Electrochemical Half-Cells and Half-Reactions

Electrochemical Cells

Manatee Community College

Introduction

If we place a strip of Iron metal in copper(II) sulfate solution we can see evidence that a chemical reaction is occurring. The blue solution caused by the copper(II) sulfate begins to fade and the iron metal has a deposit of metallic copper on it. The reaction that is taking place can be seen in Equation 1.

(1) Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)

In Equation 1 the iron has lost 2 electrons and the copper has gained 2 electrons. These types of reactions are called oxidation-reduction or redox reactions. In redox reactions we can break up the total reaction into two parts, the oxidation half and the reduction half. Equation 2 shows the oxidation half and Equation 3 shows the reduction half.

(2)Fe(s)  Fe2+(aq) + 2 electrons

(3)Cu2+(aq) + 2 electrons  Cu(s)

In the oxidation half reaction (Equation 2) Fe(s) loses 2 electrons to become Fe2+(aq). Oxidation results in the oxidation number of the substance increasing. When we set up a half reaction we have to balance for the number of electrons for the reactant side with respect to the product side. If we look at the periodic table we see that Fe(s) has 26 electrons, this means that Fe2+(aq) has only 24 electrons. To balance the number of electrons we must add 2 electrons to the product side. You can only add electrons, never subtract electrons!

In the reduction half reaction (Equation 3) Cu2+(aq) gains 2 electrons to become Cu(s). Reduction results in a decrease in the oxidation number of a substance. When we set up the half reaction, the electrons on the reactant side and the product side have to be equal to each other. If we look at the periodic table Cu(s) has 29 electrons which means that Cu2+ has 27 electrons. To get the reactant and product side to have equal number of electrons we must add 2 electrons to the reactant side.

If we place a strip of copper in Iron sulfate solution (FeSO4) we observe no evidence of a reaction. This shows that the reverse of the reaction in Equation 1 does not occur spontaneously. We may conclude that Cu2+ ion has a greater tendency to accept electrons from Fe than Fe2+ ion does to accept electrons from Cu. In other words, Cu2+ ion has a greater reduction potential than Fe2+ ion.

Table 1 ranks selected metal ions by standard reduction potentials, determined under standard conditions: 25C, 1 atm, 1 mol/L ion concentration. The standard reduction potential, Ered, measured in volts (V) provides a quantitative measure of a metal ion’s tendency to accept electrons. The greater the reduction potential the greater the tendency for reduction.

Table 2 Selected standard reduction potentials for half-reactions

Reduction Half-Equation / Ered, V
Ag+(aq) + 1 electron  Ag(s) / .80
Cu2+(aq) + 2 electrons  Cu(s) / .34
2 H+(aq) + 2 electrons  H2(g) / .00
Pb2+(aq) + 2 electrons  Pb(s) / -.13
Sn2+(aq) + 2 electrons  Sn(s) / -.14
Ni2+(aq) + 2 electrons  Ni(s) / -.26
Fe2+(aq) + 2 electrons  Fe(s) / -.44
Zn2+(aq) + 2 electrons  Zn(s) / -.76
Cr2+(aq) + 2 electrons  Cr(s) / -.91
Al3+(aq) + 3 electrons  Al(s) / -1.66
Mg2+(aq) + 2 electrons  Mg(s) / -2.37

The half-reactions that represent the oxidation of the metals in Table 1 to their respective metal ions are the reverse of the half-reactions in Table 1. The standard oxidation potential, Eox, of an oxidation half-reaction has the same numerical value, but the opposite sign. An example of this with respect to Tin can be seen in Equation 4 & 5.

(4) Sn2+(aq) + 2 electrons  Sn(s)Ered = -.14 V

(5) Sn(s)  Sn2+(aq) + 2 electronsEox = .14 V

We can use the standard oxidation and reduction potentials to predict if a reaction will be spontaneous or non-spontaneous. Let’s take a look at the reaction of Zn2+(aq) with Ni(s). First let’s set up a general reaction, which can be seen in Equation 6.

(6) Zn2+(aq) + Ni(s)  Zn(s) + Ni2+(aq)

Now we must separate the whole reaction into two half-reactions, one for oxidation (Equation 7) and one for reduction (Equation 8). When we set these up we need to include their respective standard oxidation and standard reduction potentials.

(7) Ni(s)  Ni2+(aq) + 2 electronsEox = .26 V

(8) Zn2+(aq) + 2 electrons  Zn(s)Ered = -.76

(9) Ni(s) + Zn2+(aq)  Ni2+(aq) + Zn(s) Enet = .26+(-.76) = -.50 V

To obtain the net equation we have to add the two half-reactions (see Equation 9). This results in the electrons canceling out and Enet = -.50 V. Because the standard net potential (Enet), or sum of the standard reduction and oxidation potentials, is a negative number we can predict that this reaction would not occur. Remember that in a galvanic cell/redox reaction there must be a positive voltage. Having a negative voltage means that we would have to supply a current to get this reaction to proceed.

