Mr. Bouyer
Day 1 | Day 2 - 3 | Day 4 | Lab | Skills Test

  • Identify the cathode and the anode in an electrochemical cell.
  • Properly use the shorthand representation for electrochemical cells.
  • Use the Standard Reduction Potential Table to calculate the voltage of an electrochemical cell.
  • Use voltabe calculations to predict the products of electrolysis.
  • Use Faraday's Laws to calculate the mass of products formed during electrolysis.

Electric Current: the movement of charged particles.

Metals are generally excellent conductors of electricity because the electrons of their atoms are delocalized. If the potential energy of the electrons in the metal is raised, these electrons will flow toward a point where their potential energy is lower. This movement of electrons in metals is known as metallic conduction.

Substances that produce ions in solution are electrolytes. Ions with a positive charge are called cations and ions with a negative charge are called anions. Ions in solution will move from an area of high energy toward an area of low energy producing, a current. This movement of ions in solution is known as electrolytic conduction.

If the electrodes of a battery are placed in an ionic solution, a current flows between the electrodes just as the movement of electrons in a metal results in a current. This produces a Redox reaction at each electrode by a process known as electrolysis.

The electrolytic device used for electrolysis has three essential parts:

A source of direct current to provide the energy for the reaction to occur.
Two electrodes - the cathode and the anode. The cathode provides a surface where contact can be made between electrons from the power source and the reactant in solution that is to be reduced. The cathode has a negative charge and attracts the cations in the electrolyte. The anode provides a surface where the reactant being oxidized can deposit the electrons it loses. The anode has a positive charge and attracts the anions in the electrolyte.
An electrolyte. This is the substance in a dissolved or liquid phase that permits ions to move between the electrodes, which balances the flow of electrons in the external circuit.

To remember which Redox half-reaction occurs at which electrode:

Oxidation and Anode both begin with vowles.
Reduction and Cathode both begin with consonants.

This is the same for both electrolytic devices and electrochemical cells.

 

The electrode charge is not the same in electrolytic devices and electrochemical cells:

Electrolytic Device
Electrochemical Cell
  • positive anode
  • negative cathode
  • negative anode
  • positive cathode

A generator produces an electric current by electromagnetic induction. link to a local webpage A generator does not produce electrons - it pumps electrons toward the cathode and away from the anode. The role of a generator is to raise the potential energy of the electrons on the cathode and to reduce the potential energy of the electrons on the anode.

More terms used with electricity:

Electric potential difference: the relative tendency of two substances to take on electrons.

During electrolysis, a Redox reaction takes place only when energy is provided by an external power source. A Redox reaction occurs spontaneously in electrochemical cells because of the electric potential difference between the two electrodes.

For a substance to take on electrons, another substance must give them up. These two make up the two halves of an electrochemical cell. A comparison of the electron attraction of these two half-cells becomes the electrode potentials. The sum of the two electrode potentials will be the potential (voltage) for the cell.

An example: the zinc-copper cell  (chem lab 295)
The Redox reactions involved in a cell
This cell produces 1.10 volts

The cell works because the valence electrons of zinc have a higher energy than the valence electrons of copper. This gives zinc a greater tendency to give up electrons than copper. In the external circuit, electrons will always move from the positive anode to the negative cathode.

Zinc is the anode:

Copper is the cathode:

This cell operates because:

The shorthand for the zinc-copper cell is: Zn|Zn+2||Cu+2|Cu

The anode (oxidation reaction) is always on your left.
The cathode (reduction reaction) is always on your right.
The two vertical lines between the half-cells represent the "salt bridge".

This standard reduction potential tablelink to a local webpage was used to find the potentials for the half-cells and calculate the voltage for the zinc-copper cell above. The table and calculations will be covered in detail beginning tomorrow.


