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

2 Answers
Feb 4, 2018

Answer:

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....

Feb 4, 2018

Answer:

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")#