When SO_2 is passed through an acidic solutio | vedclass

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(C) The balanced chemical equation for the reaction is:$K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$In $K_2Cr_2O_7$,the oxi

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$+6$ to $+3$

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(C) The balanced chemical equation for the reaction is:$K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$In $K_2Cr_2O_7$,the oxidation state of chromium $(Cr)$ is calculated as:$2(+1) + 2(x) + 7(-2) = 0$ $\Rightarrow 2 + 2x - 14 = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$In $Cr_2(SO_4)_3$,the oxidation state of chromium $(Cr)$ is $+3$ because the sulphate ion $(SO_4^{2-})$ has a charge of $-2$.Thus,the oxidation state of chromium changes from $+6$ to $+3$.

When $SO_2$ is passed through an acidic solution of potassium dichromate,chromium sulphate is formed. What is the change in the oxidation state of chromium?

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