Final answer:
In the given redox reaction, zinc is being reduced from +2 to 0. No oxidation occurs since tin's oxidation state remains +4 in both the reactant and product. Zn(OH)2 acts as the reducing agent, and SnO22- is the oxidizing agent.
Step-by-step explanation:
To identify what is being oxidized and what is being reduced in a redox reaction, oxidation numbers for each atom in the reactants and products are assigned. The element that has an increase in oxidation number is being oxidized and generally serves as the reducing agent, since it is losing electrons. Conversely, the element that has a decrease in oxidation number is being reduced and acts as the oxidizing agent, since it is gaining electrons.
In the provided reaction (Zn(OH)2 + SnO22- → Zn + SnO32- + H2O), let's analyze the oxidation states:
- Zn in Zn(OH)2 has an oxidation number of +2.
- Sn in SnO22- has an oxidation number of +4.
- Zn in metallic form (Zn) has an oxidation number of 0.
- Sn in SnO32- has an oxidation number of +4.
Zinc's oxidation number changes from +2 to 0, being reduced, and tin's oxidation number remains +4, indicating no change. Therefore, in this equation, no oxidation nor reduction is actually taking place for tin; only zinc is being reduced. However, for the purpose of the question, if we consider Zn(OH)2 being oxidized, then Zn(OH)2 would be the reducing agent and SnO22- would be the oxidizing agent.