Question

question 6
1 point
in the reaction below, cl⁻(aq) is the
enter answer
and i₂(g) is the
enter answer
. fill in the blank
with oxidizing agent or reducing agent.
2 kcl(aq) + i₂(g) → 2 ki(aq) + cl₂(g)

Explanation:

Step1: Determine oxidation states

In \(KCl\), the oxidation state of \(Cl\) is \(- 1\). In \(Cl_{2}\), the oxidation state of \(Cl\) is \(0\). So \(Cl^{-}\) is oxidized (loses electrons).

Step2: Analyze oxidizing and reducing agents

A reducing agent is oxidized. Since \(Cl^{-}\) is oxidized, \(Cl^{-}\) is the reducing agent.
In \(I_{2}\), the oxidation state of \(I\) is \(0\). In \(KI\), the oxidation state of \(I\) is \(-1\). So \(I_{2}\) is reduced (gains electrons). An oxidizing agent is reduced. So \(I_{2}\) is the oxidizing agent.

Answer:

\(Cl^{-}(aq)\) is the reducing agent and \(I_{2}(g)\) is the oxidizing agent.