To determine which element is undergoing reduction in the given reaction, we need to analyze the oxidation states of the elements before and after the reaction. Here is the balanced chemical equation:
[tex]\[ Zn (s) + 2 AgNO _3(aq) \rightarrow Zn \left( NO_3 \right)_2 (aq) + 2 Ag (s) \][/tex]
We'll examine the oxidation states of the relevant elements in each compound:
1. Zinc:
- Before the reaction: [tex]\(Zn (s)\)[/tex] - Zinc in its elemental form has an oxidation state of 0.
- After the reaction: [tex]\(Zn \left( NO_3 \right)_2(aq)\)[/tex] - In [tex]\(Zn(NO_3)_2\)[/tex], the zinc ion has an oxidation state of +2 (since each [tex]\(NO_3^-\)[/tex] anion is -1 and there are two of them, the zinc must be +2 to balance the charge).
2. Silver:
- Before the reaction: [tex]\(AgNO_3 (aq)\)[/tex] - In [tex]\(AgNO_3\)[/tex], the silver ion ([tex]\(Ag^+\)[/tex]) has an oxidation state of +1.
- After the reaction: [tex]\(Ag (s)\)[/tex] - Silver in its elemental form has an oxidation state of 0.
3. Elements in [tex]\(NO_3^-\)[/tex] (Nitrate ion):
- The nitrogen (N) in [tex]\(NO_3^-\)[/tex] does not change its oxidation state in this reaction (it remains +5 both before and after the reaction).
- The oxygen (O) in [tex]\(NO_3^-\)[/tex] also does not change its oxidation state (it remains -2 both before and after the reaction).
Now that we’ve established the oxidation states:
- Zinc goes from 0 to +2 (which means it loses electrons and is oxidized).
- Silver goes from +1 to 0 (which means it gains electrons and is reduced).
Therefore, the element undergoing reduction in this reaction is Silver (Ag).
