# Question 92bc1

Nov 29, 2015

$6 \text{H"^(+) + "I"_2"O"_7 + 8"e"^(-) -> 2"IO"_2^(-) + 3"H"_2"O}$

#### Explanation:

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

$\stackrel{\textcolor{b l u e}{+ 7}}{{\text{I"_2) stackrel(color(blue)(-2))("O"_7) -> stackrel(color(blue)(+3))("I") stackrel(color(blue)(-2))("O}}_{2}^{-}}$

You're dealing with a reduction half-reaction in which iodine's oxidation state changes from $\textcolor{b l u e}{+ 7}$ on the reactants' side, to $\textcolor{b l u e}{+ 3}$ 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 ($\textcolor{b l u e}{+ 7} \to \textcolor{b l u e}{+ 3}$), 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, ${\text{H}}^{+}$, 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

$6 {\text{H"^(+) + stackrel(color(blue)(+7))("I"_2)"O"_7 + 8"e"^(-) -> 2stackrel(color(blue)(+3))("I")"O}}_{2}^{-}$

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 $\left(- 2\right)$.

On the products' side, you have $2$ iodite anions, ${\text{IO}}_{2}^{-}$, for a total net charge of $\left(- 2\right)$.

The balanced half-reaction will thus be

$6 \text{H"^(+) + "I"_2"O"_7 + 8"e"^(-) -> 2"IO"_2^(-) + 3"H"_2"O}$