Bryan Maliken
Dr. Pietrovito
Chemistry 054: 120
April 21, 2005
Lab #1025
1. A Simple Daniell Cell
Complete the diagram by writing in the formulas for the composition of the electrodes and of the
ions in the solutions. Place these formulas near the appropriate arrows leading into the solutions.
Also indicate the electric charge which is created by a + or – sign near the top of each electrode.
What constitutes an electric current in a wire?
Current is a measure of the rate at which charge is flowing past some point in a wire. The
wire provides a conducting path for the current of electrons to pass through. When a voltage is
applied across the ends of a wire, an electric field is created inside the wire allowing for the flow
of electricity.
What constitutes an electric current in a solution?
The free movement of electrons from the ions in solution constitute an electric current.
The electrons flow from the anode to the cathode as the ions are oxidized (at the anode) and
reduced (at the cathode). The current of electrons goes from the anode to the cathode and the salt
bridge completes the circuit by allowing cations and anions to flow.
Why must there be actual contact of the two solutions?
There must be actual contact between the solution through the salt bridge (or semipermeable membrane or other way) so that the electrical circuit will be complete and will allow
counter ions to flow into the two cell compartments to maintain electrical neutrality. However,
the compartments must still be separated or the cell will go to equilibrium without having created
any current (no useful work).
The half-reaction taking place at the negative electrode (cathode) is:
Cu2+ + 2e-  Cu : Ered = 0.34 V
The half-reaction taking place at the positive electrode (anode) is:
Zn  Zn2+ + 2e- : Eoxi = 0.76 V
The total cell reaction is: Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
Experimental voltage for this cell is 1.04 V.
2. Standard Electrode Potentials
Calculated cell potential = 0.34 V + 0.76 V = 1.10 V
Potential Series
(reduction potentials given, oxidation potentials are opposite to these)
Couple
Volts
Zn-Zn2+
-0.76 V
Fe-Fe2+
-0.44 V
Sn-Sn2+
-0.14 V
Pb-Pb2+
-0.13 V
H2-2H+
0.00 V
Cell Reaction
Cu-Cu2+
0.34 V
2I—I2
0.54 V
Fe2+-Fe3+
0.77 V
Ag-Ag+
0.80 V
2Br—Br2
1.09 V
2Cl---Cl2
1.36 V
Mn2+-MnO4-
1.51 V
Positive Electrode
Negative Electrode
Experimental Voltage
Calculated Voltage
Fe2+(aq) + Zn(s)  Fe(s) + Zn2+(aq)
Zn
Fe
0.44 V
-0.44 V + 0.76 V = 0.33 V
Cu (aq) + Fe(s)  Cu(s) + Fe (aq)
Fe
Cu
0.55 V
0.34 V + 0.44 V = 0.78 V
2+
2+
3. The Effect of Concentration
Voltage with NH4OH added: 0.8 V
Net ionic equation: CuSO45H2O + 4NH4OH  Cu(NH3)4SO4H2O + 8H2O
Voltage with Na2S added: 0.7 V
Net ionic equation: CuSO4 + Na2S  Na2SO4 + CuS
Explain the reasons for the effects on the cell voltage you observed when these reagents were
added to cupric sulfate in the cell.
According to the Nernst equation:
E = E - (0.0591/n) log ([anode]/[cathode])
In this experiment different solutions were added only to the cupric sulfate cell which is
where the anode attractions are taking place. By adding more solution to this cell the Cu2+ ions
are taken away when they form other compounds, and the amount of Cu2+ ions that are attracted
to the cathode are reduced in concentration. Thus, the cell potential is changed. By lowering the
concentration of the ion at the cathode the ([anode]/[cathode]) fraction is no longer 1 (as
concentrations were equal), but instead becomes greater than 1. The log value for a number
greater than 1 is a positive number (instead of zero as it is for a value of 1). Thus when the
(0.0591/n) log ([anode]/[cathode]) value is subtracted from the standard cell potential, the value
for experimental cell potential decreases, and is less than the standard cell potential. This is seen
from the data as the original cell potential was 1.04 V and this decreased to 0.8 V and then 0.7 V
when more solution was added to the anode cell (it is especially effective because the NH4OH is
a 6 M concentration, so there are many moles to react with Cu2+ ions).
How would you adjust the concentrations of the Cu2+ and of the Zn2+ in the Daniell cell in order
to obtain the maximum voltage possible?
According to the Nernst equation:
E = E - (0.0591/n) log ([anode]/[cathode])
Therefore, the greatest negative value of log ([anode]/[cathode]) would produce the
largest E value. This means that the smaller the fraction of [anode]/[cathode] is the higher the
voltage will be. This would mean that a higher concentration of Cu2+ and a lower concentration
of Zn2+ would produce the highest voltage.
