Question

For the reaction :
$I^{-}+Cl{O}_3^{-}+H_{2}S{O}_4\to C{l}^{-}+HS{O}_4^{-}+I_2$

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

Answer 1

Answer: (C)

sulphur is reduced

Given reaction:

$I^{-}+Cl{O}_3^{-}+H_{2}S{O}_4\to C{l}^{-}+HS{O}_4^{-}+I_2$

Assigning the oxidation number of each element in the above reaction:

$\stackrel{-1}{I^{-}}+\stackrel{+5}{Cl}{\stackrel{-2}{O_3}}^{-}+\stackrel{+1}{H_2}\stackrel{+6}{S}\stackrel{-2}{O_4}\to \stackrel{-1}{C{l}^{-}}+\stackrel{+1}{H}\stackrel{+6}{S}{\stackrel{-2}{O_4}}^{-}+\stackrel{0}{I_2}$

Balancing the exchange of electrons as:

$6I^{-}+Cl{O}_3^{-}\to C{l}^{-}+3I_2$

Balancing O by adding $H_{2}O$ as

$6I^{-}+Cl{O}_3^{-}\to C{l}^{-}+3I_2+3H_{2}O$

Balance H by adding $H_{2}S{O}_4$

$6I^{-}+Cl{O}_3^{-}+3H_{2}S{O}_4\to C{l}^{-}+3I_2+3H_{2}O$

balance H and $S{O}_4$ as:

$6I^{-}+Cl{O}_3^{-}+6H_{2}S{O}_4\to C{l}^{-}+3I_2+3H_{2}O+3HS{O}_4^{-}$

Thus stoichiometric coefficient of $HS{O}_4^{-}$ is 3

the oxidation state of I changes from -1 to 0. Thus losses 1 electron and undergoes oxidation.

the oxidation state of $Cl$ changes from +5 to -1. Thus gains 6 electrons and undergoes oxidation.

$H,S,O$ has same oxidation number, thus no reduction or oxidation occurs.

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