Question #c9eb6

1 Answer
Nov 4, 2015

Answer:

#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.