Galvanic Cells

Oxidation-reduction reactions can occur between reactants in separate containers. In this situation electrons are transferred from the reactants being oxidized to the reactant being reduced through a wire connecting the two containers.

For example, suppose we place a Ni strip in an aqueous Ni2+ salt solution. Then we place a Cu strip in an aqueous Cu2+ salt solution in a separate container. (the metal strips are called electrodes) Next, we connect one end of a wire to the Ni strip and the other end to the Cu strip. This wire allows electrons to pass from the Ni strip to the Cu strip, provided that we make an additional connection between the containers to complete the electrical circuit. This second connection is called a salt bridge. This assembly is called a galvanic or voltaic cell. A galvanic cell is a system that utilizes a spontaneous oxidation-reduction reaction to pump electrons through an electrical circuit. In this cell we use a voltmeter to measure the net cell potential. This is the same as the Standard net potential except that it’s reactants are not necessarily in a standard state condition.

Electrolytic Cells

A second type of electrochemical cell is the electrolytic cell. This is one in which an external electric current is used to drive a non-spontaneous chemical reaction (Figure 1). This forms the basis for all electroplating processes.

Figure 1

The layout of this cell is the same as a galvanic cell except the anode is now positive and the cathode negative. Since the power source is driving electrons into the cathode, the metal cations in solution will be reduced to form the metal on the cathode. As cations are taken out of solution by this process additional metal atoms on the anode are oxidized to fresh cations. The net result is that the cathode gains mass and the anode loses mass.

In this lab you will plate copper from a solution of copper sulfate onto a copper cathode. The overall reaction is:

(10) Cu+2(aq) + 2e-  Cu(s)

Michael Faraday showed the quantitative relationship between the number of electrons flowing into the cathode and the amount of substance produced. It requires two “moles” of electrons to produce one mole of copper. There are three quantities you will measure: the current flow, time and the mass of copper. Current flow, amperes or amps, is a measure of the charge passing through the system. The units are Coulombs per second. So multiplying amps by time gives coulombs.

(11) Coulombs = amps x time (seconds)

Faraday’s Constant is the total charge on a mole of electrons.

(12) F = 96,500 Coulombs/mole

So we can get moles of electrons by dividing Coulombs by the Faraday Constant. But since it takes two moles of electrons to make a mole of copper we divide the resultant by two:

(13) Moles Copper = amps x time

2 F

Procedure

I. Oxidation-Reduction Reactions

1. The Zn-Cu System

Obtain and place a Zinc strip in a small test tube.

a)Add enough 0.1 M CuSO4 solution into the zinc test tube to

completely cover the zinc.

b)Wait 5 minutes, record your observations.

c)Take the zinc strip out of the solution and discard the zinc, keep the solution for the next step.

1. The Pb-Cu System

Obtain a lead strip and place it in the CuSO4 solution from step 1

a)Wait 5 minutes, record your observations

b)Place the metal strip and the solution into the designated waste containers.

1. The Zn-Pb System

Obtain and place a zinc strip in a small test tube

a)Add enough 0.1M Pb(NO3)2 solution into the zinc test tube to completely cover the zinc.

b)Wait 5 minutes, record your observations.

c)Place the metal strip and the solution in the designated waste containers.

1. Galvanic Cells

1. The Zn-Cu Galvanic Cell

a)Label a 20 mL beaker “Zn” and place 15 mL of 0.1 M ZnSO4 in it

b)Label a 20 mL beaker “Cu” and place 15 mL of 0.1 M CuSO4 in it

c)Fold a piece of filter paper to about ½ inch in width, fold this piece in half and place ½ in the copper solution and ½ in the zinc solution

d)Using the KNO3 solution completely soak the filter paper (this is the salt bridge)

e)Connect one end of the voltmeter, via the alligator clamps, to a Zinc strip

f)Connect the other end of the voltmeter to a copper strip

g)Make sure the voltmeter is on and set to the right setting

h)Place the copper strip in the copper solution and the zinc strip in the zinc solution, Make sure to hold them both steady and not let the alligator clips to touch the solution

i)Record the voltage on the voltmeter

j)Disconnect the alligator clips from the metal strips and dispose of all metal and solution in the appropriate waste containers

