To study the process of corrosion in metals and to determine of solubility product constants, K sp
, using a potentiometric method.
Facts:

For corrosion to occur, normally it is necessary for oxygen and water to be present.

Corrosion usually increases with an increase in strong electrolyte concentration.

If dissimilar metals are in contact, the one that will corrode will be the metal that's higher in the activity series.

The use of a sacrificial anode is a deliberate use of the dissimilar metal theory.

Corrosion can occur on a single metal - usually at 'stress' points - or at surface imperfections.

There are two main types of corrosion: Galvanic (due to difference in metals); Local (due to differences in electrolyte concentration).

Corrosion processes occur in cells usually with the metal (or metals), electrolyte, and oxygen present.

Corrosion processes are redox processes.
Definitions:

Corrosion is oxidation and occurs at the anode.

Reduction occurs at the cathode.
 Corrosion is a combination of an oxidation ½ reaction and a reduction ½ reaction. One cannot occur without the other.
Predictions:
There are various and sometimes sophisticated methods used to determine whether a specific reaction will occur spontaneously or not. Some reactions that do not occur spontaneously can be forced to using an external source of energy (i.e. electrolysis).
In electrochemistry, the Table of Standard Reduction Potentials is used to predict whether the reaction is spontaneous or not. All the ions are one molar concentration and gases are one atmosphere pressure, at 25°C. The Table of Standard Reduction Potentials shows the reduction
½ reaction and its potential. If the reaction is reversed, then it becomes an oxidation ½ reaction and the sign of the potential is reversed (see below and find it in the Standard Table).
Example:
Oxidation ½ reaction Zn(s)  Zn 2+ (aq) + 2e -
Reduction ½ reaction
Cl
2
(aq) + 2e  2Cl (aq)
Net reaction
E° = + 0.76 V
E° = + 1.36 V
Zn(s) + Cl
2
(aq)  Zn 2+ (aq) + 2Cl (aq) E° = +2.12 V
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The cell potential of the above reaction is +2.12 V and is spontaneous. Any negative cell potential calculated for a cell will not give a spontaneous reaction. If the cell potential is negative then the reaction will not occur to any great extent.
Example: Will a reaction occur if a solution of Cu 2+ (aq) is mixed with an aqueous solution of I -
(aq) to form I
2
(s) and Cu(s)?
Oxidation ½ reaction 2I (aq)  I
2
(s) + 2e -
Reduction ½ reaction Cu 2+ (aq) + 2e  Cu(s)
Net reaction
E° = - 0.53 V
E° = + 0.34 V
Cu 2+ (aq) + 2I (aq)  I
2
(s) + Cu(s) E° = - 0.19 V
For this example, the cell potential is negative and this reaction will not occur to any great extent.
The cell potential is related to the Gibbs free energy where

G = -nFE. Combining this with

G =
 G° +RT ln(Q) given the Nernst equation for the cell potential.

o 
ln(
) (
)
Concentration Cells
A concentration cell can be constructed using the same ½ cells but with different concentrations of the same electrolyte.
We can convey an electrochemical cell by using the cell notation rather than drawing a picture or an actual diagram of the cell. Cell notation lists the metals and ions involved in the reaction. A vertical line, |, denotes a phase boundary. A double line, ||, represents the salt bridge. The anode is always written on the left and the cathode on the right. The general form is: anode | electrolyte of anode || electrolyte of cathode | cathode
For example, the cell notation for the concentration cell made with Cu metal dipped into different concentration of Cu 2+ solutions can be represented as follows:
Cu | Cu 2+ (0.10 M) || Cu 2+ (1.0 M) | Cu
The half-cell written to the left of the double line, ||, is the anode.
Cu(s)  Cu 2+ (0.10 M) + 2e (6-1)
The half-cell written to the right of the double line, ||, is the cathode.
Cu 2+ (1.0 M) + 2e  Cu(s) (6-2)
Electrons will flow from the ½ cell that has the 0.10 M Cu 2+ solution to the ½ cell with the 1.0 M
Cu 2+ solution. Basically nature will try to equalize the concentrations of the copper ions in the two compartments. In a concentration cell, E° has a value of zero. Setting E° = 0 and in the
Nernst equation gives the cell potential.
E
 