4. Electrolysis: Oxidation-Reduction by Means of the Electric Current
Sketch a neat diagram for the electrolysis of a potassium iodide solution, indicating the
cathode, the anode, the direction the various ions move in the solution, and the products formed
at each electrode.
The yellow color at one electrode is due to the formation of:
This is due to a small amount of iodine being dissolved in the water.
This appeared at which electrode?
This appeared around the anode (the positive electrode).
Is this electrode the oxidizing or the reducing agent?
This electrode is the reducing agent.
The equation for the half-reaction at this electrode is:
2I- (aq) I2(s) + 2e- Oxidation potential = -0.54
The equation for the half-reaction at the other electrode (H2(g) is liberated; see experiment
directions) is:
2H2O(l) + 2e-  2OH-(aq) + H2(g) Reduction potential = -0.83 V
The net ionic equation for the electrolysis of a potassium iodide solution therefore is:
2I-(aq) + 2H2O(l)  I2(s) + 2OH-(aq) + H2(g)
Calculate the minimum external voltage needed to cause electrolysis of a 1 F KI solution (based
on the E values of the above half-reactions).
Cell potential = -0.54 V + -0.83 V = -1.37 V
Therefore, the minimal external voltage needed to cause electrolysis of this solution is 1.37 V.
Would a single Daniell cell (Zn and Cu2+) be able to bring about this oxidation-reduction, and if
so, at a reasonable rate? Explain.
No a single Daniell cell would not be able to bring about this oxidation-reduction because
the Daniell cell relies on a spontaneous reaction. The cell potential of this reaction is negative
and therefore requires energy, so it would not happen without an input of energy (which would
no longer be a Daniell cell).
5. The Corrosion of Iron
Summarize by comment, by suitable sketches, and by equations, your observations on the
rusting of iron exposed to moisture and to air when in contact with other metals you studied—
tin, copper, and zinc, respectively.
Iron (only the nail):
- The nail fully rusted (reddish-brown color) because it was easily oxidized by air.
Anode reaction: Fe(s)  Fe2+(aq) + 2eCathode reaction: O2(g) + 2H2O(l) + 4e-  4OH-(aq)
Fe2+ ions travel to the cathodic region through moisture on steel, and react with oxygen to form
rust: 4Fe2+(aq) + O2(g) + (4 + 2n)H2O(l)  2Fe2O3nH2O(s) + 8H+
- 2Fe2O3nH2O(s) is the rust compound
Iron with tin wrapped around:
- The nail rusted on the iron around the tin, but did not rust underneath the tin.
- The iron of the nail was still able to rust because the oxidation potential of tin is less than that
of iron, so the iron is oxidized by the air before the tin (Fe = 0.44 V and Sn = 0.14 V – negative
reduction potentials).
- The reason that it wouldn’t have rusted underneath the tin is because the tin was very closely
wrapped around the iron and wouldn’t allow much air and moisture to reach the iron. Also, the
oxidation potential of tin is fairly close to that of iron so it might be able to prevent the reaction
of air and iron under these closely wrapped circumstances.
- Iron still forms rust by the same reaction given above.
Iron with copper wrapped around:
- The iron with the copper wrapped around it was completely rusted and seemed to be the most
rusted of all the nails. There seemed to be large blotches near the copper that seemed to be
heavily rusted and the nail was rusted along all of the surface that wasn’t pressed against the
bottom of the beaker.
- The oxidation potential of copper is less than that off tin and is actually negative (Cu = -0.34 V)
so compared to tin, it is obvious that this copper would not prevent the oxidation of iron by
reacting with air.
Iron with zinc wrapped around:
- The iron with the zinc wrapped around it was not rusted at all, and was the only of the nails that
seemed to prevent rusting at all. Around the zinc a precipitant/gel seemed to have formed, rather
than the iron rusting.
- The reason that the zinc prevented the rusting of the iron is because its oxidation potential is
greater than that of iron, and it gets oxidized instead of iron (Zn = 0.76 V and Fe = 0.44 V). Thus
a compound of zinc oxide (ZnO) is formed which is what formed around the zinc on the nail.
2Zn(s) + O2(g) 2ZnO(s)
- The zinc acts as a “sacrificial” coating on steel.
Utilizing these observations, and the potential data, explain why a “tin” can (iron plated with tin)
rusts more easily than a galvanized pipe (iron coated with zinc), when both are exposed to
weathering.
Even though both are exposed to weathering, and therefore the iron of the “tin” can and
the galvanized pipe are both exposed to the air, the “tin” can rusts much more easily compared to
the pipe because the zinc on the pipe acts as a “sacrificial” coating. The zinc which has a higher
oxidation potential than iron will therefore react with oxygen in the air instead of the iron (or will
prevent a great deal of the iron reaction) to form a zinc oxide compound. Tin however, has a
lower oxidation potential than iron and therefore does not prevent the iron from rusting once the
iron is exposed to air. Obviously, it will prevent the iron from rusting when it completely coats it,
but otherwise the iron is much more likely to be oxidized by the air and the tin cannot prevent it
from rusting by creating its own reaction.
