Answer :
To understand the oxidation and reduction half-reactions in the given chemical reaction:
[tex]\[ \text{Pt} + 2 \text{Cl}_2 \rightarrow \text{PtCl}_4 \][/tex]
let’s break it down step-by-step:
### Step 1: Identify the elements undergoing oxidation and reduction.
1. Platinum (Pt):
- On the reactant side: Pt is in its elemental form, hence the oxidation state is 0.
- On the product side: In PtCl₄, Pt is associated with four chloride ions (Cl⁻). Since each Cl has an oxidation state of -1, the oxidation state of Pt in PtCl₄ must be +4 to balance the charges.
2. Chlorine (Cl₂):
- On the reactant side: Cl is in its diatomic molecular form (Cl₂), hence the oxidation state of each Cl is 0.
- On the product side: Each Cl in PtCl₄ is in the form of Cl⁻, thus the oxidation state is -1.
### Step 2: Write the half-reactions for the oxidation and reduction processes.
1. Oxidation Half-Reaction:
Here, Pt goes from an oxidation state of 0 to +4. Since Pt is not being oxidized nor reduced in this reaction and remains as Pt, the oxidation half-reaction would be depicted as the oxidation state changing but no actual net change for Pt:
[tex]\[ \text{Pt} \rightarrow \text{Pt} \][/tex]
2. Reduction Half-Reaction:
Here, Cl₂ is being reduced from an oxidation state of 0 to -1. Each Cl₂ molecule gains two electrons to form two Cl⁻ ions:
[tex]\[ \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \][/tex]
### Conclusion:
The balanced half-reactions describing the oxidation and reduction that occur in the given reaction are:
- Oxidation Half-Reaction:
[tex]\[ \text{Pt} \rightarrow \text{Pt} \][/tex]
- Reduction Half-Reaction:
[tex]\[ \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \][/tex]
These half-reactions correctly represent the processes occurring within the overall reaction.
[tex]\[ \text{Pt} + 2 \text{Cl}_2 \rightarrow \text{PtCl}_4 \][/tex]
let’s break it down step-by-step:
### Step 1: Identify the elements undergoing oxidation and reduction.
1. Platinum (Pt):
- On the reactant side: Pt is in its elemental form, hence the oxidation state is 0.
- On the product side: In PtCl₄, Pt is associated with four chloride ions (Cl⁻). Since each Cl has an oxidation state of -1, the oxidation state of Pt in PtCl₄ must be +4 to balance the charges.
2. Chlorine (Cl₂):
- On the reactant side: Cl is in its diatomic molecular form (Cl₂), hence the oxidation state of each Cl is 0.
- On the product side: Each Cl in PtCl₄ is in the form of Cl⁻, thus the oxidation state is -1.
### Step 2: Write the half-reactions for the oxidation and reduction processes.
1. Oxidation Half-Reaction:
Here, Pt goes from an oxidation state of 0 to +4. Since Pt is not being oxidized nor reduced in this reaction and remains as Pt, the oxidation half-reaction would be depicted as the oxidation state changing but no actual net change for Pt:
[tex]\[ \text{Pt} \rightarrow \text{Pt} \][/tex]
2. Reduction Half-Reaction:
Here, Cl₂ is being reduced from an oxidation state of 0 to -1. Each Cl₂ molecule gains two electrons to form two Cl⁻ ions:
[tex]\[ \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \][/tex]
### Conclusion:
The balanced half-reactions describing the oxidation and reduction that occur in the given reaction are:
- Oxidation Half-Reaction:
[tex]\[ \text{Pt} \rightarrow \text{Pt} \][/tex]
- Reduction Half-Reaction:
[tex]\[ \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \][/tex]
These half-reactions correctly represent the processes occurring within the overall reaction.