Answer:
The given chemical equation is:
I^- + S ==> S^-2 + I^2
To balance the reduction half-reaction of this equation, we need to identify which species is undergoing reduction, i.e., which species is gaining electrons. In this case, sulfur (S) is being reduced to sulfide (S^-2), so the reduction half-reaction involves sulfur.
The unbalanced reduction half-reaction is:
S ==> S^-2
To balance this half-reaction, we need to add electrons (e^-) to the left-hand side to balance the charge. The number of electrons added should be equal to the difference in oxidation states of sulfur between the reactant and product sides of the equation. In this case, sulfur is going from an oxidation state of 0 to -2, so it is gaining two electrons. The balanced reduction half-reaction is:
S + 2 e^- ==> S^-2
Therefore, the balanced reduction half-reaction of the given chemical equation is:
S + 2 e^- ==> S^-2