Question

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

Answer 1

This is a so-called `"comproportionation reaction"`.

Explanation:

`"Iodic acid"` is reduced to elemental iodine:

`HI^(+V)O_3 + 5H^(+)+5e^(-) rarr 1/2I_2 +3H_2O` `(i)`

`"Iodide"` is oxidized to elemental iodine:

`I^(-) rarr 1/2I_2+e^(-)` `(ii)`

So `(i)+5xx(ii)=`

`HI^(+V)O_3 + 5H^(+)+5I^(-) rarr 3I_2 +3H_2O`

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

Answer 2

WARNING! Long answer! The balanced equation is

`"HIO"_3 + "5HI" → "3I"_2 + "3H"_2"O"`

Explanation:

We start with the unbalanced equation:

`"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 `"I"` in `"HIO"_3` is reduced to 0 in `"I"_2` and the oxidation number if `"I"` in `"HI"` is increased to 0 in `"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:

`"I: +5 → 0";color(white)(l) "Change" =color(white)(l) "-5 (reduction)"`
`"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 `"I"` in `"HIO"_3` for every 5 atoms of `"I"` in `"HI"`.

That also means that we need a total of 6 `"I"` atoms.

Step 3. Insert coefficients to get these numbers

`color(red)(1)"HIO"_3 + color(red)(5)"HI" → color(red)(3)"I"_2 + "H"_2"O"`

Step 4. Balance `"O"`

We have fixed 3 `"O"` atoms on the left, so we need 3 `"O"` atoms on the right. Put a 3 before `"H"_2"O"`.

`color(red)(1)"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.

`bb("On the left"color(white)(l) "On the right")`
`color(white)(mm)"6 H"color(white)(mmmml) "6 H"`
`color(white)(mm)"6 I"color(white)(mmmmll) 6 color(white)(l)"I"`
`color(white)(mm)"3 O"color(white)(mmmml) "3 O"`

The balanced equation is

`color(red)("HIO"_3 + "5HI" → "3I"_2 + "3H"_2"O")`