Answer :
To determine which half-reaction correctly describes the oxidation taking place in the given reaction:
[tex]\[ \text{Zn (s)} + \text{Cu}^{2+} \text{(aq)} \longrightarrow \text{Zn}^{2+} \text{(aq)} + \text{Cu (s)} \][/tex]
We need to understand what oxidation is. Oxidation is the process where a species loses electrons, resulting in an increase in its oxidation state.
Here are the steps to find the correct oxidative half-reaction:
1. Identify the species involved:
- Zinc ([tex]\(\text{Zn}\)[/tex]) in solid form.
- Copper ion ([tex]\(\text{Cu}^{2+}\)[/tex]) in aqueous form.
2. Determine the oxidation state changes:
- For Zinc: Initially, Zinc is in the solid state ([tex]\( \text{Zn} \)[/tex]), which has an oxidation state of [tex]\(0\)[/tex]. After the reaction, it becomes [tex]\(\text{Zn}^{2+}\)[/tex] (aqueous), which has an oxidation state of [tex]\(+2\)[/tex].
- For Copper: Initially, Copper is a [tex]\( \text{Cu}^{2+} \)[/tex] ion (aqueous) with an oxidation state of [tex]\(+2\)[/tex]. After the reaction, Copper is in the solid form ([tex]\( \text{Cu} \)[/tex]), which has an oxidation state of [tex]\(0\)[/tex].
3. Identify the species undergoing oxidation:
- Zinc ([tex]\(\text{Zn}\)[/tex]) is going from an oxidation state of [tex]\(0\)[/tex] to [tex]\(+2\)[/tex]. This is an increase in the oxidation state, indicating oxidation.
4. Write the oxidation half-reaction:
- The oxidation half-reaction will indicate the species losing electrons. For Zinc:
[tex]\[ \text{Zn} (s) \longrightarrow \text{Zn}^{2+} \text{(aq)} + 2e^- \][/tex]
This equation shows that Zinc loses two electrons (indicated by [tex]\(2e^-\)[/tex]) when it oxidizes from a neutral atom to a [tex]\( \text{Zn}^{2+} \)[/tex] ion.
Therefore, the correct half-reaction describing the oxidation that is taking place is:
[tex]\[ \text{Zn} (s) \longrightarrow \text{Zn}^{2+} \text{(aq)} + 2e^- \][/tex]
So, the answer is:
[tex]\[ \boxed{\text{Zn (s) } \longrightarrow \text{Zn}^{2+} \text{(aq)} + 2e^-} \][/tex]
[tex]\[ \text{Zn (s)} + \text{Cu}^{2+} \text{(aq)} \longrightarrow \text{Zn}^{2+} \text{(aq)} + \text{Cu (s)} \][/tex]
We need to understand what oxidation is. Oxidation is the process where a species loses electrons, resulting in an increase in its oxidation state.
Here are the steps to find the correct oxidative half-reaction:
1. Identify the species involved:
- Zinc ([tex]\(\text{Zn}\)[/tex]) in solid form.
- Copper ion ([tex]\(\text{Cu}^{2+}\)[/tex]) in aqueous form.
2. Determine the oxidation state changes:
- For Zinc: Initially, Zinc is in the solid state ([tex]\( \text{Zn} \)[/tex]), which has an oxidation state of [tex]\(0\)[/tex]. After the reaction, it becomes [tex]\(\text{Zn}^{2+}\)[/tex] (aqueous), which has an oxidation state of [tex]\(+2\)[/tex].
- For Copper: Initially, Copper is a [tex]\( \text{Cu}^{2+} \)[/tex] ion (aqueous) with an oxidation state of [tex]\(+2\)[/tex]. After the reaction, Copper is in the solid form ([tex]\( \text{Cu} \)[/tex]), which has an oxidation state of [tex]\(0\)[/tex].
3. Identify the species undergoing oxidation:
- Zinc ([tex]\(\text{Zn}\)[/tex]) is going from an oxidation state of [tex]\(0\)[/tex] to [tex]\(+2\)[/tex]. This is an increase in the oxidation state, indicating oxidation.
4. Write the oxidation half-reaction:
- The oxidation half-reaction will indicate the species losing electrons. For Zinc:
[tex]\[ \text{Zn} (s) \longrightarrow \text{Zn}^{2+} \text{(aq)} + 2e^- \][/tex]
This equation shows that Zinc loses two electrons (indicated by [tex]\(2e^-\)[/tex]) when it oxidizes from a neutral atom to a [tex]\( \text{Zn}^{2+} \)[/tex] ion.
Therefore, the correct half-reaction describing the oxidation that is taking place is:
[tex]\[ \text{Zn} (s) \longrightarrow \text{Zn}^{2+} \text{(aq)} + 2e^- \][/tex]
So, the answer is:
[tex]\[ \boxed{\text{Zn (s) } \longrightarrow \text{Zn}^{2+} \text{(aq)} + 2e^-} \][/tex]
