To determine the reducing agent in the redox reaction:
[tex]\[
\text{MnO}_2(s) + 4 \text{H}^+(aq) + 2\text{Cl}^-(aq) \longrightarrow \text{Mn}^{2+}(aq) + 2 \text{H}_2\text{O}(l) + \text{Cl}_2(g)
\][/tex]
we follow these steps:
1. Identify the oxidation states of the elements involved in the reaction:
- Manganese (Mn) in \(\text{MnO}_2\): The oxidation state of oxygen (O) is -2. Since there are two oxygens, the total oxidation state for oxygen is \(-2 \times 2 = -4\). To balance this in \(\text{MnO}_2\), manganese must have an oxidation state of +4.
- Manganese (Mn) in \(\text{Mn}^{2+}(aq)\): The oxidation state is +2.
- Chlorine (Cl) in \(\text{Cl}^-\): The oxidation state is -1.
- Chlorine (Cl) in \(\text{Cl}_2(g)\): The oxidation state is 0 since it is in its elemental form.
2. Determine the changes in oxidation states:
- Manganese (Mn):
- Initial state: +4 (in \(\text{MnO}_2\))
- Final state: +2 (in \(\text{Mn}^{2+}\))
- Change: +4 to +2 (Mn is reduced, meaning it gains electrons).
- Chlorine (Cl):
- Initial state: -1 (in \(\text{Cl}^-\))
- Final state: 0 (in \(\text{Cl}_2\))
- Change: -1 to 0 (Cl is oxidized, meaning it loses electrons).
3. Identify which substance is oxidized:
Chlorine (\(\text{Cl}^-\)) is oxidized because it goes from an oxidation state of -1 in \(\text{Cl}^-\) to 0 in \(\text{Cl}_2\).
4. Determine the reducing agent:
The reducing agent is the substance that is oxidized in the reaction because it donates electrons to another substance. In this case, \(\text{Cl}^-\) is oxidized to \(\text{Cl}_2\), thus it acts as the reducing agent.
Therefore, the reducing agent in this reaction is [tex]\(\text{Cl}^-\)[/tex].
