Chapter 22 Notes II

Chapter 22 Notes IIBalancing Redox

Reactions

Steps to balancing a Redox Steps to balancing a Redox reaction:reaction:Lets balance the following

equation to show the steps:

HNO3 + I2 HIO3 + NO2 + H2O

1) Write the oxidation number 1) Write the oxidation number above each element in the above each element in the equation.equation.

HNO3 + I2 HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2 +4 -2 +1 -2

2) Draw a line connecting 2) Draw a line connecting elements whose oxidation elements whose oxidation numbers change.numbers change.

HNO3 + I2 HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2 +4 -2 +1 -2

3) Determine how much the oxidation 3) Determine how much the oxidation number changed for each and write number changed for each and write this number on the connecting line.this number on the connecting line.

HNO3 + I2 HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2 +4 -2 +1 -2

+1e-

-5e-

The positive in front of the 1 means it is gaining The positive in front of the 1 means it is gaining 1 electron and the negative in front of the 5 1 electron and the negative in front of the 5

means it is losing electrons.means it is losing electrons.

4) Multiply the number on the 4) Multiply the number on the connecting line by the highest connecting line by the highest subscript on the element changing.subscript on the element changing.

HNO3 + I2 HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2 +4 -2 +1 -2

+1e-+1e-

-5e-

x 1

x 2

5) Multiply the top and bottom line by 5) Multiply the top and bottom line by numbers so that the electrons numbers so that the electrons gained equal electrons lost.gained equal electrons lost.

HNO3 + I2 HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2 +4 -2 +1 -2

+1e-+1e-

-5e-

x 1

x 2

Right now, the top line multiplies to equal 1 and the bottom line multiplies to equal 10. What is the

least common multiple of 1 and 10?

5) Multiply the top and bottom line by 5) Multiply the top and bottom line by numbers so that the electrons numbers so that the electrons gained equal electrons lost.gained equal electrons lost.

HNO3 + I2 HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2 +4 -2 +1 -2

+1e-+1e-

-5e-

x 1

x 2

10!So we’ll need to multiply each line by a number

that makes it equal to ten (the two numbers don’t have to be the same).

x 10

x 1

6) Circle the last two numbers on each line. 6) Circle the last two numbers on each line. The product represents how many The product represents how many atoms of that element you need on each atoms of that element you need on each side.side.

HNO3 + I2 HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2 +4 -2 +1 -2

+1e-+1e-

-5e-

x 1

x 2

x 10

x 1

So we’ll need 10 nitrogens and 2 iodines on each side of the equation.

=2

=10

7) Place the coefficients in the 7) Place the coefficients in the equation which will give you the equation which will give you the determined number of atoms.determined number of atoms.

HNO3 + I2 HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2 +4 -2 +1 -2

+1e-+1e-

-5e-

x 1

x 2

x 10

x 1

Since there is already a two subscript on I2, it gets a one coefficient to bring the total number to two.

=2

=10

1010 2

8) Check all atoms to see if 8) Check all atoms to see if everything is balanced. If hydrogen everything is balanced. If hydrogen and oxygen are both unbalanced, and oxygen are both unbalanced, try hydrogen first.try hydrogen first.

HNO3 + I2 HIO3 + NO2 + H2O

Make sure not to disturb the ratio of elements oxidized to elements reduced that you just set up!

1010 2 4

Balancing the following using the steps:

I2 + HClO + H2O HIO3 + HCl

5 52