Step 1: Recall that Cl2 reacts differently with NaOH at different temperatures.
Chlorine undergoes disproportionation with NaOH (Cl2 is simultaneously oxidised and reduced). The product depends on whether the reaction is carried out in cold dilute or hot concentrated conditions.
Step 2: Write the reaction with cold, dilute NaOH.
\[ \text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O} \]
One Cl is reduced to Cl- (oxidation state -1, in NaCl) and the other is oxidised to Cl+ (oxidation state +1, in NaOCl, sodium hypochlorite). This reaction occurs below about 15 degrees C.
Step 3: Identify the products of the cold dilute reaction.
Products: NaCl (sodium chloride) and NaOCl (sodium hypochlorite). NaOCl is the bleaching agent in household bleach and is responsible for the disinfectant property of chlorine water.
Step 4: Write the reaction with hot, concentrated NaOH.
\[ 3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O} \]
Here, 5 Cl atoms are reduced to -1 (NaCl) and 1 Cl atom is oxidised to +5 (NaClO3, sodium chlorate). This reaction occurs above about 70 degrees C.
Step 5: Explain why temperature changes the product.
At higher temperatures, the initially formed NaOCl undergoes further disproportionation: \(3\text{NaOCl} \rightarrow 2\text{NaCl} + \text{NaClO}_3\). The hypochlorite ion is converted to chlorate under hot conditions because this disproportionation has a positive entropy change that becomes favourable at high temperatures.
Step 6: State the final answer.
Cold dilute NaOH gives NaCl and NaOCl; hot concentrated NaOH gives NaCl and NaClO3.
\[ \boxed{\text{Cold: NaCl + NaOCl};\ \text{Hot: NaCl + NaClO}_3} \]