When SO_2 gas is passed through an acidic sol | vedclass

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(B) The chemical reaction is: $K_2Cr_2O_7 + H_2SO_4 + 3SO_2 	o K_2SO_4 + Cr_2(SO_4)_3 + H_2O$. In $K_2Cr_2O_7$,the oxidation state of $Cr$ is calcula

Vedclass · Accepted Answer

$+6$ to $+3$

Vedclass · Answer

(B) The chemical reaction is: $K_2Cr_2O_7 + H_2SO_4 + 3SO_2 	o K_2SO_4 + Cr_2(SO_4)_3 + H_2O$. In $K_2Cr_2O_7$,the oxidation state of $Cr$ is calculated as: $2(+1) + 2(x) + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$. In $Cr_2(SO_4)_3$,the oxidation state of $Cr$ is calculated as: $2(x) + 3(-2) = 0 \implies 2x = 6 \implies x = +3$. Thus,the oxidation state of chromium changes from $+6$ to $+3$.

When $SO_2$ gas is passed through an acidic solution of $K_2Cr_2O_7$,what is the change in the oxidation state of chromium?

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