When sulphur dioxide is passed in an acidifie | vedclass

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(C) Acidified $K_{2}Cr_{2}O_{7}$ acts as a strong oxidizing agent and oxidizes $SO_{2}$ to $SO_{4}^{2-}$ ions (present in $Cr_{2}(SO_{4})_{3}$).The ba

Vedclass · Accepted Answer

$+4$ to $+6$

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(C) Acidified $K_{2}Cr_{2}O_{7}$ acts as a strong oxidizing agent and oxidizes $SO_{2}$ to $SO_{4}^{2-}$ ions (present in $Cr_{2}(SO_{4})_{3}$).The balanced chemical equation is:$3SO_{2} + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} \longrightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$In $SO_{2}$,the oxidation state of sulphur is $+4$.In $Cr_{2}(SO_{4})_{3}$,the sulphate ion $(SO_{4}^{2-})$ contains sulphur in the $+6$ oxidation state.Thus,the oxidation state of sulphur changes from $+4$ to $+6$.

When sulphur dioxide is passed in an acidified $K_{2}Cr_{2}O_{7}$ solution,the oxidation state of sulphur is changed from

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