Student Behavioral Objectives
- The students will describe the conditions necessary for a cell to be standard cell.
- Given a table of standard reduction potentials and a diagram or description of a Galvanic cell, the students will balance the redox equation, calculate the standard cell potential, and determine the direction of electron flow in the external circuit.
Timing, Standards, Activities
Timing and California Standards
Number of 60 min periods
Activities for Lesson 5
1. An Activity Series Lab
1. Electrochemical Cells Worksheet
1. Dependence of Cell Potential on Concentration
Answers for Galvanic Cells (L5) Review Questions
Sample answers to these questions are available upon request. Please send an email to email@example.com to request sample answers.
Multimedia Resources for Chapter 23
Gummy Bear Terminator Demo.
Electrochenical Cell animation.
This web site offers text material on the demise of the phlogiston theory and some history about some early chemists.
This website offers a short animated video on electron transfer during oxidation-reduction.
This web site offers more information on oxidation-reduction and other aspects of electrochemistry.
This video shows the process in balancing a redox equation.
The following website provides more information in balancing redox reactions.
This website provides an interactive animation of electrolysis.
This animation shows the function of a galvanic cell.
The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos. You are required to register before you can watch the videos but there is no charge. The website has one video that relates to this lesson called “The Busy Electron” that explains the principles of electrochemical cell design through batteries, sensors, and a solar-powered car.
The following provides text and animations about electrochemical cells and batteries.
Laboratory Activities for Chapter 23
Teacher’s Pages for An Activity Series Lab
Metal sheets can be purchased from chemical supply companies. For this lab, only the thinnest sheets are necessary. Metal foils can be cut with sturdy scissors. Thicker sheets may require tin snips. Any surface corrosion must be cleaned off for this lab. That can be accomplished with emery cloth. It’s much easier to sand the sheets before cutting them into squares. After sanding the metal sheets, rinse and dry them.
Nitrate compounds are often purchased as hydrates. Make sure you note if your , , , are hydrates when calculating masses for making solutions.
You can, of course, make water, water, and water, but it is much more convenient to purchase the amount you will need previous to the lab. These solutions do not store well. Purchase the amount you need year by year to guarantee strength.
If you think your students will not have enough time to complete both parts of the lab, you can have them do part I on one day, and part II on the next day. Another way to shorten the time for the lab is to make the reference solutions for part II yourself and have them on display. Just follow the first four steps in the Procedure for Part II.
Answers to Pre-Lab Questions
1. Write oxidation half-reactions for the four metals, , , , .
2. Write reduction half-reactions for the three halogens, Cl2, Br2, and I2.
Answers to Post-Lab Questions
1. Write a balanced net ionic equation for all the reactions that occurred with the metals.
2. List the four oxidation half-reactions written in the pre-lab question 1 in order of decreasing ease of oxidation. That is, the one easiest to oxidize should be first and the most difficult to oxidize should be last.
3. Write reduction half-reactions for the four ions in the experiment and put them in order of easiest to reduce first and most difficult ion to reduce last. This list should be in reverse order of your activity series in question 2.
3. Find a list of standard reduction potentials (in your textbook or elsewhere). How does your list in question 3 compare to the standard reduction potential list?
The metal ions are in the same order of ease of reduction.
4. Suppose a piece of shiny silver metal had been placed into separate solutions of each of the metallic ions in the lab and did not react with any of them. Write an oxidation half-reaction for silver and place in its proper place in your metals activity series.
Post-Lab Questions for Part II
1. Why is it assumed that the halide ions will not dissolve in mineral oil?
Halide ions are charged and will dissolve readily in water but not in a non-polar solvent like mineral oil.
2. Write net ionic equations for all the reactions that occurred with the halogens.
3. List the reduction half-reactions of the halogens in order of decreasing activity.
An Activity Series Lab
A ranking of elements according to their reactivity is called an activity series. For example, an activity series containing the elements calcium, gold, and iron would put the most reactive element of the three, calcium, at the top of the series; iron in the middle; and the least reactive, gold, at the bottom. The more reactive metal is the one that is most easily oxidized, so half-reactions for an activity series of metals would be written in terms of oxidation. The activity series for the three metals mentioned earlier would look like this.
This activity series could be determined experimentally by placing each of the metals in solutions containing the ions of the other two substances and observing which ones react. For example, we could place pieces of calcium metal into separate solutions containing ions and ions. If the reactions occur, the equations would be:
In this case, both reactions occur. The reactions would be detected by the observer because the shiny surface of the calcium metal would darken. In reaction 1, there exists a competition to give up electrons between the calcium atoms in the reactants and the iron atoms in the products. Since this reaction does occur, we know the calcium atoms are stronger at giving up electrons than iron atoms and the reaction runs forward. If the iron atoms were stronger at giving up electrons, the reaction would not occur. The calcium atoms could not transfer electrons to iron ions if the iron atoms so formed were stronger at giving up electrons. The argument for reaction 2 is the same. Since the reaction runs forward, the calcium atoms are stronger at giving up electrons than gold atoms. The fact that calcium atoms gives electrons to both these ions is the reason that calcium is placed on the top of the activity series.
