balancing redox equations: many redox equations can be balanced through trial and error! you have...

Balancing Redox Equations:

Many redox equations can be balanced through trial and error! You have been doing this for a least two years without even knowing it! : )

Do the following:

___Al + ____HCl ____ AlCl3 + ___ H2

2 6 2 3

Done!

Balancing Redox Reactions:

Another Example: (spectator ions have not been included)

Sn + Ag+ Sn2+ + Ag

1. Assign oxidation numbers.

0 +1 +2 0

Sn + Ag+ Sn2+ + Ag

2. Oxidation occurs when the oxidation number increases. Reduction occurs when the oxidation number decreases. Write the two half reactions.

Oxidation: Sn Sn2+

Reduction: Ag+ Ag

3. Use electrons to balance the charges in the half reactions. In oxidation the electrons appear on the right. In reduction the electrons appear on the left.

Oxidation: Sn Sn2+ + 2e-

Reduction: Ag+ + 1e- Ag

4. If the number of electrons transferred is not equal multiple by a whole number so that the number of electrons lost equals the number gained.

Oxidation: Sn Sn2+ + 2e-

Reduction: (Ag+ + 1e- Ag) x2

5. Add the half reactions:

Oxidation: Sn Sn2+ + 2e-

Reduction: 2Ag+ + 2e- 2Ag

Net Balanced Redox Reaction (notice that the electrons cancel!)

Sn + 2 Ag+ Sn2+ + 2Ag

Steps for Balancing More:

Step 1: Identify and write the two half-reactions.

Step 2: Balance the elements and charges for each half-reaction: 2 a) Balance all the elements except hydrogen and oxygen.

2b) Balance the oxygen atoms by adding H2O to the appropriate side.

2c) Balance the hydrogen by adding H+ to the appropriate side.

2d) Balance the charges by adding electrons to the appropriate side.

Step 3: Multiply one or both of the half-reactions by a whole number so that the number of electrons gained and lost is equal.

Step 4: Combine the half-reactions. Eliminate anything common to the product and reactant sides.

Step 5: Check that all of the elements are balanced and that the total charge on each side is the same.

Example: Balance: MnO4

- + Fe2+ Mn2+ + Fe3+

Step 1:

Reduction: MnO4- Mn2+ Oxidation: Fe2+ Fe3+

+7, -2 + 2 +2 +3

Step 2:

b) MnO4- Mn2+ + 4H2O

c) MnO4- + 8H+ Mn2+ + 4H2O

d) MnO4- + 8H+ + 5e- Mn2+ + 4H2O

Oxidation:

Fe2+ Fe3+ +1e-

Step 3:

MnO4- + 8H+ + 5e- Mn2+ + 4H2O

(Fe2+ Fe3+ +1e-) x5

Step 4:

Reduction: MnO4- + 8H+ + 5e- Mn2+ + 4H2O

Oxidation: 5Fe2+ 5 Fe3+ +5e-

Net Reaction: MnO4- + 8H+ +5Fe2+ Mn2+ + 4H2O + 5 Fe3+

(electrons are cancelled!)

Step 5:

Notice that the final redox equation is balanced by atom and by charge.

Redox reactions are usually too complex to use trial and error method.

Balance the following example that is in an acidic solution (assume the presence of H2O and H+):

HNO3 + Fe2+ Fe3+ + NO2

1. Assign oxidation numbers.

+1+5-2 +2 +3 +4 -2

HNO3 + Fe2+ Fe3+ + NO2

2. Write the half reactions:

Oxidation: Fe 2+ Fe3+

Reduction: HNO3 NO2

3. Balance the half reactions. Use water to balance the oxygen (a), then hydrogen ions to balance the hydrogen(b), then electrons to balance the charges(c).

a)

Oxidation: Fe 2+ Fe3+

Reduction: HNO3 NO2 + H2O

b)

Oxidation: Fe 2+ Fe3+

Reduction: HNO3 + H+ NO2 + H2O

c)

Oxidation: Fe 2+ Fe3+ + 1e-

Reduction: HNO3 + H+ + 1e- NO2 + H2O

4. Multiply the oxidation and reduction equations by whole numbers so that the number of electrons transferred is equal.

Oxidation: Fe 2+ Fe3+ + 1e-

Reduction: HNO3 + H+ + 1e- NO2 + H2O

5. Add the reactions so that electrons cancel.

Net: Fe 2++ HNO3 + H+ Fe3+ + NO2 + H2O

6. Check the equation is balance by charge and atom.

Videos to review the concept:

Watch the following videos to help you balance redox equations:

http://www.youtube.com/watch?v=KIyGr-1snMY&feature=relmfu

http://www.youtube.com/watch?v=-B3RWeC_7oI

http://www.youtube.com/watch?v=TBmwhTzc41o&feature=fvwrel

Chapter 20.3- Read pages 645 654

Page 647 #17 & #18

Page 649 #19 & #20

Page 652 #21

Page 654 #22,23,24 & 25

CHAPTER 20

Review the study guide on page 656

Complete the odd numbered questions on page 658

Complete the Standardized Test Prep

Worksheet Redox - 3