1. The Fe-Cu Galvanic cell

a)Follow the same procedure given in the previous cell while changing the metal strips and the solutions used

b)Use 15 mL of 0.1 M FeSO4 and 15 mL of 0.1 M CuSO4

c)Place the Iron strip in the FeSO4 solution and the Copper strip in the CuSO4 solution

d)Take the voltage

e)Disconnect the alligator clips from the metal strips and dispose of all metal and solution in the appropriate waste containers

III. The Copper Electrolytic Cell

1. Obtain 2 copper strips and remove the oxide from the surface with sandpaper or steel wool.
2. Label one strip A for anode and the other C for cathode and weigh them.
3. Place the strips in a 50 mL beaker connected to a power supply and ammeter as shown in the schematic on the next page.
4. Add 40 mL of 1.0M CuSO4/0.1M H2SO4 and a magnetic stirrer. Stir the solution gently.
5. Raise the voltage dial on the power supply until you get a reading of about 0.25 amps on the ammeter (approximately 4 volts).
6. Time the process for ten minutes and carefully note the current flow.
7. Turn off the power, remove the copper strips and rinse them in a beaker of water. Dry them carefully and weigh the strips.
8. Determine the mass of copper plated, the moles of copper plated and compare the results with the expected value from the current flow and time.

Electrolytic Cell Schematic

+ -

Power SupplyAmmeter

Cu Anode (+) (-) Cu Cathode

Data Sheet

I.Oxidation-Reduction Reactions

1. The Zn-Cu System

Observations:

1. The Pb-Cu System

Observations:

1. The Zn-Pb System

Observations:

II.Galvanic Cells

1. The Zn-Cu Galvanic Cell2. The Fe-Cu Galvanic Cell

Cell voltage: ___ Cell voltage:

Copper Electrolytic Cell Data Table

Current (amps):______Time( seconds):______

Coulombs:______

Calculated moles of copper produced (Show calculations):______

Initial Anode Mass:______Initial Cathode Mass:______

Final Anode Mass:______Final Cathode Mass:______

Mass of Cu Lost:______Mass of Cu Formed:______

Average Copper Mass Produced:______

Moles of Copper Produced:______
Post-Lab Questions

1. Oxidation-Reduction Reactions

1. The Zn-Cu System

Oxidation ½ reaction: Eoxid = ____

Reduction ½ reaction:Ered = ____

Net Equation: Enet = _____

Prediction of spontaneity:

Comparison of predicted spontaneity with experimental observations:

1. The Pb-Cu System

Oxidation ½ reaction: Eoxid = ____

Reduction ½ reaction: Ered =_____

Net Equation: Enet = ___

Prediction of spontaneity: .

Comparison of predicted spontaneity with experimental observations:

1. The Zn-Pb System

Oxidation ½ reaction: Eoxid = ____

Reduction ½ reaction: Ered =_____

Net Equation: Enet =_____

Prediction of spontaneity:

Comparison of predicted spontaneity with experimental observations:

1. Galvanic Cells

1. The Zn-Cu Galvanic Cell

Cu ½ reaction: E = ______

Zn ½ reaction: E = ______

Net cell reaction: Enet = ______

Comparison of observed and calculated cell voltages:

_

1. The Fe-Cu Galvanic Cell

Cu ½ reaction: E = ______

Fe ½ reaction: E = ______

Net cell reaction:Enet = ______

Comparison of observed and calculated cell voltages:

III Electrolytic Cells

1. Aluminum is manufactured by the Hall-Heroult Process which produces aluminum from aluminum oxide in an electrolytic cell. The net reaction is:

Al+3 + 3e-  Al(s)

You need to design a cell that will produce Al at the rate of 1 kg per hour. How many amps must your power supply produce?

Pre-Lab Questions:

1. Answer the following questions about a galvanic cell that would have the following net reaction

Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)

a)Write the oxidation and reduction half reactions, net reaction, include the cell potential of the oxidation, reduction, and net cell potential

Oxidation ½: Eoxid =_____

Reduction ½: Ered =_____

Net reaction: Enet =_____

b)Is this reaction spontaneous? Why or why not?

1. Rank the following metals by ease of reduction, starting with the easiest to reduce

Ag+, Ni2+, Al3+, Pb2+, Mg2+

1. Rank the following metals by ease of oxidation, starting with the easiest to oxidize

Mg, Ag, Fe, Cu, Cr

1. For the following reactions state if the reactions are favored in the forward or reverse direction

a) Ag+(aq) + Cu(s)  Cu2+(aq) + Ag(s)

b) Ni2+(aq) + Sn(s)  Ni(s) + Sn2+(aq)

c)Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)

d)Pb(s) + Zn2+(aq)  Zn(s) + Pb2+(aq)

1

Electrochemical Cells