When SO_{2} is passed through acidified K_{2} | vedclass

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(D) The reaction of acidified potassium dichromate with $SO_{2}$ is as follows:$K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3SO_{2} \longrightarrow K_{2}SO_{4} +

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the solution turns green

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(D) The reaction of acidified potassium dichromate with $SO_{2}$ is as follows:$K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3SO_{2} \longrightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$In this reaction,$SO_{2}$ acts as a reducing agent and is oxidized to sulphate ions,while the orange-colored dichromate ion $(Cr_{2}O_{7}^{2-})$ is reduced to green-colored chromic ion $(Cr^{3+})$ in the form of chromic sulphate $(Cr_{2}(SO_{4})_{3})$.Therefore,the solution turns green.

When $SO_{2}$ is passed through acidified $K_{2}Cr_{2}O_{7}$ solution:

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