In balancing the half-reaction S_2O_3^{2-} to | vedclass

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Question

(A) The half-reaction is $S_2O_3^{2-} 	o S_{(s)}$.First,balance the sulfur atoms: $S_2O_3^{2-} 	o 2S_{(s)}$.Next,calculate the oxidation state chang

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$4$ on the left

Vedclass · Answer

(A) The half-reaction is $S_2O_3^{2-} 	o S_{(s)}$.First,balance the sulfur atoms: $S_2O_3^{2-} 	o 2S_{(s)}$.Next,calculate the oxidation state change: In $S_2O_3^{2-}$,the average oxidation state of $S$ is $+2$. In $S_{(s)}$,the oxidation state is $0$.Since there are $2$ sulfur atoms,the total change in oxidation state is $2 	imes (2 - 0) = 4$.Therefore,$4$ electrons must be added to the left side to balance the charge: $S_2O_3^{2-} + 4e^- 	o 2S_{(s)} + 3O^{2-}$.Thus,$4$ electrons are added on the left.

In balancing the half-reaction $S_2O_3^{2-} \to S_{(s)}$,the number of electrons that must be added is:

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