Question #92bc1

1 Answer
Nov 29, 2015

Answer:

#6"H"^(+) + "I"_2"O"_7 + 8"e"^(-) -> 2"IO"_2^(-) + 3"H"_2"O"#

Explanation:

Start by assigning oxidation numbers to the atoms that take part in the half-reaction

#stackrel(color(blue)(+7))("I"_2) stackrel(color(blue)(-2))("O"_7) -> stackrel(color(blue)(+3))("I") stackrel(color(blue)(-2))("O"_2^(-))#

You're dealing with a reduction half-reaction in which iodine's oxidation state changes from #color(blue)(+7)# on the reactants' side, to #color(blue)(+3)# on the products' side.

The problem tells you to note the charges because you need to make sure that once you balance the half-reaction, the overall charge on the reactants' side must be equal to the overall charge on the products' side.

So, start by balancing the iodine atoms

#stackrel(color(blue)(+7))("I"_2)"O"_7 -> 2stackrel(color(blue)(+3))("I")"O"_2^(-)#

Now, each iodine atom will gain #4# electrons (#color(blue)(+7) -> color(blue)(+3)#), which means that two iodine atoms will gain a total of #8# electrons.

#stackrel(color(blue)(+7))("I"_2)"O"_7 + 8"e"^(-) -> 2stackrel(color(blue)(+3))("I")"O"_2^(-)#

Since you're in acidic solution, you can balance the oxygen atoms by adding water molecules to the side that needs oxygen, and the hydrogen atoms by adding protons, #"H"^(+)#, to the side that needs hydrogen.

In this case, you have #7# atoms of oxygen on the reactants' side, but only #4# on the products' side. Add #3# molecules of water to get

#stackrel(color(blue)(+7))("I"_2)"O"_7 + 8"e"^(-) -> 2stackrel(color(blue)(+3))("I")"O"_2^(-) + 3"H"_2"O"#

To balance the hydrogen atoms, add #6# protons to the reactants' side

#6"H"^(+) + stackrel(color(blue)(+7))("I"_2)"O"_7 + 8"e"^(-) -> 2stackrel(color(blue)(+3))("I")"O"_2^(-)#

Now make sure that the charge is balanced. On the reactants' side, you have #6# protons and #8# electrons, which is equivalent to a total net charge of #(-2)#.

On the products' side, you have #2# iodite anions, #"IO"_2^(-)#, for a total net charge of #(-2)#.

The balanced half-reaction will thus be

#6"H"^(+) + "I"_2"O"_7 + 8"e"^(-) -> 2"IO"_2^(-) + 3"H"_2"O"#