Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109	General Chemistry	Summer 2002
Lecture Notes:: 18 June	© R. Paselk 2002
PREVIOUS		NEXT

Aqueous Solutions, cont.

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:
  1. Balance atoms other than O & H by inspection.
  2. Balance O by adding H₂O to the opposite side.
  3. Balance H by addding H⁺ as appropriate.
  4. Balance the charge by adding electrons (e^-) - add to same side as excess of positive charge, or opposite side if excess negative charge.
  5. 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^- Æ Mn²⁺ + Cl₂

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

MnO4- Æ Mn2+ MnO4- Æ Mn2+ MnO4- Æ Mn2+ + 4 H2O 8 H+ + MnO4- Æ Mn2+ + 4 H2O 5 e- + 8 H+ + MnO4- Æ Mn2+ + 4 H2O 10 e- + 16 H+ + 2 MnO8- Æ 2 Mn2+ + 8 H2O Cl- Æ Cl2 2 Cl- Æ Cl2 ... ... 2 Cl- Æ Cl2 + 2 e- 10 Cl- Æ 5 Cl2 + 10 e-

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

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

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:

Gases

Gases: Briefly discussed overall properties of gases (fills container, compressible, lo density, lo viscosity).

• Gas particles exert pressure.
  • implication - gas particles have momentum (mass and velocity).

What is Pressure? Pressure is the force/unit area. Due to collisions of particle with walls of container etc.

Units of Pressure:

• mmHg - based on manometers. Two types:
  • open tube - measures pressure relative to current atmospheric pressure.
  • closed tube - measures pressure relative to contents of the enclosed volume at the closed end. (A barometer is an example where the enclosed space is "empty", that is it contains a vacuum since the vapor pressure of mercury is very low.)

• atm = 760 mmHg at 1 gravity = 1.01 Bar
• others include: psi (pounds/square inch), pascal, Torr (= 1 mmHg), millibar, etc.

Gases are characterized by four properties

• Amount of substance, n  (in moles)
• Volume, V (in Liters)
• Pressure, P (in atm, though often measured in mmHg)
• Temperature (in K)

Gas Laws

Gas Laws describe the relationships between the four properties characterizing any gas:

• Amount of substance, n  (in moles)
• Volume, V (in Liters)
• Pressure, P (in atm, though often measured in mmHg)
• Temperature (in K)

Boyle's Law: Boyle's Law describes the relationship between pressure and volume when the temperature and amount of substance are held constant.

Plotting pressure volume data (keeping n and T constant) gives a graph for a hyperbola (xy = c), as seen below:

Notice that we can rearrange this equation to give a straight-line relationship:

Thus, "At constant temperature the volume of any quantity of gas is inversely proportional to its pressure." V = k (1/P) & P₁V₁ = P₂V₂.

(Aside on straight-line plots: Very popular in science. Traditionally, we will do almost anything to get a straight line. Why? Because straight lines easy to recognize and evaluate. Also easy to evaluate statistically.)

Charles' Law: The relationship between volume and temperature was determined much later because accurate thermometers had to be developed first. But once thermometers were available a number of workers determined that volume is directly proportional to temperature. Plotting data for the relation of volume of a gas to temperature between 0° C and 100 ° C gives a plot similar to that below:

Extrapolating this data to V = 0 we can find an absolute minimum value of temperature on the assumption that negative volumes can't exist:

The intercept on the volume axis is then taken as absolute zero = -273.15 °C = 0 K for an ideal or "perfect" gas with particles of zero volume and no interactions other than collisions.

Algebraically we then find that V = k'T, & V₁/T₁ = V₂/T₂.

Combined Gas Law: We can combine Boyle's and Charles' relationships (T was part of the constant for Boyle's Law and P is part of the constant for Charles' Law) to give:

Avogadro's Law: V = an, where n = moles of stuff. So we have a linear relation between volume and moles; and V₁/n₁ = V₂/n₂.

Ideal Gas Law

Ideal Gas Law ("Perfect Gas Law"): The constant for the combined law includes amount of stuff, and breaking that out we then get

I will base all of my examples on this equation because that requires a minimum of memorization. However you may find it easier to memorize a series of equations such as the "combined gas law equation" etc.

As an example, let's find the molar volume of a gas under standard conditions of temperature and pressure (STP). STP are defined as: P = 1 atm and T = 0° C. Thus we need to solve the gas equation for 1 mole of gas at 273.15 K (= 0°C+ 273.15) and 1 atm:

Syllabus / Schedule

C109 Home

C109 Lecture Notes

© R A Paselk

Last modified 18 June 2002