Q.The reaction of cold, dilute $\text{NaOH}$ with $\text{Cl}_2$ is $\text{Cl}_2 + 2\text{NaOH} \to \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O}$. Show, using oxidation numbers, that this is a disproportionation reaction, and explain what the term means.

[!TLDR] Cl starts at $0$ in $\text{Cl}_2$ and ends at $-1$ (in NaCl, reduced) and $+1$ (in NaOCl, oxidized) — the same element is simultaneously oxidized and reduced. [!ANSWER] This is a disproportion …

In $\text{Cl}_2 + 2\text{NaOH} \to \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O}$, chlorine in the reactant $\text{Cl}_2$ is at oxidation number $0$ (a free element). In the products, chlorine appears in two different compounds: in $\text{NaCl}$, Cl is $-1$ (a decrease from $0$, so this chlorine atom is reduced); in $\text{NaOCl}$, using O$=-2$ and Na$=+1$, Cl must be $+1$ (an increase from $0$, so this chlorine atom is oxidized). Since a single element, starting from a single oxidation state in $\text{Cl}_2$, ends up split between one product where it is reduced and another where it is oxidized, this reaction is classified as a disproportionation — chlorine acts as both the oxidizing agent and the reducing agent for itself. This is p …

Assign oxidation numbers to chlorine in the reactant and in each product separately, then check whether the same starting elemen …

Assuming disproportionation requires two different elements to be involved, rather than recognizing it as a single ele …