RT nF ln( Q )
For the cell reaction to be spontaneous, the cell potential must be positive, E > 0. Therefore the reaction quotient must be less than 1, Q <1.
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Part A - Corrosion and Electrodes
1. A technician will have prepared four petri dishes containing an agar-salt solution. Once the agar cools, this is an electrolyte supported in gel matrix. It contains 7.5 g NaCl, 5 g powdered agar, 5 mL of 5% potassium ferricyanide and 1 mL of 1% phenolphthalein solution.
2. The petri dishes will have the following metal or metals inserted in the agar gel.
Dish 1 Dish 2 Dish 3 Dish 4
Fe Fe / Zn Fe / Cu Cu / Fe / Zn
3. Record your observations of the petri dishes provided. Record the location of the various metals and the observed colours.
You may observe one or more of the following results:

Oxidation of Fe to Fe
2+
: In the agar-salt solution any iron that is oxidized will react with the 5% potassium ferricyanide and form a blue colour described as "Prussian Blue". If iron is not oxidized , there is no blue colour in the agar gel.

Oxidation of Zn to Zn 2+ : In the agar-salt solution any zinc metal that is oxidized will give a white-grey colour (maybe hard to see).

Reduction of O
2
to OH : The agar-salt solution contains phenophthalein, an acid-base indicator that is pink in basic conditions. If oxygen is reduced then the hydroxide that is formed will cause the agar to turn pink. The reaction is O
2
+ 2H
2
O + 4e  4OH .
Remember that neither oxidation nor reduction can occur alone.
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Part B - Activities of Metals
A voltmeter (pH meter) is used to measure the potential differences between two metals.

A positive potential value (+) means that the metal on the reference lead is oxidizing.

A negative potential value (-) means that the metal on the input lead is oxidizing.

