# How does "iodic acid", HIO_3, react with iodide anion, I^-, to give elemental iodine?

Mar 3, 2017

This is a so-called $\text{comproportionation reaction}$.

#### Explanation:

$\text{Iodic acid}$ is reduced to elemental iodine:

$H {I}^{+ V} {O}_{3} + 5 {H}^{+} + 5 {e}^{-} \rightarrow \frac{1}{2} {I}_{2} + 3 {H}_{2} O$ $\left(i\right)$

$\text{Iodide}$ is oxidized to elemental iodine:

${I}^{-} \rightarrow \frac{1}{2} {I}_{2} + {e}^{-}$ $\left(i i\right)$

So $\left(i\right) + 5 \times \left(i i\right) =$

$H {I}^{+ V} {O}_{3} + 5 {H}^{+} + 5 {I}^{-} \rightarrow 3 {I}_{2} + 3 {H}_{2} O$

This is AFAIK balanced with respect to mass and charge, and therefore a reasonable representation of chemical reality.

Mar 4, 2017

WARNING! Long answer! The balanced equation is

$\text{HIO"_3 + "5HI" → "3I"_2 + "3H"_2"O}$

#### Explanation:

$\text{HIO"_3 + "HI" → "I"_2 + "H"_2"O}$

Step 1. Identify the atoms that change oxidation number

We start by determining the oxidation numbers of every atom in the equation.

stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("I")stackrelcolor(blue)("-2")("O")_3 + stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("-1")("I") → stackrelcolor(blue)(0)("I")_2 + stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O")
color(white)(stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("I")stackrelcolor(blue)("-6")("O")_3 + stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("-1")("I") → stackrelcolor(blue)(0)("I")_2 + stackrelcolor(blue)("+2")("H")_2stackrelcolor(blue)("-2")("O"))

We see that the oxidation number of $\text{I}$ in ${\text{HIO}}_{3}$ is reduced to 0 in ${\text{I}}_{2}$ and the oxidation number if $\text{I}$ in $\text{HI}$ is increased to 0 in ${\text{I}}_{2}$.

This is a comproportionation reaction, a reaction in which an element in a higher oxidation state reacts with the same element in a lower oxidation state to give the element in an intermediate oxidation state.

The changes in oxidation number are:

$\text{I: +5 → 0";color(white)(l) "Change" =color(white)(l) "-5 (reduction)}$
$\text{I: -1 → 0"; color(white)(ll)"Change ="color(white)(l) "+1 (oxidation)}$

Step 2. Equalize the changes in oxidation number

We need 1 atom of $\text{I}$ in ${\text{HIO}}_{3}$ for every 5 atoms of $\text{I}$ in $\text{HI}$.

That also means that we need a total of 6 $\text{I}$ atoms.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{1} \text{HIO"_3 + color(red)(5)"HI" → color(red)(3)"I"_2 + "H"_2"O}$

Step 4. Balance $\text{O}$

We have fixed 3 $\text{O}$ atoms on the left, so we need 3 $\text{O}$ atoms on the right. Put a 3 before $\text{H"_2"O}$.

$\textcolor{red}{1} \text{HIO"_3 + color(red)(5)"HI" → color(red)(3)"I"_2 + color(blue)(3)"H"_2"O}$

Every formula now has a coefficient. The equation should be balanced.

Step 7. Check that all atoms are balanced.

$\boldsymbol{\text{On the left"color(white)(l) "On the right}}$
$\textcolor{w h i t e}{m m} \text{6 H"color(white)(mmmml) "6 H}$
$\textcolor{w h i t e}{m m} \text{6 I"color(white)(mmmmll) 6 color(white)(l)"I}$
$\textcolor{w h i t e}{m m} \text{3 O"color(white)(mmmml) "3 O}$

The balanced equation is

$\textcolor{red}{\text{HIO"_3 + "5HI" → "3I"_2 + "3H"_2"O}}$