Aqueous chlorine is added to a solution of aluminum bromide in a single replacement reaction which forms no precipitate and bromine liquid. Determine the net ionic equation.

Since this is a single replacement reaction, it will take the form of $$\ce{A + B-C -> A-C + B}$$

Since this reaction forms no precipitate, the new compound will not be a solid.

$$\begin{multline} \ce{aqueous chlorine + solution of aluminum bromide ->}\\ \ce{ non-solid compound + liquid bromine} \end{multline}$$ \begin{align} \ce{Cl2(aq) + Al^{3+}Br3^-(aq) &-> \text{non-solid compound} + Br2(l)}\\ \ce{Cl2(aq) + AlBr3(aq) &-> \text{non-solid compound} + Br2(l)}\\ \ce{Cl2(aq) + AlBr3(aq) &-> Al^{3+}Cl3- + Br2(l)}\\ \ce{Cl2(aq) + AlBr3(aq) &-> AlCl3 + Br2(l)} \end{align}

Since most chloride salts are soluble (except for silver halides), $\ce{AlCl3}$ is predicted to be aqueous.

\begin{align} \ce{Cl2(aq) + AlBr3(aq) &-> AlCl3(aq) + Br2(l)}\\ \ce{3Cl2(aq) + 2AlBr3(aq) &-> 2AlCl3(aq) + 3Br2(l)} \end{align}

Here is the complete ionic equation:

$$\ce{6Cl⁻(aq) + 2Al^{3+}(aq) + 6Br-(aq) -> 2Al^{3+}(aq) + 6Cl-(aq) + 3Br2(l)}$$

Cancelling out the spectator ions, we get

$$\ce{6Br-(aq) -> 3Br2(l)}$$

In lowest terms:

$$\ce{2Br-(aq) -> Br2(l)}$$


2 Answers 2


You are making a mistake splitting up your reaction equation from \eqref{full} to \eqref{wrong} in order to filter out the spectator ions.

\begin{align} \ce{3Cl2(aq) + 2AlBr3(aq) &-> 2AlCl3(aq) + 3Br2(l)}\tag1\label{full}\\ \ce{6Cl⁻(aq) + 2Al^{3+}(aq) + 6Br-(aq) &-> 2Al^{3+}(aq) + 6Cl-(aq) + 3Br2(l)}\tag{wrong}\label{wrong} \end{align}

What you are essentially doing is not conserving the charges. Dichlorine $\ce{Cl2}$ is a neutral molecule, and the following is what your transition implies, but is also wrong: $$\ce{Cl2 -> 2Cl-}\tag{also wrong}$$

The correct way to transform \eqref{full} is to split (dissociate) only the ionic compounds $\ce{AlBr3}$ and \ce{AlCl3}, and then cancel the spectator ions:

\begin{align} \ce{3 Cl2 (aq) + 2 AlBr3 (aq) &-> 2 AlCl3 (aq) + 3 Br2 (l)}\tag1\\ \ce{3 Cl2 (aq) + 2 Al^3+ (aq) + 6 Br- (aq) &-> 2 Al^3+ (aq) + 6 Cl- (aq) + 3 Br2 (l)}\tag2\label{split}\\ \end{align}

Next step is to cancel the aluminium ions on each side, and reduce the equation to \eqref{net-ionic}:

\begin{align} \ce{3 Cl2 (aq) + 6 Br- (aq) &-> 6 Cl- (aq) + 3 Br2 (l)}\tag3\\ \ce{Cl2 (aq) + 2 Br- (aq) &-> 2 Cl- (aq) + Br2 (l)}\tag4\label{net-ionic} \end{align}

I don't think the reaction will actually produce liquid bromine, but aqueous bromine instead; that doesn't matter to the solution though.


You do not render the chlorine correctly. Keep the chlorine as diatomic molecules on the reactant side and note that because of the charges the chlorine atoms do not cancel out. Your net ionic equation should therefore include both halogens.


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