Zn(s) + CuSO4(aq) $\to$ ZnSO4(aq) + Cu(s) In the reaction above, the oxidizing agent is

Analysis

To identify the oxidizing agent in this redox reaction, we need to track the changes in the oxidation states of the elements involved.

Concept: Redox Reactions

• Oxidation: The loss of electrons (increase in oxidation number). The substance that undergoes oxidation is the reducing agent.
• Reduction: The gain of electrons (decrease in oxidation number). The substance that undergoes reduction is the oxidizing agent.

Step-by-Step Solution

1. Assign oxidation numbers to each element: • Reactants: • Zn(s) is an element in its standard state, so its oxidation number = 0. • In CuSO$_4$: Cu is +2, SO$_4$ is -2. So, Cu = +2. • Products: • In ZnSO$_4$: Zn is +2, SO$_4$ is -2. So, Zn = +2. • Cu(s) is an element in its standard state, so its oxidation number = 0.

2. Identify the changes: • Zinc (Zn): Oxidation number increases from 0 to +2. This is oxidation. Since Zn caused the reduction of copper, it is the reducing agent. • Copper (Cu): Oxidation number decreases from +2 to 0. This is reduction. Since the copper ion (in CuSO$_4$) gained electrons, it is the oxidizing agent.

3. Conclusion: The substance that is reduced is the oxidizing agent. In this reaction, the copper ion in CuSO$_4$ is reduced to metallic copper. Therefore, the entire compound CuSO$_4$ acts as the oxidizing agent.

Exam Tip

A simple way to remember this is the acronym OIL RIG:

• Oxidation Is Loss (of electrons)
• Reduction Is Gain (of electrons) The oxidizing agent is the one that "gains" the electrons, causing the other substance to be oxidized.