To find the oxidation half-reaction for the given reaction:
[tex]\[
Mg (s) + ZnCl_2 (aq) \rightarrow MgCl_2 (aq) + Zn (s)
\][/tex]
We need to analyze the changes in oxidation states of the elements involved.
1. Identify the oxidation states of elements before and after the reaction:
- Magnesium ([tex]\( Mg \)[/tex]) in its solid state has an oxidation state of 0.
- Zinc ([tex]\( Zn \)[/tex]) in [tex]\( ZnCl_2 \)[/tex] starts with an oxidation state of +2 (since [tex]\( Cl \)[/tex] has an oxidation state of -1 and there are two chloride ions).
- In [tex]\( MgCl_2 \)[/tex], magnesium ends up with an oxidation state of +2 (since there are two chloride ions each having a -1 charge, summing to -2, magnesium must be +2 to balance the charge).
- Zinc in its solid state has an oxidation state of 0.
2. Determine which element is oxidized and which is reduced:
- Oxidation is the process of losing electrons.
- Reduction is the process of gaining electrons.
For Magnesium (Mg):
[tex]\[
Mg (s) \rightarrow Mg^{2+} + 2e^-
\][/tex]
Magnesium starts in the 0 oxidation state and ends up in the +2 oxidation state, indicating it has lost 2 electrons. This is oxidation.
For Zinc (Zn):
[tex]\[
Zn^{2+} + 2e^- \rightarrow Zn (s)
\][/tex]
Zinc starts in the +2 oxidation state and ends up in the 0 oxidation state, indicating it has gained 2 electrons. This is reduction.
3. Identify the oxidation half-reaction:
Given the options:
A. [tex]\(Mg (s) \rightarrow Mg ^{2+} + 2e ^{-}\)[/tex]
B. [tex]\(Zn^{2+} + 2e ^{-} \rightarrow Zn (s)\)[/tex]
C. [tex]\(Zn (s) \rightarrow Zn^{2+} + 2e ^{-}\)[/tex]
D. [tex]\(Mg^{2+} + 2e ^{-} \rightarrow Mg (s)\)[/tex]
The oxidation half-reaction involves the species being oxidized, which here is magnesium.
Therefore, the correct oxidation half-reaction is:
[tex]\[
Mg (s) \rightarrow Mg^{2+} + 2e^-
\][/tex]
So the answer is:
A. [tex]\(Mg (s) \rightarrow Mg^{2+} + 2e^-\)[/tex]
