Question

Question #92bc1

Answer 1

`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"`