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7 Electrochemistry

ObjectivesThe objectives of this experiment are to learn to:

• Define the terms oxidation and reduction.• Experimentally identify oxidation and reduction reactions.• Classify half-cell reactions in order of their tendency to gain or lose electrons.• Predict electrochemical cell potentials, given the identity of the component half-

cells and the electrochemical series.

BackgroundThe small voltages generated by a chemical reaction offer a great deal of informa- tion concerning the behavior of the reaction, as well as valuable clues concerning the concentration of certain chemical species within the reaction mixture. Many chemical reactions involve gain or loss of electrons by an atom or an ion. Chemists use the term “reduction” to indicate a gain of electrons (when electrons are added to an atom or ion, the charge is increased in a negative direction, or “reduced”). Conversely, the term “oxidation” is used to indicate a loss of electrons.

Part I of this experiment will illustrate the movement of electrons in a forced electro- chemical reaction; the other parts of the experiment involve spontaneous electrochemical reactions. Our study of spontaneous electrochemical reactions will involve three gen- eral problems: (1) Can metals and ions be ordered according to their general tendency to gain or lose electrons? (2) Can this “electrochemical series” be used to predict the potential of new electrochemical cells? (3) What can electrochemistry tell us regarding the choice and protection of building materials?

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Zn0 metal

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xidation–ReductionMany chemical reactions fall in the family of “oxidation–reduction” reactions. Con- sider the case, shown in Figure 1, of a sample of metallic zinc immersed in a solution of zinc nitrate. There are two reactions possible: (1) The metallic zinc may cast off two electrons to become Zn2

(Equation 1); or, (2) the Zn2 ion may accept two electrons to become metallic zinc (Equation 2):

Zn0 (metal) → Zn2 2e– (1)

Zn2 2e– → Zn0 (metal) (2)

Zn++ solution

Figure 1. A zinc half-cell.

The reaction shown in Equation 1 is an oxidation reaction since it involves a loss of electrons; that is, the reactant (Zn0) gave up electrons. The reaction shown in Equation 2 is a reduction reaction because it involves a gain of electrons by the reactant, the Zn2 ion. As soon as the zinc plate is immersed in the solution, both reactions take place in equilibrium (Equation 3):

Zn0 Zn2 2e– (3)

The overall process can be seen to be an exchange between the ions of the solution and the zinc atoms of the metal plate. However, since this is an equilibrium situation, no extra electrons are available, and from the point of view of an external observer, no reaction appears to take place. An observable reaction will not take place unless an ad- ditional ion is present to either contribute or accept the extra electrons.

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Consider the similar case of metallic lead placed into a solution of lead nitrate, which contains the lead(II) ion (Pb2). Again, two reactions are possible in this situation. The metallic lead could give up electrons in a manner analogous to the metallic zinc in Equation 1, or the lead ion could accept electrons in a manner analogous to the zinc ion in Equation 2. In other words, the metallic lead could be oxidized and the lead ion could be reduced. As before, an equilibrium analogous to Equation 3 exists between these two reactions, and no observable reaction occurs.

Suppose, however, that the two reactions just discussed were coupled by wires that would conduct electrons from one reaction to the other (Figure 2). Such an arrangement is called an electrochemical cell, and the two reactions are called half reactions or half- cells. Which way would the electrons flow? Would the zinc oxidize to Zn2 giving up two electrons which could pass through the wire to the lead half-cell, where the Pb2 ion could be reduced to lead? Or would the lead have the greater tendency to oxidize to a Pb2 ion and give up two electrons, which could pass through the wire to reduce the Zn2 ion to zinc metal?

voltmeterelectrons

Zn0 2e- + Zn2+

oxidation

Zn0 metal

KNO3 salt bridge

Zn++ solution

Pb0 metal

Pb2+ + 2e- Pb0

reduction

Pb++ solution

a

Figure 2. An o idation–reduction resulting from the combination of the zinc half-cell and the lead half- cell. The potassium nitrate salt bridge is placed

between the two cells to permit migration of positive or negative ions, thus preserving the equality of charge in each beaker.

