# Question 79236

Sep 12, 2015

Here's how the half-reactions look like.

#### Explanation:

Chlorinegas can react with water to form hypochlorous acid, $\text{HClO}$, and hydrochloric acid, $\text{HCl}$.

This reaction is actually a disproportionation reaction because chlorine is being oxidized and reduced at the same time.

The balanced chemical equation - with oxidation numbers included- looks like this

$\stackrel{\textcolor{b l u e}{0}}{\text{Cl"_2) + stackrel(color(blue)(+1))("H"_2) stackrel(color(blue)(-2))("O") -> stackrel(color(blue)(+1))("H") stackrel(color(blue)(+1))"Cl" stackrel(color(blue)(-2)) ("O") + stackrel(color(blue)(+1)) ("H") stackrel(color(blue)(-1)) ("Cl}}$

Notice that chlorine is being oxidized to a +1 oxidation state in hypochlorous acid and reduced to a -1 oxidation state in hydrochloric acid.

The half-reactions will be

• reduction half-reaction

$\stackrel{\textcolor{b l u e}{0}}{\text{Cl"_2) + 2e^(-) -> 2"H"stackrel(color(blue)(-1))("Cl}}$

Use protons to balance the hydrogen atoms

2"H"^(+) + stackrel(color(blue)(0))("Cl"_2) + 2e^(-) -> 2"H"stackrel(color(blue)(-1))("Cl")

• reduction half-reaction

stackrel(color(blue)(0))("Cl"_2) -> 2"H"stackrel(color(blue)(+1))("Cl")"O" + 2e^(-)#

Balance the oxygen atoms by adding water molecules and the hydrogen atoms by adding protons

$2 {\text{H"_2"O" + stackrel(color(blue)(0))("Cl"_2) -> 2"H"stackrel(color(blue)(+1))("Cl")"O" + 2e^(-) + 2"H}}^{+}$

The number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction, which means that you can add these two half-reactions to get

$2 {\text{H"^(+) + "Cl"_2 + 2e^(-) + 2"H"_2"O" + "Cl"_2 -> 2"HCl" + 2e^(-) + "2HClO" + 2"H}}^{+}$

This is reduced to

$\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {\text{H"^(+)))) + "Cl"_2 + color(red)(cancel(color(black)(2e^(-)))) + 2"H"_2"O" + "Cl"_2 -> 2"HCl" + 2"HClO" + color(red)(cancel(color(black)(2e^(-)))) + color(red)(cancel(color(black)(2"H}}^{+}}}}$

$2 \text{Cl"_2 + 2"H"_2"O" -> 2"HClO" + 2"HCl}$

Finally, you have

${\text{Cl"_(2(aq)) + "H"_2"O"_text((l]) -> "HClO"_text((aq]) + "HCl}}_{\textrm{\left(a q\right]}}$