# Question d28a4

Jun 22, 2017

$2 {\text{Br"_ ((aq))^(-) + "F"_ (2(aq)) -> 2"F"_ ((aq))^(-) + "Br}}_{2 \left(a q\right)}$

#### Explanation:

The idea here is that fluorine is more reactive than bromine, so it will displace bromine from sodium bromide.

Keep in mind that this reaction takes place in aqueous solution, so the fluorine gas is first dissolved in water

${\text{F"_ (2(g)) rightleftharpoons "F}}_{2 \left(a q\right)}$

The balanced chemical equation that describes this single displacement reaction looks like this

$2 {\text{NaBr"_ ((aq)) + "F"_ (2(aq)) -> 2"NaF"_ ((aq)) + "Br}}_{2 \left(a q\right)}$

The complete ionic equation looks like this

$2 {\text{Na"_ ((aq))^(+) + 2"Br"_ ((aq))^(-) + "F"_ (2(aq)) -> 2"Na"_ ((aq))^(+) + 2"F"_ ((aq))^(-) + "Br}}_{2 \left(a q\right)}$

In order to get the net ionic equation, you must eliminate the spectator ions, i.e. the ions that are present on both sides of the equation.

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

You will end up with

$2 {\text{Br"_ ((aq))^(-) + "F"_ (2(aq)) -> 2"F"_ ((aq))^(-) + "Br}}_{2 \left(a q\right)}$

At this point, the fact that this reaction is also a redox reaction becomes obvious. Fluorine will oxidize the bromide anions to bromine and be reduced to fluoride anions in the process.

Keep in mind that fluorine gas is extremely reactive and toxic!