Answer :
To determine the correct half-reaction that describes the oxidation taking place in the given redox reaction:
[tex]\[ Zn (s) + 2 HCl (aq) \rightarrow ZnCl_2 (aq) + H_2 (g) \][/tex]
We need to break down the roles of the reactants and products in terms of oxidation and reduction. A redox reaction comprises two half-reactions: oxidation and reduction. Oxidation involves the loss of electrons, while reduction involves the gain of electrons.
In the given reaction, Zinc ([tex]\(Zn\)[/tex]) starts as a solid ([tex]\(s\)[/tex]) and ends up in the ionic form as [tex]\(Zn^{2+}\)[/tex] in [tex]\(ZnCl_2\)[/tex] (aq). On the other side, the hydrogen ions [tex]\(H^+\)[/tex] (from [tex]\(HCl\)[/tex]) gain electrons to form hydrogen gas ([tex]\(H_2\)[/tex]).
Let's specifically check what happens to the zinc:
- Zinc starts as a neutral atom [tex]\(Zn (s)\)[/tex].
- It loses 2 electrons during the reaction to become [tex]\(Zn^{2+}\)[/tex] ion.
The oxidation half-reaction for the zinc is:
[tex]\[ Zn (s) \rightarrow Zn^{2+} (aq) + 2 e^- \][/tex]
This shows the zinc atom losing electrons, which is the definition of oxidation.
Among the options provided:
1. [tex]\(Zn^{2+}(s)+2 e^{-}(aq) \longrightarrow Zn (s)\)[/tex] – This is a reduction half-reaction and is not correct.
2. [tex]\(Zn (s) \longrightarrow Zn^{2+}(aq)+2 e^{-}\)[/tex] – This correctly represents the oxidation half-reaction.
3. [tex]\(2 H^{+}+2 e^{-} \longrightarrow H_2\)[/tex] – This is a reduction half-reaction for hydrogen.
4. [tex]\(H_2+2 e^{-} \longrightarrow 2 H^{+}\)[/tex] – This is not applicable in the context given and is a form of oxidation but not relevant here.
Therefore, the correct half-reaction that describes the oxidation taking place is:
[tex]\[ Zn (s) \longrightarrow Zn^{2+}(aq) + 2 e^- \][/tex]
[tex]\[ Zn (s) + 2 HCl (aq) \rightarrow ZnCl_2 (aq) + H_2 (g) \][/tex]
We need to break down the roles of the reactants and products in terms of oxidation and reduction. A redox reaction comprises two half-reactions: oxidation and reduction. Oxidation involves the loss of electrons, while reduction involves the gain of electrons.
In the given reaction, Zinc ([tex]\(Zn\)[/tex]) starts as a solid ([tex]\(s\)[/tex]) and ends up in the ionic form as [tex]\(Zn^{2+}\)[/tex] in [tex]\(ZnCl_2\)[/tex] (aq). On the other side, the hydrogen ions [tex]\(H^+\)[/tex] (from [tex]\(HCl\)[/tex]) gain electrons to form hydrogen gas ([tex]\(H_2\)[/tex]).
Let's specifically check what happens to the zinc:
- Zinc starts as a neutral atom [tex]\(Zn (s)\)[/tex].
- It loses 2 electrons during the reaction to become [tex]\(Zn^{2+}\)[/tex] ion.
The oxidation half-reaction for the zinc is:
[tex]\[ Zn (s) \rightarrow Zn^{2+} (aq) + 2 e^- \][/tex]
This shows the zinc atom losing electrons, which is the definition of oxidation.
Among the options provided:
1. [tex]\(Zn^{2+}(s)+2 e^{-}(aq) \longrightarrow Zn (s)\)[/tex] – This is a reduction half-reaction and is not correct.
2. [tex]\(Zn (s) \longrightarrow Zn^{2+}(aq)+2 e^{-}\)[/tex] – This correctly represents the oxidation half-reaction.
3. [tex]\(2 H^{+}+2 e^{-} \longrightarrow H_2\)[/tex] – This is a reduction half-reaction for hydrogen.
4. [tex]\(H_2+2 e^{-} \longrightarrow 2 H^{+}\)[/tex] – This is not applicable in the context given and is a form of oxidation but not relevant here.
Therefore, the correct half-reaction that describes the oxidation taking place is:
[tex]\[ Zn (s) \longrightarrow Zn^{2+}(aq) + 2 e^- \][/tex]