Answer :
To determine the reducing agent in the reaction:
[tex]\[ Cl_2(aq) + 2Br^-(aq) \longrightarrow 2Cl^-(aq) + Br_2(aq) \][/tex]
we need to examine the changes in oxidation states of the elements involved. The reducing agent is the substance that loses electrons (is oxidized) in the reaction.
Step-by-Step Solution:
1. Determine the oxidation states:
- In [tex]\(Cl_2\)[/tex], chlorine has an oxidation state of 0 because it is in its elemental form.
- In [tex]\(Br^-\)[/tex], bromine has an oxidation state of -1 because it is an ion with a single negative charge.
- In [tex]\(Cl^-\)[/tex], chlorine has an oxidation state of -1 because it is an ion with a single negative charge.
- In [tex]\(Br_2\)[/tex], bromine has an oxidation state of 0 because it is in its elemental form.
2. Identify the changes in oxidation states:
- Chlorine: [tex]\(Cl_2 (0) \rightarrow 2Cl^- (-1)\)[/tex], which means chlorine gains electrons (is reduced).
- Bromine: [tex]\(2Br^- (-1) \rightarrow Br_2 (0)\)[/tex], which means bromine loses electrons (is oxidized).
3. Determine the reducing agent:
- The reducing agent is the substance that loses electrons. Here, bromine [tex]\(Br^-\)[/tex] goes from -1 to 0, meaning it loses an electron and is thus oxidized.
Therefore, bromine [tex]\( (Br^-) \)[/tex] loses an electron, so it is the reducing agent. The correct description is:
Bromine [tex]$(Br^-)$[/tex] loses an electron, so it is the reducing agent.
[tex]\[ Cl_2(aq) + 2Br^-(aq) \longrightarrow 2Cl^-(aq) + Br_2(aq) \][/tex]
we need to examine the changes in oxidation states of the elements involved. The reducing agent is the substance that loses electrons (is oxidized) in the reaction.
Step-by-Step Solution:
1. Determine the oxidation states:
- In [tex]\(Cl_2\)[/tex], chlorine has an oxidation state of 0 because it is in its elemental form.
- In [tex]\(Br^-\)[/tex], bromine has an oxidation state of -1 because it is an ion with a single negative charge.
- In [tex]\(Cl^-\)[/tex], chlorine has an oxidation state of -1 because it is an ion with a single negative charge.
- In [tex]\(Br_2\)[/tex], bromine has an oxidation state of 0 because it is in its elemental form.
2. Identify the changes in oxidation states:
- Chlorine: [tex]\(Cl_2 (0) \rightarrow 2Cl^- (-1)\)[/tex], which means chlorine gains electrons (is reduced).
- Bromine: [tex]\(2Br^- (-1) \rightarrow Br_2 (0)\)[/tex], which means bromine loses electrons (is oxidized).
3. Determine the reducing agent:
- The reducing agent is the substance that loses electrons. Here, bromine [tex]\(Br^-\)[/tex] goes from -1 to 0, meaning it loses an electron and is thus oxidized.
Therefore, bromine [tex]\( (Br^-) \)[/tex] loses an electron, so it is the reducing agent. The correct description is:
Bromine [tex]$(Br^-)$[/tex] loses an electron, so it is the reducing agent.
