When SO_2 is passed through acidified K_2Cr_2 | vedclass

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(D) When $SO_2$ gas is passed through an acidified $K_2Cr_2O_7$ solution,it acts as a reducing agent and reduces $Cr(VI)$ to $Cr(III)$.The balanced ch

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Green $Cr_2(SO_4)_3$ is formed

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(D) When $SO_2$ gas is passed through an acidified $K_2Cr_2O_7$ solution,it acts as a reducing agent and reduces $Cr(VI)$ to $Cr(III)$.The balanced chemical equation is:$K_2Cr_2O_7 + H_2SO_4 + 3SO_2 	o K_2SO_4 + Cr_2(SO_4)_3 + H_2O$In this reaction,the orange-colored $K_2Cr_2O_7$ solution turns green due to the formation of chromium$(III)$ sulfate,$Cr_2(SO_4)_3$.

When $SO_2$ is passed through acidified $K_2Cr_2O_7$ solution:

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