In the reaction \( \text{Zn(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)} \), what is the oxidation state of zinc in the products?}
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In redox reactions, oxidation corresponds to an increase in oxidation state, while reduction corresponds to a decrease in oxidation state. Always identify the oxidation states of elements in both reactants and products to determine the changes.
The given reaction is:
\[
\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}
\]
1. Identify oxidation states:
- In \( \text{Zn(s)} \), zinc is in its elemental form, so its oxidation state is \( 0 \).
- In \( \text{Cu}^{2+}(aq) \), the oxidation state of copper is \( +2 \).
2. Oxidation and reduction:
- Zinc goes from \( 0 \) to \( +2 \) (oxidation), and copper goes from \( +2 \) to \( 0 \) (reduction).
3. Oxidation state of zinc in the product:
- In the product, \( \text{Zn}^{2+}(aq) \), zinc has an oxidation state of \( +2 \).
Thus, the oxidation state of zinc in the products is \( +2 \).
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