Question

How do you balance redox equations using the oxidation number change method?

For example:
`ClO_3^(-) (aq) + I^(-) (aq) -> Cl^(-) (aq) + I_2 (aq)` [acidic solution

Answer 1

You introduce electrons as virtual particles and assign oxidation states...according to given rules.

Explanation:

Chlorate is REDUCED to chloride....`Cl(+V)rarrCl(-I)`..and we account for the difference in oxidation number `+V-(-I)=6` by the addition of electrons....

`ClO_3^(-) +6H^+ + underbrace(6e^(-))_"i.e. +V/-I" rarr Cl^(-) +3H_2O(l)` `(i)`

Iodide is oxidized to iodine...`I(-I)rarrI_2(0)`

`I^(-) rarr 1/2I_2+underbrace(e^(-))_"-I/0"` `(ii)`

In each case we account for the difference in oxidation number by the addition (reduction) or subtraction (oxidation) of electrons....

We take `(i)+6xx(ii):`

`ClO_3^(-) + 6HI(aq) +6e^(-)rarr3I_2(s) +6e^(-)+Cl^(-)+3H_2O`

....to give finally after cancellation...

`ClO_3^(-) + 6HI(aq) rarr3I_2(s) +Cl^(-)+3H_2O`

The which is balanced with respect to mass and charge....as indeed it must be if we purport to represent chemical reality....

Answer 2

WARNING! Long answer! The balanced equation is

`"ClO"_3^"-" + "6I"^"-" + "6H"^"+" → "Cl"^"-" + "3I"_2 + "3H"_2"O"`

Explanation:

The unbalanced equation is

`"ClO"_3^"-" + "I"^"-" → "Cl"^"-" + "I"_2`

Step 1. Identify the atoms that change oxidation number

Start by determining the oxidation numbers of every atom in the equation.

Ignore `"O"`. It comes in automatically during the balancing procedure.

`stackrelcolor(blue)("+5")("Cl")"O"_3^"-" + "I"^"-" → "Cl"^"-" + stackrelcolor(blue)(0)("I")_2`

We see that the oxidation number of `"Cl"` decreases from +5 to -1 and the oxidation number of `"I"` increases from -1 to 0.

The changes in oxidation number are:

`"Cl: +5 → -1"; "Change ="color(white)(m) "-6 (reducttion)"`
`"I: -1 → 0"; color(white)(mll)"Change ="color(white)(ll) "+1 (oxidation)"`

Step 2. Equalize the changes in oxidation number

We need 6 atoms of `"I"` for every 1 atom of `"Cl"`.

Step 3. Insert coefficients to get these numbers

`color(red)(1)"ClO"_3^"-" + color(red)(6)"I"^"-" → color(red)(1)"Cl"^"-" + color(red)(3)"I"_2`

Step 4. Balance `"O"`

Add `"H"_2"O"` molecules to the deficient side.

`color(red)(1)"ClO"_3^"-" + color(red)(6)"I"^"-" → color(red)(1)"Cl"^"-" + color(red)(3)"I"_2 + color(blue)(3)"H"_2"O"`

Step 5. Balance `"H"`

Add `"H"^"+"` ions to the deficient side.

`color(red)(1)"ClO"_3^"-" + color(red)(6)"I"^"-" + color(brown)(6)"H"^"+" → color(red)(1)"Cl"^"-" + color(red)(3)"I"_2 + color(blue)(3)"H"_2"O"`

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

Step 6. Check that all atoms are balanced.

`ulbb("Atom"color(white)(m)"On the left"color(white)(m) "On the right")`
`color(white)(m)"Cl"color(white)(mmmmm)1color(white)(mmmmmmll) 1`
`color(white)(m)"O"color(white)(mmmmml)3color(white)(mmmmmmll) 3`
`color(white)(m)"I"color(white)(mmmmmll)6color(white)(mmmmmmll) 6`
`color(white)(m)"H"color(white)(mmmmml)6color(white)(mmmmmmll) 6`

Step 7. Check that charge is balanced.

`ulbb("On the left"color(white)(m) "On the right")`
`color(white)(mmm)"1-"color(white)(mmmmmm)"1-"`

The balanced equation is

`color(blue)("ClO"_3^"-" + "6I"^"-" + "6H"^"+" → "Cl"^"-" + "3I"_2 + "3H"_2"O")`