Acidified K_2Cr_2O_7 solution turns green whe | vedclass

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Question

(A) The reaction between acidified $K_2Cr_2O_7$ and $Na_2SO_3$ is a redox reaction where $Cr_2O_7^{2-}$ acts as an oxidizing agent and $SO_3^{2-}$ act

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$Cr_2(SO_4)_3$

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(A) The reaction between acidified $K_2Cr_2O_7$ and $Na_2SO_3$ is a redox reaction where $Cr_2O_7^{2-}$ acts as an oxidizing agent and $SO_3^{2-}$ acts as a reducing agent.The balanced chemical equation is:$K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2SO_4 \rightarrow K_2SO_4 + 3Na_2SO_4 + Cr_2(SO_4)_3 + 4H_2O$In this reaction,the orange-colored dichromate ion $(Cr_2O_7^{2-})$ is reduced to the green-colored chromium$(III)$ sulfate $(Cr_2(SO_4)_3)$,which causes the solution to turn green.

Acidified $K_2Cr_2O_7$ solution turns green when $Na_2SO_3$ is added to it. This is due to the formation of

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