Calculations
look like this		 Zn Zn+2 + 2e- ......... +0.76 Volts
Cu+2 + 2e- Cu ........ +0.34 Volts
                          Total = +1.10 Volts

Your level of understanding of this material will determine your success with the electrochemistry concept. Study the information until you are comfortable with it. Be prepared to answer the following questions in class tomorrow:

  1. What causes metallic conduction?
  2. How are metallic and electrolytic conduction different?
  3. At which electrode does reduction occur during electrolysis?
  4. What is the purpose of a generator?
  5. What does voltage measure?
  6. Are cations positive or negative?
  7. Why can't the potential energy of a single half-cell be measured?
  8. Oxidation always takes place at which electrode?
  9. What would cause electrons to stop flowing in a cell?
  10. What does a single vertical line represent in the shorthand for a voltaic cell?
  11. What does a double vertical line represent in the shorthand for a voltaic cell?
  12. How is the voltage of a voltaic cell determined?
  13. What direction do electrons flow in the external circuit of a cell?

Day 2 - 3

Redox Reaction Calculator link to an Internet Website

Standard Reduction Potential Table:
shows the voltage of reduction half-reactions.

  • All reactions on the table are written as reductions. The substance on the left side of the arrow is the oxidizing agent, an electron acceptor. The substance on the right side of the arrow is the reducing agent, an electron donor.
    • In other words, all reactions on the table represent cathode reactions.
  • To obtain the anode reaction, reverse the direction of the reaction on the table. This makes the reaction oxidation.
    • For the oxidation reaction, the voltage is the same magnitude, but with the opposite sign.
  • Electrons must flow from reducting agent to oxidizing agent, that is toward the more positive potential on the chart.

A basic rule of thumb:

When comparing two half-cell reactions, the lowest one will be the cathode.

Standard Reduction Potential Table (at 25oC, 101kPa, 1M)
 
strong reducing agents Half-Reactionvolts strong oxidizing agents
Li+ + e- Li - 3.04
Al+3 + 3e- Al - 1.68
Zn+2 + 2e- Zn - 0.76
Fe+2 + 2e- Fe - 0.44
2H2O + 2e- H2 + 2OH - - 0.41
Ni+2 + 2e- Ni - 0.26
Pb+2 + 2e- Pb - 0.13
2H+ + 2e- H2 0.00
Cu+2 + 2e- Cu 0.34
Cu+ + e- Cu 0.52
Fe+3 + e- Fe+2 0.77
Ag+ + e- Ag 0.80
O2 + 4H+2 + 4e- 2H2O 0.82
Br2 + 2e- 2Br - 1.07
Cl2 + 2e- 2Cl - 1.36
Au+3 + 3e- Au 1.52

The standard reduction potential table can be used to:

1. Predict the direction chemical reactions will go.

2. Determine the maximum voltage produced by an electrochemical cell.

3. Predict the products of electrolysis.
 
  • For an electric current to pass through an aqueous salt solution, a chemical reaction must take place at both electrodes.
  • At each electrode, water will be in competition with the salt. This means there are always two possibilities for the product at each electrode.
  • Compare the voltages of the two possibilities at each electrode. The least negative voltage will identify the product.

Sample Problem Using the Reduction Table:

Predict the products of the electrolysis of 1M CuCl2 (aq)
    The two possible reduction reactions at the cathode are:
    Cu+2 + 2e- Cu . . . . 0.34V
    2H2O + 2e- H2 + 2OH - . . . . - 0.41V

    The least negative value requires less energy, and will be the reaction to take place. Cu (s) will be produced at the cathode.

    The two possible oxidation reactions at the anode are:
    2Cl- Cl2 + 2e- . . . . -1.36
    2H2O O2 + 4H+ + 4e- . . . . - 0.82

    The least negative, O2 (g) will be produced at the anode.

    Cu (s) and O2 (g) are produced by electrolysis of 1M CuCl2 (aq)
 

 

Practice Problems:
Use the standard reduction potential table to work these problems.link to a local webpage

Day 4

Some practical applications of electrochemical cells:

Dry cell - acid form:

This is the source of power for an ordinary flashlight. Most "flashlight batteries" produce 1.5 volts. The case of the cell is zinc metal acting as the anode. At the center of the cell is a stick of graphite for the cathode. The graphite stick is surrounded by a paste of MnO2 and NH4Cl. This is a dry cell. It is NOT a battery

Half-reactions for this cell are:

Anode - Zn (s) Zn+2 + 2e -

Cathode - 2NH4+ (aq) + 2MnO2 (s) + 2e - Mn2O3 (s) + 2NH3 (aq) + H2O (l)

Remember that the term "battery" refers to two or more cells connected together. The 9-volt transistor battery is a true battery. It contains six individual 1.5 volt cells connected in series to produce 9 volts.

