What are the spectator ions and balanced net ionic equation for this reaction: #Cl_2(g) + NaBr(aq) -> Br_2(l) + NaCl(aq)#?

1 Answer
Mar 12, 2017

#"Cl"_ (2(g)) + 2"Br"_ ((aq))^(-) -> "Br"_ (2(g)) + 2"Cl"_ ((aq))^(-)#


The reaction between chlorine gas and sodium bromide produces liquid bromine and aqueous sodium chloride, as shown by the unbalanced chemical equation

#"Cl"_ (2(g)) + "NaBr"_ ((aq)) -> "Br"_ (2(l)) + "NaCl"_ ((aq))#

The first thing to do here is to balance the chemical equation. To do that, add a coefficient of #2# to sodium bromide and a coefficient of #2# to sodium chloride.

This will get you

#"Cl"_ (2(g)) + 2"NaBr"_ ((aq)) -> "Br"_ (2(l)) + 2"NaCl"_ ((aq))#

Now, sodium bromide and sodium chloride are soluble ionic compounds, which means that they dissociate completely in aqueous solution, i.e. they exist as ions

#"NaBr"_ ((aq)) -> "Na"_ ((aq))^(+) + "Br"_ ((aq))^(-)#

#"NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#

This means that you can rewrite the balanced chemical equation as -- remember that when you write an ionic compound as dissociated ions, the stoichiometric coefficient gets distributed to both ions!

You will have

#"Cl"_ (2(g)) + 2xx ["Na"_ ((aq))^(+) + "Br"_ ((aq))^(-)] -> "Br"_ (2(l)) + 2xx ["Na"_ ((aq))^(+) + "Cl"_ ((aq))^(-)]#

which is equivalent to

#"Cl"_ (2(g)) + 2"Na"_ ((aq))^(+) + 2"Br"_ ((aq))^(-) -> "Br"_ (2(g)) + 2"Na"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-)#

Now, the spectator ions are the ions that are present on both sides of the equation. In this case, the sodium cations will be spectator ions.

In order to get the net ionic equation, you must eliminate the spectator ions from the balanced chemical equation

#"Cl"_ (2(g)) + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + 2"Br"_ ((aq))^(-) -> "Br"_ (2(g)) + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + 2"Cl"_ ((aq))^(-)#

This will get you

#"Cl"_ (2(g)) + 2"Br"_ ((aq))^(-) -> "Br"_ (2(g)) + 2"Cl"_ ((aq))^(-)#