Question

identify the oxidizing agent in the following reaction: kclo₂→kcl + o₂
o chlorine in kclo₂
o potassium in kclo₂
o oxygen in kclo₂
o potassium in kcl

Explanation:

Step1: Determine oxidation states

Oxidation state of \(K\) is \(+ 1\) (group 1 metal). Let oxidation state of \(Cl\) in \(KClO_2\) be \(x\). Using formula for neutral compound \(+1+x + 2(-2)=0\), so \(x = + 3\). In \(KCl\), oxidation state of \(Cl\) is \(-1\). Oxidation state of \(O\) in \(KClO_2\) is \(-2\), in \(O_2\) is \(0\).

Step2: Analyze oxidation - reduction

Chlorine in \(KClO_2\) (\(+3\)) is reduced to \(Cl\) in \(KCl\) (\(-1\)). Substance that gets reduced is oxidizing agent. Oxygen in \(KClO_2\) (\(-2\)) is oxidized to \(O\) in \(O_2\) (\(0\)). Potassium has oxidation state \(+1\) in both \(KClO_2\) and \(KCl\) (no change).

Answer:

Chlorine in \(KClO_2\)