Dry cell - alkaline form:

The NH4Cl in the "acid form" cell is replaced by KOH and the zinc is in powder form rather than a solid piece of metal. The graphite cathode is eliminated and acid corrosion of the container does not occur. The alkaline cell is more efficient and can be miniaturized to fit more varied applications.
Explore this Internet Website about Chemistry Lead storage battery - acid form:

The normal "car battery" consists of six 2-volt cells connected in series to produce 12 volts. Each cell uses a plate of lead for the anode. The cathode is PbO2 powder formed into a conducting grid. The electrodes are immersed in dilute H2SO4.

Half-reactions for each cell are:

Anode - Pb (s) + SO4 -2 (aq) PbSO4 (s) + 2e -

Cathode - PbO2 (s) + SO4 -2 (aq) + 4H + (aq) + 2e - PbSO4 (s) + 2H2O (l)

Pb (s) and PbO2 (s) are converted to PbSO4 (s) while the battery is being used. When the battery is recharged, the overall reaction proceeds in the reverse direction, restoring the reactants. This allows the battery to continue to be used. Most car batteries become useless after 3 to 5 years because side reactions occur that produce a sludge that interferes with the battery's operation.

The amount of charge in this type of battery can be determined by measuring the density of the electrolyte. Sulfuric acid has a density greater than water. As the battery operates, sulfuric acid is consumed, lowering the density of the electrolyte. If the density falls below 1.2 g/cm3, the battery is ready for recharging.

Rechargeable nickel-cadmium cell:

The products of the nickel-cadmium cell are all solids that adhere to the electrodes, making the reactions easily reversed during recharging.

Half-reactions for this cell are:

Anode - Cd (s) + 2OH - (aq) Cd(OH)2 (s) + 2e -

Cathode - NiO2 (s) + 2H2O (l) + 2e - Ni(OH)2 (s) + 2OH - (aq)

Corrosion of metals is an electrochemical process.

Corrosion refers to the deterioration of metals. The most common example is the rusting of iron. A weak electrochemical cell is formed between the metal and some metallic impurities in the presence of water. The iron atoms are oxidized in the presence of water and oxygen and go into solution as ions.

the key points to useful information on this page

Faraday's Laws:  (chem lab 284)
  1. The mass of an element produced at an electrode during electrolysis varies directly as the quantity of electricity that is passed through the solution.
  2. The quantity of different elements that can be deposited by the same amount of electricity depends on the equivalent masses of these elements.

Coulomb: the quantity of electricity produced by a current of one ampere flowing for one second.

  • Coulombs = (amperes) (seconds)
  • 96 485 coulombs = 1 mole e-

A Sample Problem Using Faraday's Laws:

How many grams of aluminum are produced if 31500 coulombs of electricity pass through an aluminum nitrate solution?

  1. Write a balanced equation for the reaction.
    Al+ + 3e- Al
  2. Convert coulombs to moles of electrons, then to moles of aluminum, then to grams of aluminum.

    Notice that the mole ratio between electrons and product, aluminum in this case, uses the coefficients from the balanced equation. This is what Faraday is referring to with his statement of "equivalent masses".


    Calculation using Faraday's Law

Practice Problems:
  1. What must happen for a "battery" to be recharged?
  2. Write the chemical equation representing the "rusting" of iron.
  3. If 10 amperes of current flow for 30 minutes through a solution of copper (II) nitrate, how many moles of copper metal are produced?
  4. How many seconds are required to deposit 2.5 grams of iron on an object using 15.5A of current passing through a iron (III) nitrate solution?
  5. Use the electrode reactions for a lead storage battery given above to answer the question: How many grams of PbO2 are used when a current of 50A is drawn from the battery during one hour?

Research Links:

Chemistry Class