Question 97adb

Jun 10, 2016

Here's what I got.

Explanation:

You're dealing with a single replacement reaction in which chlorine, ${\text{Cl}}_{2}$, displaces the *bromide anion, ${\text{Br}}^{-}$, from potassium bromide, $\text{KBr}$, to form potassium chloride, $\text{KCl}$, and liquid bromine, ${\text{Br}}_{2}$.

${\text{Cl"_ (2(aq)) + 2"KBr"_ ((aq)) -> 2"KCl"_ ((aq)) + "Br}}_{2 \left(a q\right)}$

Now, potassium bromide and potassium chloride are both soluble salts, which means that they are completely dissociated in aqueous solution.

You can thus write them as

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

and

${\text{KCl"_ ((aq)) -> "K"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

This means that the balanced chemical equation can be rewritten as

${\text{Cl"_ (2(aq)) + 2 xx overbrace(["K"_ ((aq))^(+) + "Br"_ ((aq))^(-)])^(color(blue)("KBr"_ ((aq)))) -> 2 xx overbrace(["K"_ ((aq))^(+) + "Cl"_ ((aq))^(-)])^(color(darkgreen)("KCl"_ ((aq)))) + "Br}}_{2 \left(a q\right)}$

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

This represents the complete ionic equation for the reaction. In order to get the net ionic equation, eliminate the spectator ions, i.e. the ions that are present on both sides of the equation.

In this case, you have

${\text{Cl"_ (2(aq)) + 2color(red)(cancel(color(black)("K"_ ((aq))^(+)))) + 2"Br"_ ((aq))^(-) -> 2color(red)(cancel(color(black)("K"_ ((aq))^(+)))) + 2"Cl"_ ((aq))^(-) + "Br}}_{2 \left(a q\right)}$

and thus the net ionic equation is

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{Cl"_ (2(aq)) + 2"Br"_ ((aq))^(-) -> 2"Cl"_ ((aq))^(-) + "Br}}_{2 \left(a q\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, this is also a redox reaction in which chlorine is being reduced to chloride anions and bromine is being oxidized to molecular bromine.

You can see this be assigning oxidation states to the atoms that take part in the reaction

${\stackrel{\textcolor{b l u e}{0}}{\text{Cl")_ (2(aq)) + 2 stackrel(color(blue)(-1))("Br") ""^(-)"" _ ((aq)) -> 2stackrel(color(blue)(-1))("Cl")""^(-)""_ ((aq)) + stackrel(color(blue)(0))("Br}}}_{2 \left(a q\right)}$

The oxidation number of chlorine goes from $\textcolor{b l u e}{0}$ on the reactants' side to $\textcolor{b l u e}{- 1}$ on the products' side, which means that it is being reduced.

On the other hand, the oxidation number of bromine goes from $\textcolor{b l u e}{- 1}$ on the reactants' side to $\textcolor{b l u e}{0}$ on the products' side, which means that it is being oxidized.

The oxidation half-reaction looks like this

2stackrel(color(blue)(-1))("Br") ""^(-)""_ ((aq)) -> stackrel(color(blue)(0))("Br") _(2(aq)) + 2"e"^(-)

Here each bromine atom loses one electron because its oxidation state is increasing by $1$, so two bromine atoms will lose $2$ electrons.

The reduction half-reaction looks like this

stackrel(color(blue)(0))("Cl")_ (2(aq)) + 2"e"^(-) -> 2stackrel(color(blue)(-1))("Cl") ""^(-)""_ ((aq))#

Here each chlorine atom gains one electron because its oxidation state is decreasing by $1$, so two chlorine atoms will gain $2$ electrons.