Question

question 8
consider the following reaction and select the reducing agent.
cl₂ + 2 nabr → 2 nacl + br₂
br⁻ in nabr
cl⁻ in nacl
cl₂
br₂

Explanation:

Step1: Determine oxidation states

• In \(NaBr\), \(Na\) has an oxidation state of \(+1\), so \(Br\) has an oxidation state of \(- 1\).
• In \(Br_{2}\), \(Br\) has an oxidation state of \(0\).
• In \(Cl_{2}\), \(Cl\) has an oxidation state of \(0\).
• In \(NaCl\), \(Na\) has an oxidation state of \(+1\), so \(Cl\) has an oxidation state of \(-1\).

Step2: Identify oxidation and reduction

• \(Br^{-}\) in \(NaBr\) is oxidized (from \(-1\) to \(0\)).
• \(Cl_{2}\) is reduced (from \(0\) to \(-1\)).

Step3: Define reducing agent

• A reducing agent is a substance that donates electrons (gets oxidized). Since \(Br^{-}\) in \(NaBr\) is oxidized, it is the reducing agent.

Answer:

Br⁻ in NaBr