Question

For the reaction, $I^{-}+Cl{O}_3^{-}+H_{2}S{O}_4\to C{l}^{-}+HS{O}_4^{-}+I_2$ the correct statement (s) in the balanced equation is/are.

• A. stoichiometric coefficient of $HS{O}_4^{-}$ is 6
• B. Iodide is oxidised
• C. Sulphur is reduced
• D. $H_{2}O$ is one of the products

Answer 1

Answer: (A), (B), (D)

$H_{2}O$ is one of the products

1, 2, and 4
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 H_2 SO_4.
Oxidation half – reaction, $2I^{-}\to I_2+2e^{-}$ … (i)
Here, $I^{-}$ is converted into $I_2$. Oxidation number of I is increasing from -1 t 0 hence, this is a type of oxidation reaction.
Reduction half - reaction
$6H^{+}+Cl{O}_3^{-}+6e^{-}\to C{l}^{-}+3H_{2}O$ … (ii)
. Here, $H_{2}O$ releases as a product. Hence, option (4) is correct.
Multiplying equation (i) by 3 and adding in equation (ii)
$6I^{-}+Cl{O}_3^{-}+6H^{+}\to C{l}^{-}+3I_2+3H_{2}O$
$6I^{-}+Cl{O}_3^{-}+6H_{2}S{O}_4\to C{l}^{-}+3I_2+3H_{2}O+6HS{O}_4^{-}$
. Stoichiometric coefficient of $HS{O}_4^{-}$ is 6.
Hence, option (1), (2) and (4) are correct.