Answer :
To identify the correct half-reaction for the reduction taking place in the given reaction:
[tex]\[ I^{-}(aq) + Cl_2(g) \longrightarrow 2Cl^{-}(aq) + I_2(aq) \][/tex]
we need to determine which species is being reduced. Reduction refers to the gain of electrons.
First, let's assign oxidation states for each element in the reactants and products:
1. \( I^- \) has an oxidation state of -1.
2. \( Cl_2 \) in its elemental form has an oxidation state of 0.
3. \( Cl^- \) in the product has an oxidation state of -1.
4. \( I_2 \) in its diatomic form has an oxidation state of 0.
In the reaction, \( I^- \) changes from -1 to 0, indicating it is being oxidized. Meanwhile, \( Cl_2 \) changes from 0 to -1, indicating it is being reduced.
Now, we need to write the half-reactions:
1. Oxidation half-reaction:
[tex]\[ I^- \longrightarrow I_2 + e^- \][/tex]
Here, two iodide ions each lose one electron to form one molecule of \( I_2 \).
2. Reduction half-reaction:
[tex]\[ Cl_2 + 2e^- \longrightarrow 2Cl^- \][/tex]
Here, a molecule of \( Cl_2 \) gains two electrons to form two chloride ions.
Given the options:
- \( Cl_2(g) + 2e^- \longrightarrow 2Cl^-(aq) \) (Correct)
- \( Cl_2(g) + e^- \longrightarrow 2Cl^-(aq) \) (Incorrect, not balanced)
- \( I^- \longrightarrow I_2(aq) + 2e^- \) (Part of the oxidation half-reaction)
- \( I^- \longrightarrow I_2(aq) + e^- \) (Incorrect, not balanced)
The half-reaction correctly describing the reduction is:
[tex]\[ Cl_2(g) + 2e^- \longrightarrow 2Cl^-(aq) \][/tex]
[tex]\[ I^{-}(aq) + Cl_2(g) \longrightarrow 2Cl^{-}(aq) + I_2(aq) \][/tex]
we need to determine which species is being reduced. Reduction refers to the gain of electrons.
First, let's assign oxidation states for each element in the reactants and products:
1. \( I^- \) has an oxidation state of -1.
2. \( Cl_2 \) in its elemental form has an oxidation state of 0.
3. \( Cl^- \) in the product has an oxidation state of -1.
4. \( I_2 \) in its diatomic form has an oxidation state of 0.
In the reaction, \( I^- \) changes from -1 to 0, indicating it is being oxidized. Meanwhile, \( Cl_2 \) changes from 0 to -1, indicating it is being reduced.
Now, we need to write the half-reactions:
1. Oxidation half-reaction:
[tex]\[ I^- \longrightarrow I_2 + e^- \][/tex]
Here, two iodide ions each lose one electron to form one molecule of \( I_2 \).
2. Reduction half-reaction:
[tex]\[ Cl_2 + 2e^- \longrightarrow 2Cl^- \][/tex]
Here, a molecule of \( Cl_2 \) gains two electrons to form two chloride ions.
Given the options:
- \( Cl_2(g) + 2e^- \longrightarrow 2Cl^-(aq) \) (Correct)
- \( Cl_2(g) + e^- \longrightarrow 2Cl^-(aq) \) (Incorrect, not balanced)
- \( I^- \longrightarrow I_2(aq) + 2e^- \) (Part of the oxidation half-reaction)
- \( I^- \longrightarrow I_2(aq) + e^- \) (Incorrect, not balanced)
The half-reaction correctly describing the reduction is:
[tex]\[ Cl_2(g) + 2e^- \longrightarrow 2Cl^-(aq) \][/tex]
