Redox Reactions: Question 9

Worked solution

Part (a): Oxidation number of chlorine

Chlorine in $\text{Cl}_2$ is an uncombined element, so its oxidation number is $0$.

In $\text{NaCl}$, sodium (a Group I metal) has an oxidation number of $+1$, and the compound is neutral overall, so chlorine must be $-1$.

Part (b): Oxidation number of bromine

In $\text{NaBr}$, by the same reasoning, sodium is $+1$ and the compound is neutral, so bromine must be $-1$.

Bromine in $\text{Br}_2$ is an uncombined element, so its oxidation number is $0$.

Part (c): Identifying oxidation and reduction

• Chlorine’s oxidation number changes from $0$ (in $\text{Cl}_2$) to $-1$ (in $\text{NaCl}$). A decrease, so chlorine is reduced.
• Bromine’s oxidation number changes from $-1$ (in $\text{NaBr}$) to $0$ (in $\text{Br}_2$). An increase, so bromine is oxidised.

Since both an increase and a decrease in oxidation number occur in the same reaction, this confirms it is a redox reaction.

Part (d): Identifying the oxidising agent

An oxidising agent is the species that gains electrons (and is itself reduced), causing another species to lose electrons.

Chlorine’s oxidation number falls from $0$ to $-1$, which corresponds to each chlorine atom gaining one electron: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$. In gaining these electrons, chlorine causes the bromide ions to lose electrons and become bromine. So chlorine, $\text{Cl}_2$, is the oxidising agent.

Final answers

• (a) Chlorine: $0$ in $\text{Cl}_2$, $-1$ in $\text{NaCl}$.
• (b) Bromine: $-1$ in $\text{NaBr}$, $0$ in $\text{Br}_2$.
• (c) Chlorine is reduced (0 to −1); bromine is oxidised (−1 to 0).
• (d) Chlorine, $\text{Cl}_2$, is the oxidising agent, it gains electrons.