Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110	General Chemistry	Fall 2003
Lecture Notes::Lec 12_22 September	© R. Paselk 2003
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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.

Review terms: Look briefly at some terms used in 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:

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^- 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^-

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 in 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?

• 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 o 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:

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? Separate the reaction into its two half-reactions by putting the components of each half reaction into its own container. Thus we 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.

Before we go further, let's take a look at the units and instruments for measuring electricity.

Measuring Electricity

Electrical units. There are four different units we need to be familiar with:

1. Volt (J/C) - This is the unit of electrical potential difference. It is analogous to height in a gravitational field. It is the potential difference needed for the flow of one coulomb (C) of charge to produce one joule (J) of work.
2. Coulomb (As) - This is the unit of electrical charge. There are 9.65 x 10⁴ coulombs (one Faraday's constant) in one mole of electrons.
3. Amp (A) - This is the unit of electrical current. It is the SI base unit for electricity (all other electrical units may be constructed from the amp and other SI base units). An Amp is equal to the flow of one coulomb of electrons in one second.
4. Ohm (W) - This is the unit of electrical resistance. It is defined as the amount of resistance which will require one volt of potential to give a current of one amp.

Measuring devices. Electrical "meters" are commonly used to measure volts and amps.

• The difficulty in measuring voltage is that any flow of current during the measurement process will lower the voltage!
  • Since the common mechanical meter (a D'Arsonval meter) relies on the production of a force due to a flow of current though a coil in a magnetic field, they generally do not give a high accuracy measurement. The quality of such meters is commonly given as xxx W/V, since the greater the resistance the lower the current flow and thus the more accurate the meter.
  • For much of the past century or so the best voltages were thus measured with a potentiometer and null meter arrangement in which measurements were made under conditions where no current flows. These devices give very precise voltages, but are inconvenient and require trained users.
  • Fortunately, modern electronics have enables the creation of volt meters based on solid state devices (such as FETs) that have input resistances of millions of ohms and greater. Digital meters can thus give very accurate voltages with virtually no current loss.

If we now connect the two metal strips in our set-up with a wire with a meter on it what will occur? If we watch very carefully with a sensitive meter we should see 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.

© R A Paselk

Last modified 23 September 2003