What is the oxidation state of S in the sulph | vedclass

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(A) This is a redox reaction in which sulphur is oxidized and bromine is reduced according to the following balanced equation:
$SO_3^{2-}{_{	ext{(aq

Vedclass · Accepted Answer

$+6$

Vedclass · Answer

(A) This is a redox reaction in which sulphur is oxidized and bromine is reduced according to the following balanced equation:
$SO_3^{2-}{_{	ext{(aq)}}} + Br_{2	ext{(l)}} + H_2O_{	ext{(l)}} ightarrow SO_4^{2-}{_{	ext{(aq)}}} + 2Br^{-}{_{	ext{(aq)}}} + 2H^{+}{_{	ext{(aq)}}}$
The sulphur-containing product is the sulphate ion,$SO_4^{2-}$.
To find the oxidation state of $S$ in $SO_4^{2-}$,let it be $x$.
The sum of oxidation states of all atoms in an ion is equal to the charge on the ion:
$x + (4 	imes -2) = -2$
$x - 8 = -2$
$x = +6$
Therefore,the oxidation state of $S$ in the product is $+6$.

What is the oxidation state of $S$ in the sulphur-containing product of the following reaction?
$SO_3^{2-}{_{\text{(aq)}}} + Br_{2\text{(l)}} + H_2O_{\text{(l)}} \rightarrow $

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What is the oxidation state of $S$ in the sulphur-containing product of the following reaction? $SO_3^{2-}{_{\text{(aq)}}} + Br_{2\text{(l)}} + H_2O_{\text{(l)}} \rightarrow $