Suppose that the metallic zinc has a greater tendency to oxidize than the metallic lead. This is indeed the case, and the reaction shown in Equation 1 will occur at the zinc plate. The two electrons produced will travel through the wire, pass through the meter, and end up in the lead half reaction. There they will combine with a Pb2 ion to form Pb2. Thus,

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zinc is oxidized in one beaker and lead ion is reduced in the other. The zinc plate will be the negative electrical pole, since electrons are leaving at this point.

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The lead plate will be the positive pole, since electrons are accepted by the Pb2 ions at this plate. A voltmeter can be used to determine the direction of electron flow and the relative charge of each pole. In this way we can determine the tendency and direction for electrons to flow between the half-cells.

The force pushing the electrons away from the zinc plate and the attractive force ex- erted on the electrons as they approach the lead plate provide an excellent measure of the tendency for oxidation and reduction to take place in this reaction. In the case of zinc and lead, the sum of the forces moving the electron from the zinc to the lead is about 0.52 volts. By trying other half-cells of metals and their ions with a lead or zinc half-cell, we should be able to develop a table listing the various elements and ions in order of their tendency to gain or lose electrons.

Safety PrecautionsObserve all normal laboratory precautions. Exercise care in the handling of solutions and equipment. Wash your hands before leaving the lab, since some of the solutions are poisonous if ingested.

MaterialsE uipment• Beakers (from your locker)• Battery board, test leads with clips, metal samples (check out from stockroom)

upplies• Filter paper strips, steel wool, sandpaper, plastic document protector (supplied in

lab)

Reagents• 0.1 M solutions of Zn, Pb, Fe, Cu, and Ag nitrates in small dropper bottles• 0.5 M copper sulfate• 0.1 M potassium nitrate• 1.0 M, 0.01 M, and 0.001 M solutions of copper nitrate

Experimental ProceduresPart I. A Forced Electrochemical ReactionAs an introduction to this phenomenon of gain and loss of electrons, set up a simple experiment involving a beaker with a solution of copper sulfate (which provides Cu2 and (SO4)2– ions), a battery, two test leads with clips, a copper wire, and a key or other piece of metal. This experiment is shown in Figure 3. Use the copper sulfate solution from the large container in the front of the lab, not the copper nitrate solution in the small dropper bottles.

Cu wire Key

CuSO4 solution

Battery

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Figure 3. Apparatus for demonstrating electroplating.

There are two precautions you should observe while doing this experiment. (1) Do not allow the leads from the two terminals of the battery to come in direct contact with each other. This will result in rapid discharge of the battery. There is no health or safety hazard, since the voltage is quite small. (2) Do not allow the clips to come in contact with the solution, since they may react with the solution and become corroded.

Hold the key and the copper wire in the solution for a minute or two, without any elec- trical connection to either. Withdraw and observe both the wire and key closely. Record your observations in the Data Summary sheet at the end of the experiment. Repeat the experiment, but this time connect the negative pole of the battery (the black lead) to the key, and the positive pole (red lead) to the copper wire. Again, maintain immer- sion for one to two minutes, then withdraw them and observe the two metal surfaces. Record your observations.

Copper ions (Cu2) from the solution pick up electrons from the battery at the surface of the key, thus converting the copper ion into metallic copper (Equation 4):

Cu2 2e– → Cu0 (metal) (4)

Conversely, an equal number of copper atoms (Cu0) on the copper wire are stripped of two electrons each, becoming Cu2 ions, and move into solution to balance the charge of the negative sulfate ions.

Part II. The Electrochemical eriesCan elements and ions be arranged in order of their tendency to gain electrons? A simple and easy-to-use electrochemical cell can be built with the MicroLAB Multi-EChem Half-cell Module, illustrated in Figures 4 and 5.

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Reduction reaction Oxidation reaction

Salt bridge

Cu2+ + 2 electronsCu0 Zn2+ + 2 electronsZn0

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Voltmeter

Figure 4. One can e perimentally develop the electrochemical series by comparing a series of metal/ion pair half-cells against one “reference” metal/ion half-cell.

