To determine the number of electrons produced and used in the given reaction, we need to write and balance the half-reactions for the oxidation and reduction processes separately.
#### Reduction Half-Reaction:
The reduction process involves hydrogen ions ([tex]\(H^{+}\)[/tex]) gaining electrons to form hydrogen gas ([tex]\(H_2\)[/tex]). The balanced reduction half-reaction can be written as:
[tex]\[ -2 H^{+} + 2 e^{-} \rightarrow H_2 \][/tex]
Here, two hydrogen ions ([tex]\(H^{+}\)[/tex]) gain two electrons ([tex]\(e^{-}\)[/tex]) to form one molecule of hydrogen gas ([tex]\(H_2\)[/tex]).
#### Oxidation Half-Reaction:
The oxidation process involves zinc ([tex]\(Zn\)[/tex]) losing electrons to form zinc ions ([tex]\(Zn^{2+}\)[/tex]). The balanced oxidation half-reaction can be written as:
[tex]\[ Zn \rightarrow Zn^{2+} + 2 e^{-} \][/tex]
In this half-reaction, one zinc atom ([tex]\(Zn\)[/tex]) loses two electrons ([tex]\(e^{-}\)[/tex]) to form one zinc ion ([tex]\(Zn^{2+}\)[/tex]).
#### Summary:
- In the reduction half-reaction:
- Electrons produced: 0
- Electrons used: 2 (since two hydrogen ions gain two electrons)
- In the oxidation half-reaction:
- Electrons produced: 2 (since one zinc atom loses two electrons)
- Electrons used: 0
The number of electrons produced and used in these half-reactions can be summarized as follows:
- Electrons produced in the oxidation half-reaction: 2
- Electrons used in the reduction half-reaction: 2
These values ensure that the overall reaction is balanced in terms of electron flow. The balanced half-reactions are:
[tex]\[ -2 H^{+} + 2 e^{-} \rightarrow H_2 \][/tex]
[tex]\[ Zn \rightarrow Zn^{2+} + 2 e^{-} \][/tex]
Thus, 2 electrons are produced in the oxidation half-reaction and 2 electrons are used in the reduction half-reaction.
