Some Basic Concepts of Chemistry 2 Question 24
24. For the reaction, $\mathrm{I}^{-}+\mathrm{ClO}{3}^{-}+\mathrm{H}{2} \mathrm{SO}{4} \longrightarrow \mathrm{Cl}^{-}+\mathrm{HSO}{4}^{-}+\mathrm{I}_{2}$ the correct statement(s) in the balanced equation is/are
(a) stoichiometric coefficient of $\mathrm{HSO}_{4}^{-}$is 6
(2014 Adv)
(b) iodide is oxidised
(c) sulphur is reduced
(d) $\mathrm{H}_{2} \mathrm{O}$ is one of the products
Numerical Value Based Question
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Solution:
- PLAN This problem includes concept of redox reaction. A redox reaction consists of oxidation half-cell reaction and reduction half-cell reaction. Write both half-cell reactions, i.e. oxidation half-cell reaction and reduction half-cell reaction. Then balance both the equations
Now determine the correct value of stoichiometry of $\mathrm{H}{2} \mathrm{SO}{4}$.
Oxidation half-reaction, $2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_{2}+2 \mathrm{e}^{-}$
Here, $\mathrm{I}^{-}$is converted into $\mathrm{I}_{2}$. Oxidation number of $\mathrm{I}$ is increasing from -1 to 0 hence, this is a type of oxidation reaction.
Reduction half-reaction
$6 \mathrm{H}^{+}+\mathrm{ClO}{3}^{-}+6 e^{-} \longrightarrow \mathrm{Cl}^{-}+3 \mathrm{H}{2} \mathrm{O}$
- Here, $\mathrm{H}_{2} \mathrm{O}$ releases as a product. Hence, option (d) is correct.
Multiplying equation (i) by 3 and adding in equation (ii)
$$ 6 \mathrm{I}^{-}+\mathrm{ClO}{3}^{-}+6 \mathrm{H}^{+} \longrightarrow \mathrm{Cl}^{-}+3 \mathrm{I}{2}+3 \mathrm{H}_{2} \mathrm{O} $$
$6 \mathrm{I}^{-}+\mathrm{ClO}{3}^{-}+6 \mathrm{H}{2} \mathrm{SO}{4} \longrightarrow \mathrm{Cl}^{-}+3 \mathrm{I}{2}+3 \mathrm{H}{2} \mathrm{O}+6 \mathrm{HSO}{4}^{-}$
- Stoichiometric coefficient of $\mathrm{HSO}_{4}^{-}$is 6 .
Hence, option (a), (b) and (d) are correct.