Oxidation occurs at the anode and reduction at the cathode.
In this part of the lab, you will set up electrochemical cells using the half-cells indicated on the data sheet.
Salt bridges have been prepared for you.
1. On a piece of paper towel, polish the metal electrode strips with some steel wool. Wipe the electrodes clean with a damp paper towel.
2. Use the Fe | Fe 2+ (0.10 M) || Fe 2+ (0.10 M) | Fe electrochemical cell to zero the instrument.
Connect one electrode to the reference lead and the other electrode to the input lead. Set the instrument to +mV scale and read the 0-14 pH scale as 0-1400 mV. Zero the instrument.
Note: Do not immerse the alligator clips in the cell solutions. Only the electrode should come in contact with the solution in all the electrochemical cells that you prepare.
3. For each of the remaining electrochemical cells have the electrode for the Fe | Fe 2+ (0.10 M) cell connected to the reference and the electrode of the other cell connected to the input side. Record the voltage obtained.
Part C - Concentration Cells
1. Construct a galvanic cell that has two different concentrations of Cu connect the two ½ cells and a copper electrode for each ½ cell.
2+ . Use a salt bridge to
2. Cell 1: The first galvanic cell has a beaker of 1.0 M Cu 2+ versus a beaker of 0.10 M Cu 2+ .
Measure the cell potential. Record which concentration is connected to the input terminal.
3. Cell 2: The second galvanic cell has a beaker of 1.0 M Cu 2+ versus a beaker of 0.0010 M
Cu 2+ . Measure the cell potential. Record which concentration is connected to the input terminal.
Part D - Corrosion
This has been set-up by the technician.
1. Record your observations for a nail in water saturated with oxygen gas.
2. Record your observations for a nail in water saturated with nitrogen gas.
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Part E
– Determination of Solubility Product Constants, K sp
1. Polish two silver electrodes and arrange apparatus as shown below.
+ e
Voltmeter e
-
Silver Anode
Salt Bridge
Silver Cathode
Saturated AgX solution
Stir Bar
0.010 M AgNO
3
Solution
3
2
4
5
1
6
7
8
9
11
3
2
4
5
1
6
7
8
9
10
A concentration cell to determine the K sp
value of AgX.
2. Connect the electrode in the 0.010 M Ag + to the positive terminal of the pH meter and the second electrode to the negative terminal of the pH meter.
3. Add about 10 to 15 drops of AgNO
3
to the beaker containing 0.010 M sodium chloride until a AgCl precipitate starts to form (ie - until milky). Start the stirrer and record the voltage of the cell together with the temperature. Enter the data on the datasheet for Cell 1.
Describe the cell using cell notation.
Repeat this procedure using 0.010 M sodium bromide solution (Cell 2) and 0.010 M sodium iodide solution (Cell 3) in place of the sodium chloride solution.
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Part A - Corrosion and Electrodes
Sketch to record your observations, colours, location of the metals and any important results of the nails in the petri dishes.
Metal Half-cell Reaction
Anode
Dish 1
Fe
Cathode
Dish 2
Fe / Zn
Anode
Cathode
Anode
Dish 2
Fe / Cu
Cathode
Anode
Dish 4
Fe / Zn /Cu
Cathode
Why is Zn referred to as the sacrificial anode when it is paired with Fe?
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Electrochemical Cell Reference Input Voltage (mV) Anode Cathode
Fe | Fe 2+ (0.10 M) || Fe 2+ (0.10 M) | Fe
Cell
Cell 1: 1 M Cu 2+ versus 0.1 M Cu 2+
Fe Fe
Fe | Fe 2+ (0.10 M) || Cu 2+ (0.10 M) | Cu Fe Cu
Fe | Fe 2+ (0.10 M) || Zn 2+ (0.10 M) | Zn Fe Zn
Based on your experimental results arrange the metal ions in order of decreasing reduction potentials.
Use a diagram to illustrate the Fe | Fe 2+ (0.10 M) || Cu 2+ (0.10 M) | Cu electrochemical cell. Label the anode and cathode; indicate the direction of movement of electrons, anions, and cations. Write equations to describe the reaction occurring at the anode and cathode.
Voltage (mV) Concentration on Input Lead
Cell 2: 1 M Cu 2+ versus 0.001 M Cu 2+
Write the cell notation for Cell 1. Include the concentrations
For Cell 1 which solution undergoes oxidation? Include the reaction.
For Cell 1 which solution undergoes reduction? Include the reaction.
What is the experimental trend in the cell potential as the concentration of the cell with the lower concentration decreased? Is this consistent with the Nernst Equation?
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Record your observations, colours and any important results below.
Fe nail in O
2 saturated water Fe nail in N
2 saturated water
Explain your observations.
sp
Temperature ___________
K sp
Equilibrium Reaction Cell
Cell 1: Cell with NaCl solution
Voltage (mV)
Cell 2: Cell with NaBr solution
Cell 3: Cell with NaI solution
Using the Nernst equation, calculate the [Ag + ] in each of the silver halide solutions (ie - AgCl, AgBr, and AgI). Calculate K sp
and pK sp
for each of the three salts. Assume that the halide concentration in each case is 0.0100 M.
Provide the calculation of K sp
and pK sp
for AgCl as a sample calculation. pK sp
= -log(K sp
)
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Tabulate calculated values of K sp
and pK sp
of the solubility product constants and % error in the pK sp.
𝑠𝑝
𝑠𝑝 𝑒𝑥𝑝𝑡
𝑠𝑝 𝑎𝑐𝑐𝑒𝑝𝑡
𝑠𝑝 𝑎𝑐𝑐𝑒𝑝𝑡
Salt
[Ag + ]
(experimental value)
AgCl
AgBr
AgI
Comment on your results.
K sp
(experimental value) pK sp
(experimental value)
K sp
(accepted value)
1.8x10
-10
5.3x10
-13
1.5x10
-16 pK sp
(accepted value)
% error in pK sp
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