Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109
General Chemistry
Spring 2009

Lecture Notes:: 13 February
© R. Paselk 2002

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*Humboldt State University* ® *Department of Chemistry*

Richard A. Paselk
Chem 109	General Chemistry	Spring 2009
Lecture Notes:: 13 February	© R. Paselk 2002

PREVIOUS		NEXT

Oxidation/Reduction (Redox) Reactions

In these reactions we see a transfer of electrons from one atom or molecule to another. First let's look at some terms.

Oxidation refers to taking electrons away from a substance. So to oxidize means to behave like oxygen normally does and "steal" electrons.
Reduction refers to recieving electrons - it is the opposite of oxidation.
Note that oxidation and reduction always go together. When oxygen oxidizes it is itself reduced (it gains electrons).
- Consider the example of burning natural gas:

CH₄ + 2 O₂ CO₂ + 2 H₂O

Notice that the methane is oxidized by the oxygen. We say that the carbon and hydrogen are both oxidized to give the new covalent products, water and carbon dioxide, because the electrons are not evenly shared, they are pulled toward the oxygens in each bond.

Examples:

A piece of clean copper is placed in a solution of silver nitrate. Assuming the copper goes to Cu(II) write a balanced net ionic equation. Notice that in this case you must balance the charge in order to get the final equation.

Ag⁺ + NO₃^- + Cu⁰

Ag⁰ + NO₃^- + Cu²⁺

Balancing: 2Ag⁺ + Cu⁰

2Ag⁰ + Cu²⁺

Consider the reaction of calcium metal with hydrochloric acid. This is similar to the reaction of sodium and water, and hydrogen gas is given off. Write a net ionic equation for this reaction:

H⁺ + Cl^- + Ca⁰

H_{2 (g)} + Ca²⁺ + Cl^-

Balancing: 2 H⁺ + Ca⁰

H_{2 (g)} + Ca²⁺

A piece of clean iron is placed in a solution of copper(II) sulfate. Assuming iron goes to Fe(III) write a balanced equation. Again the problem is to balance the charge.

Cu²⁺ + SO₄^2- + Fe⁰

Cu⁰ + Fe³⁺ + SO₄^2-

Balancing: 3 Cu²⁺ + 2 Fe⁰

3 Cu⁰ + 2 Fe³⁺

(Additional examples of solving net-ionic equations can be found in the Discussion Module.)

Balancing Redox Equations

There are two common methods for balancing redox reactions: the oxidation number method and the half-reaction method. The half-reaction method works very well for ionic reactions, it is relatively easy to give partial credit, and it is the only method I will use in this class. If you know how to do the other method you are welcome to do so, but be careful to make sure you show your work or I won't be able to give partial credit!

The Half-Reaction Method. In the half-reaction method what we do is first break an equation into two parts and then balance the parts individually. Presented stepwise:

Acid Solution:

Separate the reaction into two half-reactions.

Balance each half-reaction separately:

Balance atoms other than O & H by inspection.
Balance O by adding H₂O to the opposite side.
Balance H by addding H⁺ as appropriate.
Balance the charge by adding electrons (e^-) - add to same side as excess of positive charge, or opposite side if excess negative charge.
Balance the charges of the two half-reactions by multiplying appropriately.

Add two equations together

Cancel items appearing on both sides.

Example. Balance the following equation as it occurs in acid solution:

MnO₄^- + Cl^-

Mn²⁺ + Cl₂

First break the equation into two half reactions, one for Mn and one for Cl

MnO₄^- Mn²⁺

MnO₄^- Mn²⁺
MnO₄^- Mn²⁺ + 4 H₂O
8 H⁺ + MnO₄^- Mn²⁺ + 4 H₂O
5 e^- + 8 H⁺ + MnO₄^- Mn²⁺ + 4 H₂O
10 e^- + 16 H⁺ + 2 MnO₄^- 2 Mn²⁺ + 8 H₂O

Cl^- Cl₂

2 Cl^- Cl₂
...
...
2 Cl^- Cl₂ + 2 e^-
10 Cl^- 5 Cl₂ + 10 e^-

10 e^- + 16 H⁺ + 2 MnO₈^-+ 10 Cl^-

2 Mn²⁺ + 8 H₂O + 5 Cl₂ + 10 e^-

16 H⁺ + 2 MnO₄^- + 10 Cl^- 2 Mn²⁺ + 8 H₂O + 5 Cl₂

(Additional examples for balancing Redox equations in acidic solution can be found in the Discussion Module.

Basic Solution:

balance as in acid, then:
1. Add enough OH^- (equal numbers to both sides) to cancel the H⁺. (This is necessary because there will not be protons present in a basic solution!). (There are a couple of other conventions for balancing in basic solutions. If you are familiar with another and prefer it, you may use it instead.)
2. Combine the H⁺ and OH^- on the appropriate side of the equation to give waters.
3. Go back and cancel waters which appear on both sides to give the final equation.

Example: Balance the equation above in basic solution (I don't believe this reaction actually occurs in basic solution, but due to our time constraints we'll pretend it does.)

First we balance as above to give:

16 H⁺ + 2 MnO₄^- + 10 Cl^-

2 Mn²⁺ + 8 H₂O + 5 Cl₂

Next add 16 OH^- to each sides to cancel H⁺

16 OH^- + 16 H⁺ + 2 MnO₄^- + 10 Cl^-

2 Mn²⁺ + 8 H₂O + 5 Cl₂ + 16 OH^-

Combine OH^- and H⁺ to give 16 H₂O

16 H₂O + 2 MnO₄^- + 10 Cl^-

2 Mn²⁺ + 8 H₂O + 5 Cl₂ + 16 OH^-

canceling waters then gives the final equation:

8 H₂O + 2 MnO₄^- + 10 Cl^- 2 Mn²⁺ + 5 Cl₂ + 16 OH^-

(Additional examples for balancing Redox equations in acidic solution can be found in the Discussion Module.)

Molarity and the Composition of Solutions

The most commonly used concentration term in chemistry is molarity = 1 mole of solute dissolved in enough solvent to give 1 L = M = mol/L. This is the most popular concentration unit at least in part because of the convenience of making molar solutions with volumetric flasks (show flask).

Note that molarity has various meanings in describing solutions. Commonly molarity refers to the amount of substance dissolved in one liter of liquid to give a solution. Thus if 158.52 g (1 mole) of strontium chloride is dissolved in enough water to give one liter we will have a 1 L solution of strontium chloride. Note however, that while the solution is 1M in Sr²⁺, it is 2 M in Cl^- and 3 M in terms of particles!

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© R A Paselk

Last modified 13 February 2009

Humboldt State University ® Department of Chemistry

Richard A. Paselk

Oxidation/Reduction (Redox) Reactions

Oxidation/Reduction (Redox) Reactions

Balancing Redox Equations

(Additional examples for balancing Redox equations in acidic solution can be found in the Discussion Module.

(Additional examples for balancing Redox equations in acidic solution can be found in the Discussion Module.)

Molarity and the Composition of Solutions