This unit has eight small wells, each connected to a central salt bridge through a small channel. The salt bridge is made of an aqueous solution of potassium nitrate, and con- tacts the half-cells through a water-soaked porous cylinder. Ions can move through this porous barrier without allowing much mixing of the solutions. The center of the cylinder is filled with potassium nitrate to keep the cylinder wet and soaked with mobile ions.

Different metal/ion half-cell pairs can be set up in each of the outer wells, and the volt- age produced by the two half-cells measured with a voltmeter as illustrated in Figures 4 and 5.

Figure 5. The Multi-EChem half-cell module can hold eight different metal/ion half-cells. If you use one as a reference, you can use the sign and magnitude

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of the voltage observed to order the elements in terms of their ability to gain or lose electrons, as compared to the reference metal/metal ion half-cell.

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Set up your MicroLAB to measure voltage. Select the Voltage measurement, the dual banana jack voltage input, the 2.5 volt range, and accept the factory calibration.

Each of the wells will hold a different ion/metal half-cell. For example, in this figure, the left cell has a copper II nitrate solution and a copper wire.

You have a black and red alligator clips connecting to the MicroLAB voltage input. If you get a positive voltage reading, electrons are running from the half-cell module into the black alligator lead, on to the MicroLAB voltmeter, and back out the red lead to the solution.

In this example (Figures 4 and 5), the voltage reads positive, indicating that the zinc/zinc ion half-cell on the right is producing electrons. The reaction (see Figure 4) is going to the left. It is an oxidation reaction. Zn2+ + 2 electrons ← Zn0. The copper/copper ion half-cell on the left in Figures 4 and 5 is accepting electrons. It is a reduction reaction. This reaction is going to the right: Cu2+ + 2 electrons → Cu0.

By convention, o idation–reduction reactions are always written from left to right as reduction reactions. A longer arrow can be used to indicate the predominant direction of the reaction.

Half-Cells and Ion PathThe lower channel from each half-cell well to the central salt bridge area provides a non-drying, low resistance aqueous path for ions up to the porous salt bridge membrane. The porous membrane is conditioned before use by soaking the unit with KNO3 solu- tion. This fills the pores in the membrane and prevents solutions from leaking from one half-cell to another. The aqueous salt bridge inside the circular porous membrane contacts all half-cells equally.

Half-Cell verflow VolumeThe milled overflow volume above each channel to the salt bridge is there to catch overflow as you fill the half-cell well. It prevents cross-contamination of half-cell solu- tions during filling. If you use a plastic pipette to deliver the ion solution to the well, it is easy to stop filling when the lower channel is full.

Experimental Development of the Electrochemical eriesPut a piece of lead metal on the lead nitrate solution and a piece of zinc metal on the zinc nitrate solution. Contact the two metals with your interface clip leads, and measure the voltage produced between the Zn and Pb plates. Which way are the electrons flowing? In which half-cell is the oxidation–reaction taking place? Repeat the measurement using a Pb/Pb2

half-cell and a Cu/Cu2 half-cell. (Be sure to use the 0.1 M copper nitrate solution, not one of the other concentrations.) Which way do the electrons flow now? Is there as great a tendency for the reaction to proceed?

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Try this again, substituting the Ag/Ag half-cell for the Cu/Cu2 half-cell. Continue to use the Pb/Pb2 half-cell as a reference. What are your conclusions in this trial?

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Chemistry 121 Experimental

Which half-cell reaction has the greatest tendency toward reduction? Which has the greatest tendency toward oxidation? Can you list the metals you worked with in order of their tendency to gain electrons? Write half-cell reaction equations for the metals you tested on the chart in the Data Summary sheet. Use lead as your reference or “0” voltage, and enter the half-cell equations for the other elements above or below lead on the chart, in order of their tendency to gain electrons.

The choice of lead as a zero on our reduction tendency chart was entirely arbitrary. The important thing is that we used the same reference half-cell throughout all of our measurements. The zero chosen by physical chemists is hydrogen gas undergoing the reversible reaction shown in Equation 5.

2 H 2e– H2 (5)

This is a difficult reaction to implement, thus we chose lead as an alternative standard. With respect to the hydrogen electrode, lead has a half-cell potential of –0.126 V.

