Chapter 5: Problem 8
Identify the oxidized reactant, the reduced reactant, the oxidizing agent, and the reducing agent in the following reactions: (a) \(\mathrm{Fe}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Cu}(s)\) (b) \(\mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{MgCl}_{2}(s)\) (c) \(2 \mathrm{Al}(s)+\mathrm{Cr}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Cr}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\)
Short Answer
Step by step solution
Determine oxidation states for Reaction (a)
Identify components for Reaction (a)
Determine oxidation states for Reaction (b)
Identify components for Reaction (b)
Determine oxidation states for Reaction (c)
Identify components for Reaction (c)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Oxidation States
- Oxidation refers to an increase in oxidation state, which corresponds to the loss of electrons.
- Reduction refers to a decrease in oxidation state, indicating the gain of electrons.
Oxidizing Agent
In redox reactions, identifying the oxidizing agent involves finding the molecule or ion that has gained electrons and, therefore, experienced a reduction in its oxidation state.
- Example: In reaction (a), \({\text{Cu}^{2+}}\) serves as the oxidizing agent because it gains 2 electrons, reducing its oxidation state from +2 to 0.
Reducing Agent
In redox reactions, the reducing agent can be recognized as the species whose oxidation state increases, as it loses electrons.
- For example, in reaction (b), magnesium (Mg) acts as the reducing agent since it loses 2 electrons, increasing its oxidation state from 0 to +2.
Electron Transfer
- An oxidation process involves the loss of electrons by a species, increasing its oxidation state.
- A reduction process, on the other hand, involves the gain of electrons, decreasing its oxidation state.
Chemical Reactions
- These reactions are essential in processes such as combustion, corrosion, and energy production.
- Balancing redox reactions includes ensuring that the number of electrons lost equals the number of electrons gained, maintaining charge balance.