Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110	General Chemistry	Summer 2006
Lecture Notes::Lec 8_8 June	© R. Paselk 2006
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K_eq and Temperature

In Chem 109 we used le Chatelier's Principle to predict the effects of changes in temperature on reaction equilibria. Thus, for example, for an exothermic reaction, if T increases heat must have been added, therefore the reaction will use up some of the extra heat, driving the reaction to the left and lowering the temperature back toward its original value. With our new thermodynamic knowledge we can determine this relationship quantitatively.

and

Substituting we can say

dividing by - RT:

Assuming deltaH° and deltaS° are constants (a reasonable approximation if deltaT is not large) this is now in the form of the equation of a straight line:

Plotting:

So we can find both deltaH and deltaS (as seen in the lab).

We can also find these values non-graphically by looking at a system at two different temperatures and doing a bit of algebra:

Subtracting (2) from (1) we get:

And using log rules:

Another common form of the van't Hoff Equation is:

Free Energy and Work

Reversible vs. irreversible processes. In order for a process to be completely efficient we say it must be reversible - that is we can get back to the starting point without energy loss. At the microscopic level (atoms and molecules or smaller) processes are frequently reversible. However at the macroscopic level there are inevitably heat losses due to friction etc.

For real processes a further complication arises in that the only energy available is the Free Energy, so we measure the efficiency of a process by how much of the free energy (deltaG) is captured.

As state above, in real processes some energy is lost to the environment as heat. Folks often lament this loss because it appears to be wasted - wouldn't it be better to capture all the deltaG to do work? If so why hasn't nature done so? After all life on Earth has had somewhere around 3.5 x 10⁹ years to get it right, and glycolysis still "wastes" about 40% of the available deltaG. Why? Unfortunately in order to capture the maximum amount of deltaG we have to accomplish things in very small equilibrium steps. Very highly simplified, if you're very efficient, its hard to generate enough working energy, and you get eaten! For human designed processes you have to be somewhat inefficient to get anything done during your lifetime.

So like many things we have to compromise and try to optimize the process between speed and completion versus efficiency. The challenge is to reach this optimum, which generally changes with circumstance -life's tough!

Introduction to Electrochemistry

Why do we care? Examples of electrochemistry all around us (batteries, plating, corrosion, fuel cells, manufacture etc.)

Electrochemistry is the interchange of electrical and chemical energy.

First let's review Redox chemistry.

Terms: Look briefly at some terms used in redox/electrochemistry.

• Oxidation: electrons are removed (note oxidation is what oxygen does).
• Reduction: electrons are gained (reduce ores to metals).
• Oxidizing agent/oxidant: takes electrons in reaction (acts like oxygen).
• Reducing agent/reductant: provides electrons in chemical reaction.

Note that in any chemical redox reaction that oxidation and reduction are always coupled getting an exchange of status:

oxidant₁ + reductant₁ reductant₂ + oxidant₂

Balancing Half-Reactions. In electrochemistry we separate a redox reaction physically into half-reactions. Thus we need to be able to separate a redox reaction into half-reactions and to balance them, as we did in Chem 109. So let's review redox balancing by 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. We will just review the method for reactions in acid solution. But remember the same method works for basic solutions as well with a few additional steps.

Presented step wise for 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 adding 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^- Mn²⁺ + Cl₂

First break the equation into two half reactions, one for Mn and one for Cl, then follow the steps above for each:

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 MnO8- 2 Mn2+ + 8 H2O Cl- Cl2 2 Cl- Cl2 ... ... 2 Cl- Cl2 + 2 e- 10 Cl- 5 Cl2 + 10 e-

Finally, combine half reactions and cancel terms:

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

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

Consider a simple Redox system consisting of a beaker full of copper sulfate and a zinc strip as an example:

When we place the zinc strip into the Cu²⁺ solution we will observe a gradual darkening of the zinc strip while the solution gradually goes from light blue to colorless. (you may recall doing this this reaction in the Ionic Reactions lab in Chem 109).

So what's going on?

• The loss of the blue color indicates that the Cu²⁺ is being lost from the solution, being converted to Cu metal.
• Of course something must replace the Cu²⁺ and its charge. There are no bubbles or other indications of solution reactions, so the most likely reactant is the zinc going to Zn²⁺.
  • This is reinforced by the appearance of the black coating on the zinc strip where it contacts the solution. Since finely divided metals in general are black, this coating is apparently the deposited copper metal, picking up electrons from zinc as it is oxidized on the surface.
• Thus we postulate the following reactions to explain the phenomena:

Cu²⁺ + 2 e^- Cu

Zn Zn²⁺ + 2 e^-

Cu²⁺ + Zn Zn²⁺ + Cu

Galvanic Cells

What we'd really like to do is capture the energy of this redox reaction as a flow of electrons. After all our half-reactions tell us electrons are being exchanged. So how do we do this? We can separate the reaction into its two half-reactions by putting the components of each half reaction into its own container. We then have:

1. A beaker filled with Zn²⁺ (for example, 1 M zinc sulfate solution) with a zinc strip immersed in the solution.
2. A beaker filled with Cu²⁺ (for example, 1 M copper sulfate solution) with a copper strip immersed in the solution.

If we now connect the two metal strips in our set-up with a wire with a meter on it what will occur?

Let's go back to our two containers each with its own half-reaction and connect them with a wire. What happens? If we watch very carefully with a sensitive current meter we should see a pulse of electricity followed by a return to zero.

Why does the current not continue? The problem is that the charges in the two beakers quickly build up until the free energy of the redox reaction is not sufficient to over come the work needed to move the charges against the potential gradients.

So what can we do to allow the flow of electrons to continue? We need to connect the solutions in the beakers so that the charges can be neutralized with a counter flow of ions. The connecting ionic fluid is referred to as a salt-bridge, as seen in the figure. Other arrangements are possible such as semi-permeable membranes etc. Such an arrangement is called a Galvanic cell.

Galvanic cells (or Voltaic cells) are cells in which the overall redox reactions occur spontaneously (equilibrium favors products) as written. They can serve as a source of electric power (as a battery).

© R A Paselk

Last modified 8 June 2006