Optional: Try iron metal and ferric ion (Fe3), again using the lead reference half-cell. Can you position this ion on your reduction potential chart?

Part III. Predicting Cell VoltageSuppose you were to make an electrochemical cell with copper and zinc half-cells. Using the information you have compiled, which way would you expect the electrons to flow? Which half-cell would contain the oxidation reaction? Which would contain the reduction reaction? Can you predict the force exerted upon the electrons flowing from one half-cell to the other? Set up this cell and check the voltage against your pre- dictions. Make predictions and measurements for cells pairing copper with silver and zinc with silver. Use the table on the Data Summary sheet to record your predictions and observations.

Part IV. ome mall Research Problems Explain your observations concerning the two small problems outlined below, using chemical equations and cell potentials to substantiate your explanation.

1. Place about 30 mL of copper sulfate solution (not copper nitrate) into a clean beaker. Dip clean strips of Zn, Cu, Pb, and Ag into the beaker for about 20 seconds each. Observe and explain. You may need to wipe each strip with a paper towel to remove fingerprints before placing it into the solution.

2. Your lab instructor will put a small coil of copper wire in a silver nitrate solution. Observe what happens after a few minutes and explain your observations.

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Level II. Effects of Concentration ( ptional)The tendency of an electrochemical reaction to proceed is determined by the relative electron-attracting abilities of the two elements or ions involved, as indicated by their positions in the electrochemical series. It is also determined by the concentration of the participating ions. This effect can be used to determine the concentration of ionic solutions. It is most commonly used to measure the concentration of hydrogen (hy- dronium) ions with a pH meter.

In Part II you measured the voltage produced by a cell consisting of a lead/lead ion half-cell and a copper/copper ion half-cell. The concentrations of both solutions were0.1 M. Measure the voltages for cells using copper metal with each of the other avail- able concentrations of copper nitrate solutions. Use lead and 0.1 M lead nitrate for the reference half-cell in each measurement. (Note that you must change filter paper for each experiment.) How much does the cell voltage change for each factor of ten change in ion concentration?

Try this experiment again using silver and 0.1 M silver nitrate for the reference half-cell. Prepare a cell and measure the voltage using copper metal with each concentration of copper nitrate. Compare the results you obtain with the Ag/Ag reference half-cell to those obtained with the Pb/Pb2 reference half-cell. Attach a table of your data and a brief discussion of your results to the Data Summary sheet.

Practical Applications: Environmental Aspects of ElectrochemistryThere are several factors one must consider when selecting building materials. Four important factors are physical strength, appearance, resistance to corrosion, and cost. For example, one might not choose the same material for a bridge structure as for a set of tableware. While physical strength is a problem for the engineer, and cost is simply an economic problem, the problem of weather resistance is an area in which electro- chemistry can provide some guidance. Corrosion of metals is most commonly due to oxidation reactions with atmospheric oxygen, water, or pollutants present in the air. Iron is oxidized slowly by reaction with oxygen to form ferric oxide, Fe2O3. Equation 6 shows this oxidation half reaction:

Fe0 → 2e– Fe2 → e– Fe3 (6)

Ferric oxide is a red-brown material commonly known as rust. Since this oxide tends to flake off, the metal beneath is continually exposed to air, and eventually the metal object is ruined. One can prevent this sort of damage by preventing contact between the iron and the atmosphere. The most common approach to this problem is to paint the surface.

The electrochemical series orders metals and their ions in terms of their tendency to gain or lose electrons. This information places us in a good

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position to predict the behavior of these materials and to choose materials that will exhibit the properties required for a certain job. Several elements are shown in Figure 6. Let us consider their tendency to gain or lose electrons and some common uses for these materials.

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+1.5 Au3+ ++1.2 Pt2+ +

+0.78 Ag+ ++0.77 Fe3+ +

+0.33 Cu2+ +0 2H+ +

-0.44 Fe2+ +

-0.76 Zn2+ +

3e Au0

2e Pt0

1e Ag0

1e Fe2+

2e Cu0

2e 0

2e Fe0

2e Zn0

-1.66 Al3+ + 3e Al0

-2.37 Mg2+ + 2e Mg0

Figure 6. Electrochemical properties of some common materials.

