O_{4} \to C l^{⊖} + 3 I_{2} + 3 H_{2} O + 6 H S O_{4}^{-}$
(R.A) (O.A)

R.A - reducing agent
O.A. - oxidising agent

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Similar questions

I^{⊖} + C l O_{3}^{⊖} + H_{2} S O_{4} \to C l^{⊖} + H S O_{4}^{-} + I_{2}

I^{-} + {C l O_{3}}^{-} + H_{2} S O_{4} \to C l^{-} + {H S O_{4}}^{-} + I_{2}

I^{-} + C l O_{3}^{-} + H_{2} S O_{4} ⟶ C l^{-} + H S O_{4}^{-} + I_{2}

I^{-} + C l O_{3}^{-} + H_{2} S O_{4} \to C l^{-} + H S O_{4}^{-} + I_{2}

I^{-} + C l O_{3}^{-} + H_{2} S O_{4} \to C l^{-} + H S O_{4}^{-} + I_{2}

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