In the equation,BiO_3^{-} + 6H^{+} + xe^{-} r | vedclass

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(A) To find the value of $x$,we need to balance the charge on both sides of the equation.In $BiO_3^{-}$,the oxidation state of $Bi$ is $x + 3(-2) = -1

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$2$

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(A) To find the value of $x$,we need to balance the charge on both sides of the equation.In $BiO_3^{-}$,the oxidation state of $Bi$ is $x + 3(-2) = -1$,so $x = +5$.In $Bi^{3+}$,the oxidation state of $Bi$ is $+3$.The change in oxidation state is $+5 - (+3) = 2$.However,we must also balance the total charge.Total charge on the left side: $(-1) + 6(+1) + x(-1) = 5 - x$.Total charge on the right side: $(+3) + 3(0) = +3$.Equating the charges: $5 - x = 3$,which gives $x = 2$.

In the equation,$BiO_3^{-} + 6H^{+} + xe^{-} \rightarrow Bi^{3+} + 3H_2O$. What is the value of $x$?

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