# Question c9eb6

Nov 4, 2015

$6 \text{H"^(+) + "ClO"_3^(-) + 6"e"^(-) -> "Cl"^(-) + 3"H"_2"O}$

#### Explanation:

The first thing to do here is assign oxidation numbers to all the atoms that take part in this half-reaction

stackrel(color(blue)(+5))("Cl") stackrel(color(blue)(-2))("O")_3""^(-) -> stackrel(color(blue)(-1))"Cl""^(-)

Notice that the oxidation state of chlorine goes from $\textcolor{b l u e}{+ 5}$ on the reactants' side, to $\textcolor{b l u e}{- 1}$ on the products' side, which means that it's being reduced.

More specifically, each chlorine atoms gains a total of $6$ electrons.

stackrel(color(blue)(+5))("Cl") "O"_3^(-) + 6"e"^(-) -> stackrel(color(blue)(-1))"Cl""^(-)

Now, in acidic solution you can balance oxygen atoms by adding water molecules and hydrogen atoms by adding protons, ${\text{H}}^{+}$.

Now, notice that your half-reaction has $3$ oxygen atoms on the reactants' side, but none on the products' side.

This means tha tyou will need to add $3$ water molecules on the products' side to balance the oxygen atoms

stackrel(color(blue)(+5))("Cl") "O"_3^(-) + 6"e"^(-) -> stackrel(color(blue)(-1))"Cl"""^(-) + 3"H"_2"O"#

Now you have $6$ hydrogen atoms on the products' side, and none on the reactants' side. This means that you must add $6$ protons on the reactants' side to get

$6 \text{H"^(+) + stackrel(color(blue)(+5))("Cl") "O"_3^(-) + 6"e"^(-) -> stackrel(color(blue)(-1))"Cl"""^(-) + 3"H"_2"O}$

And now the half-reaction is balanced.