Why are cans of tin plate, rather than of galvanized iron, used for preserving fruits and other
food products?
Cans of tin plate are used for preserving fruits and other food products because although
they are not as protective of iron as zinc, they do form a durable, effective oxide coating around
the cans, and are much less expensive to produce than galvanized iron cans. They are both cheap
and effective for preserving food because generally food cans are not subject to much weathering
or large amounts of moisture and would have no reason to be rusting (they do receive moisture
inside the can, but no oxygen so this still will not form rust).
Application of Principles, and Problems
1. When zinc dust is mixed with 1 F CuSO4, the solution becomes distinctly warm. How
would the potential of your Daniell cell (part 1) be affected if the cell were at 0C instead
of room temperature? Explain.
An equation that relates cell potential to temperature is derived from a free energy
equation and these equations are:
G = H – TS
-(n)(F)(E) = H – TS
(n)(F)(E) = -H + TS
In this equation (n)(F) can be considered a constant for any specific cell making the cell potential
alone dependent on temperature. As the temperature goes down this means that less will be
added to the H value and therefore the E value will also go down. However, if the situation of
the Daniell cell is analyzed with a different equation, you can see that this cell is not affected by
a change in temperature, simply because the anode and cathode ion concentrations are equal. The
equation that proves this is the Nernst equation E = E - ((RT)/(nF)) log ([anode]/[cathode]).
From this you can see that if the anode and cathode ratio is 1 (which it is in part 1) then the log
of 1 = 0 and therefore will eliminate any temperature dependence. If the anode and cathode ion
concentrations are different, then the cell becomes temperature dependent (the bigger the
difference the larger the effect that temperature will have).
2. a. You have previously used KMnO4 as a strong oxidizing agent and SnCl2 as a good
reducing agent, both in an acid solution. At the right, diagram and electric cell involving
these reagents. Indicate your choice of electrodes, ions in the solutions, salt bridge, and
the behavior of all parts of the cell in detail, as you did for the Daniell cell.
b. Write the equations for the electrode reactions and for the total cell reaction.
Cathode reaction: MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Anode reaction: 2Cl-(aq)  Cl2(g) + 2eOverall reaction: 10Cl-(aq) + 2MnO4-(aq) + 16H+(aq)  5Cl2(g) + 2Mn2+(aq) + 8H2O(l)
c. Calculate the potential to be expected if all ions are at 1 M concentration.
MnO4+ reduction potential = 1.51 V
Cl- oxidation potential = -1.36 V
Cell potential = 1.51 V + 1.36 V = 2.87 V
3. Suppose that a copper electrode A is dipped in 1.0 F CuSO4 and another copper electrode
B is dipped in 0.01 F CuSO4, and the two solutions are connected by a suitable salt bridge. What,
if anything would be the reading on a sensitive voltmeter connected to the two electrodes, and
which would be the positive and which the negative electrode? Can you write an equation to
express the operation of such an electric cell?
The Nernst equation: E = E - (0.0591/n) log ([anode]/[cathode]) can be used to calculate the
cell potential of this concentration cell. E would be equal to 0 in this cell because the oxidation
and reduction potentials are exact opposites. The reading of this voltmeter would be E = 0 (0.0591/2) log ([0.01 M]/[1.0 M]) = 0.0591 V. The cathode (negative) electrode of this cell
would be electrode A and the anode (positive) electrode of this cell would be electrode B.
4. a. In commercial silver plating, the article to be plated (such as a teaspoon) is connected to
which electrode of the battery?
It is connected to the cathode.
b. Is this electrode the oxidizing or the reducing agent?
Reduction takes place at the cathode, so therefore the electrode is the oxidizing agent.
c. To plate a surface of silver 0.05 mm thick on a teaspoon (estimated 40 cm2 surface)
would require about 2.00 g of silver. How long would it take to plate this much silver,
using a 0.400 ampere current?
0.400 ampere = (0.400 C/sec) X (S seconds) X (1 Faraday/96,485 C) X (1 mole Ag/1
Faraday) X (107.9 g Ag/1 mole Ag) = 2.00 g Ag
(4.473 X 10-4)(S) = 2.00 g Ag
S = (2.00 g Ag)/(4.473 X 10-4) = 4471.04 seconds to plate out 2.00 g Ag
d. Suppose you plated 2.00 g of gold on this same spoon from an AuCl3 solution, using a
0.400 ampere current. How long would it take?
0.400 ampere = (0.400 C/sec) X (S seconds) X (1 Faraday/96,485 C) X (1 mole Au/3
Faraday) X (197.0 g Au/1 mole Au) = 2.00 g Au
(2.722 X 10-4)(S) = 2.00 g Au
S = (2.00 g Au)/( 2.722 X 10-4) = 7346.57 seconds to plate out 2.00 g Au