Aluminum is a commonly used building material because of its strength, light weight, low cost, and weather resistance. One would not predict good weather resistance by observing its position in the electrochemical series. However, the oxide coating produced when aluminum reacts with oxygen sticks tightly to the metal, effectively protecting the surface from further oxidation. This oxide can rub off leaving a black deposit on hands or clothing. The metal surface will rather quickly reoxidize. Sometimes the surface of aluminum is deliberately oxidized or “anodized” to form a thick protective coat.

Zinc finds a good deal of use because of its low cost and its great tendency to lose electrons. Galvanized garbage cans and rain gutters are made of iron or steel coated with a thin layer of zinc. The zinc, when exposed to the atmosphere, will rapidly form a protective oxide layer. If the zinc layer is scratched through to expose the iron, the zinc will still preferentially oxidize and protect the iron as long as any zinc remains in the elemental state.

Iron is a widely used structural material because of its strength and low cost. Iron does oxidize (rust) rapidly when exposed to atmospheric conditions, but it is commonly protected by galvanizing or painting.

Copper is used extensively in the electrical industry because of its excellent electrical conductivity. At one time it was also used as a roofing material. Copper oxidizes more slowly than does iron, for example, and its oxide coat protects the remaining copper from further oxidation.

Silver is used in jewelry, tableware, and photography. As anyone who has cleaned silver- ware knows, silver slowly forms an oxide coating. However,

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since its tendency to oxidize is much less than the metals discussed above, a clean piece of silver will remain shiny for

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a reasonable period of time. Although silver is the best metallic electrical conductor, its principal industrial use today is in the field of photography. Here, advantage is taken of the ease with which silver may be reduced from its ionic form (Ag in compounds such as silver bromide) to metallic silver, Ag0. It is this reduced (metallic) silver that forms the image on black and white photographic film and prints.

Platinum exhibits considerably greater attraction for its electrons than silver does and is usually found in its elemental state instead of in an ore. Its great resistance to oxidation by other substances contributes to its usefulness in jewelry and in scientific apparatus.

Of the materials presented in Figure 6, gold has the tightest hold on its electrons and will remain bright after long exposure to atmospheric conditions or perspiration. For this reason it is highly prized as a material for rings and other jewelry. Gold has so little tendency to lose electrons that, unlike most of the other materials discussed earlier, it is commonly found in the elemental state in nature. Thus, one hears of “panning for nuggets of gold,” but never of panning for nuggets of zinc.

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ate of Lab Meetingartner’s Name if applicable

Course/SectionInstructor

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Data SummaryPart I. A Forced Electrochemical ReactionBriefly describe what you observed in this experiment. What part did the electrons play in this reaction?

Part II. The Electrochemical eriesEnter your voltage measurements in the table below for each half-cell when paired with a lead/lead ion reference half-cell.

Half-cell Reduction Reaction

Cell Voltage Compared to

Lead Half-cell

xidation or Reduction as Compared

to Lead Half-cell

Cu/Cu0 Cu 2e → Cu0

Zn/Zn0 Zn 2e → Zn0

Ag/Ag0 Ag 1e → Ag0

Fe3 Fe0 Fe3 ?e → Fe?

Data table for determination of the electrochemical series. The reactions shown in this table may proceed in either direction (reduction or oxidation), depending upon the identity of the reference half-cell.

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Enter the reaction equations for the metal/ion half-cells at the appropriate location on the chart below.

Reference Pb2+ + 2e- Pb0

Part III. Predicting Cell VoltageEnter your data in the table below.

Half-cell Pairs

xidation Reaction

Reduction Reaction

Predicted Voltage

bserved Voltage

Cu and Zn

Cu and Ag

Zn and Ag

Part IV. ome mall Research Problems List your observations and conclusions here.

1. The reaction between Cu2

and... Zn0

Cu0

Pb0

Ag0

2. The reaction between Ag and Cu0

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