Question

identify the oxidizing agent and the reducing agent for
4li(s) + o₂(g) → 2li₂o(s)
express your answers as chemical formulas separated by a comma.
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oxidizing agent, reducing agent

Explanation:

Step1: Determine oxidation states

• For \(Li\) in \(Li(s)\): oxidation state is \(0\).
• For \(O\) in \(O_{2}(g)\): oxidation state is \(0\).
• For \(Li\) in \(Li_{2}O(s)\): oxidation state is \(+1\).
• For \(O\) in \(Li_{2}O(s)\): oxidation state is \(- 2\).

Step2: Identify oxidation and reduction

• \(Li\) is oxidized (oxidation state from \(0\) to \(+1\)).
• \(O_{2}\) is reduced (oxidation state from \(0\) to \(-2\)).

Step3: Define oxidizing and reducing agents

• The substance that is reduced is the oxidizing agent (\(O_{2}\)).
• The substance that is oxidized is the reducing agent (\(Li\)).

Answer:

\(O_{2},Li\)