Balancing Redox Reactions

Oxidation number method

Half reaction method

Fe₂O_3(s)+CO_(g)—> Fe_(s)+CO_2(g)

Find what is being oxidized

Find what is being reduced

Find the oxidation numbers for the elements

Make sure that the numbers are balanced

Make a new balanced equation using your numbers from the oxidation and reduction

+3 -2 +2 -2 0 +4 -2

Fe₂O_(s)+CO_(g)—> Fe_(s)+CO_2(g)

Carbon is being oxidized +2(3) = 6

Iron is being reduced -3(2) = -6

For CO is carbon monoxide

+3 -2 +2 -2 0 +4 -2

Fe₂O_3(s)(3)CO_(g) —> 2Fe_(s) + (3)CO₂

Fe₂O_3(s) + 3Co_(g) —> 2Fe_(s) + 3Co_2(g)

When balancing equations:

1. Switch the signs for reduction and oxidation for the reactions

2. Add the number reaction to the products

Ex:
+3 -2 +2 -2 0 +4 -2

Fe₂O_3(s)(3)CO_(g) —> 2Fe_(s) + (3)CO₂

4 = product

+ -2 = reactant (reversed sign)

________

+2 Carbon

0 = product

+ -3 = reactants (reversed sign)

______

-3 = Iron

Find the common multiple of 2 and -3, and make sure they add to 0

2(3)=6

-3(2)=-6

6-6=0

Go back and add the 3 and 2 values to the equation.

Multiply 3 by the carbon and multiply 2 by the iron to even out the oxygen

Don’t multiply 2 to the Fe₂O₃ because it’s already balanced.

Fe₂O_3(s)+(3)CO_(g)—> (2)Fe_(s)+(3)CO_2(g)

KCIO_3(s) —> KCI_(s)+O_2(g)

+1 +5 -2 +1 -1 0

K CI O_3(s) —> KCI_(s)+O_2(g)

CI is being reduced -6(1)

Oxygen is being oxidized +2(3) = 6

2KCIO_3(s)—>2KCI_(s)+3O_2(g)

For this, instead of adding 1 to CI, add 2 to balance it out.

HNO_2(aq)+HI_(aq)—>NO_(g)+I_2(s)+H₂O_(l)

+1 +3 -2 +1 -1 +2 -2 0 +1 -2
HNO_2(aq)+ HI_(aq)—>NO_(g)+I_2(s)+H₂O_(l)

Iodine is being oxidized +1 (0 + (+1)

Nitrogen is being reduced -1 (+2 + (-3)

Remember Oxygen’s charge in a compound is always -2

And since the common multiple of the equation is 1, you don’t have to multiply it by anything.

This one is pretty hard…

KMnO_4(aq)+HCI_(aq)—>MnCI_2(aq)+CI_2(g)+H₂O_(l)+KCI_(aq)

K⁺_(aq)+MnO^-_4(aq) + H⁺_(aq) + CI^-_(aq) —> Mn ³⁺_(aq) + 2CI^-_(aq) + CI_2(g) + H₂O_(l) + K⁺_(aq) + O^-_(aq)

_{-1 0}

CI^- —> Cl₂

_{+7 +2}

MnO₄^- —> Mn²⁺

5(2CI^- —→ CI₂+2e^-)

2(5e^-+MnO₄^- +8H⁺ —> Mn²⁺ + 4H₂O)

10CI^- —> 5CI₂+10e^-

2MnO₄^- + 16H⁺ +10e^- —> 2Mn²⁺+8H₂O

10Cr + 2MnO₄^- +16H⁺ —> 5CI₂+2Mn²⁺ +8H₂O +6CI^- +2K⁺ + 6CI^- +2K⁺

2kMaO_4(aq)+16HCI_(aq)—>2MnCI_2(aq) + 5 (I₂+8H₂O+2KCI_(aq))