Question

Question #c9eb6

Answer 1

`6"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 `color(blue)(+5)` on the reactants' side, to `color(blue)(-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, `"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"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.