Question

identify the species oxidized, the species reduced, the oxidizing agent and the reducing agent in the following electron transfer reaction.

3mn²⁺ + 2al → 3mn + 2al³⁺

as the reaction proceeds, electrons are transferred from to .

Explanation:

Step1: Determine oxidation states

• For \(Al\) in \(Al\) (reactant), oxidation state is \(0\). In \(Al^{3 + }\) (product), it is \(+ 3\).
• For \(Mn^{2+}\) (reactant), oxidation state is \(+2\). In \(Mn\) (product), it is \(0\).

Step2: Identify oxidized and reduced species

• Species oxidized: The species that loses electrons (oxidation state increases). \(Al\) goes from \(0\) to \(+3\), so \(Al\) is oxidized.
• Species reduced: The species that gains electrons (oxidation state decreases). \(Mn^{2+}\) goes from \(+2\) to \(0\), so \(Mn^{2+}\) is reduced.

Step3: Identify oxidizing and reducing agents

• Oxidizing agent: The species that causes oxidation (gets reduced). \(Mn^{2+}\) is the oxidizing agent.
• Reducing agent: The species that causes reduction (gets oxidized). \(Al\) is the reducing agent.

Step4: Determine electron transfer

• Electrons are transferred from the species that is oxidized (\(Al\)) to the species that is reduced (\(Mn^{2+}\)).

Answer:

species oxidized: \(Al\)
species reduced: \(Mn^{2+}\)
oxidizing agent: \(Mn^{2+}\)
reducing agent: \(Al\)
Electrons are transferred from \(Al\) to \(Mn^{2+}\)