Question

identify the unbalanced half - reaction in which a species is reduced.
c(s)→co(g)
cr₂o₃(s)→cro₄²⁻(aq)
so₄²⁻(aq)→h₂s(g)
2i⁻(aq)→i₂(s)
al(s)→al³⁺(aq)
identify the initial and final oxidation states for the element iodine in the equation
2i⁻(aq)→i₂(s)
initial oxidation state:
final oxidation state:

Explanation:

Step1: Recall reduction definition

Reduction is a decrease in oxidation state.

Step2: Analyze each half - reaction

• For \(C(s)\to CO(g)\), carbon's oxidation state increases from 0 to +2, so it's oxidation.
• For \(Cr_2O_3(s)\to CrO_4^{2 -}(aq)\), chromium's oxidation state increases, so it's oxidation.
• For \(SO_4^{2 -}(aq)\to H_2S(g)\), sulfur's oxidation state decreases from +6 in \(SO_4^{2 -}\) to - 2 in \(H_2S\), so it's reduction.
• For \(2I^-(aq)\to I_2(s)\), iodine's oxidation state increases from - 1 to 0, so it's oxidation.
• For \(Al(s)\to Al^{3+}(aq)\), aluminum's oxidation state increases from 0 to +3, so it's oxidation.

Step3: Determine oxidation states of iodine

In \(2I^-(aq)\), the oxidation state of iodine is - 1 (the charge of the ion). In \(I_2(s)\), the oxidation state of iodine is 0 (element in its elemental form).

Answer:

The unbalanced half - reaction in which a species is reduced: \(SO_4^{2 -}(aq)\to H_2S(g)\)
Initial oxidation state of iodine: - 1
Final oxidation state of iodine: 0