Suppose we now place shiny pieces of iron metal into different solutions containing calcium ions and gold ions. If the reactions occur, the equations would be:
In this case, reaction 3 does not occur. That is, the shiny piece of iron metal remains shiny. No change occurs in this mixture. Iron atoms cannot give electrons to calcium ions, because calcium atoms are stronger at giving electrons than iron atoms. Reaction 4, on the other hand, does occur. The shiny surface of the iron metal darkens and a reaction is apparent. Iron atoms can give electrons to gold ions because iron is stronger at giving electrons than gold, iron is MORE REACTIVE. Therefore, in our activity series, iron must go below calcium but above gold.
The third situation would be to place shiny pieces of gold metal into different solutions of ions and ions.
In this case, neither reaction occurs. This is because both calcium atoms and iron atoms are stronger at donating electrons and therefore, gold cannot give electrons to either of these metals. That is the reason that gold is placed on the bottom of our activity series.
- Write oxidation half-reactions for the four metals, , , , .
- Write reduction half-reaction for the three halogen, , , and .
The purpose of this experiment is to determine an activity series for four metals. If your teacher has you carry out Part II, you will also be determining an activity series for three halogens. Part II makes use of a solvent extraction technique that will be described at the beginning of Part II.
The series of metals to be studied are copper, zinc, magnesium, and lead. Solutions of metal nitrates for each of these metals are placed in reaction wells. A piece of each metal is then placed in the other metals’ nitrate solutions and observed to see if any reaction occurs. If a metal reacts with another metal nitrate, then the solid metal has reduced the other metal ion and is, therefore, the more reactive metal of the two. By comparing the results of 16 different reactions, the five metals are ranked from most reactive to least reactive.
Materials and Materials (per lab group) for Part I
- 4 - copper foil, squares
- 4 - zinc foil, squares
- 4 - magnesium foil, squares
- 4 - lead foil, squares
of copper(II) nitrate solution,
of zinc nitrate solution,
of magnesium nitrate solution,
of lead(II) nitrate solution,
- 24-well reaction plate
- 5 – beral-type pipets
- Chemical forceps
- Stirring rod
- emery cloth
All solutions are irritating to skin, eyes, and mucous membranes. Handle solutions with care, avoid getting the material on you, and wash your hands carefully before leaving the lab.
Procedure for Part I
- Place the 24-well plate on top of a sheet of white paper (easier to see changes). Set the plate so there are 6 wells across and 4 wells down. Refer to the figure above to see how the wells are arranged and marked. Note that each well is identified by a unique combination of letter and number, where the letter refers to a horizontal row and the number to a vertical column. Make sure your well plate matches this one.
- Put one dropper full (about 15 drops) of copper(II) nitrate in wells B1, C1, and D1.
- Put one dropper full of magnesium nitrate in wells A2, C2, and D2.
- Put one dropper full of lead(II) nitrate in wells A3, B3, and D3.
- Put one dropper full of zinc nitrate in wells A4, B4, and C4.
- Put a small piece of shiny copper metal (you many need to buff it with emery cloth first) in each of the wells containing a solution in the first row (A).
- Add shiny Mg metal to the solutions in the second row (B), add shiny Pb metal to the solutions in the third row (C), and add shiny Zn metal to the solutions in the fourth row (D). Use a stirring rod to submerge the pieces of metal in the solutions (but don’t carry over solution from one well to another). Allow to stand for at least .
- Determine if reactions occurred in each well by observing if a new metal has been deposited or the surface of the metal has become coated or discolored.
The pieces of solid metal go into waste baskets. All solutions go down the sink followed by plenty of water. Wash the well plate with soap and tap water and then either dry it or rinse it in distilled water.
Data for Part I
- Write a balanced net ionic equation for all the reactions that occurred with the metals.
- List the four oxidation half-reactions written in the pre-lab question 1 in order of decreasing ease of oxidation. That is, the one easiest to oxidize should be first and the most difficult to oxidize should be last.
- Write reduction half-reactions for the four ions in the experiment and put them in order of easiest to reduce first and most difficult ion to reduce last. This list should be in reverse order of your activity series in question 2.
- Find a list of standard reduction potentials (in your textbook or elsewhere). How does your list in question 3 compare to the standard reduction potential list?
- Suppose a piece of shiny silver metal had been placed into separate solutions of each of the metallic ions in the lab and did not react with any of them. Write an oxidation half-reaction for silver and place in its proper place in your metals activity series.
Background for Part II
A similar activity series can be constructed for the halogens. The most active halogen is the most easily reduced, accepting electrons from the less active species and replacing it in the halide salt. For example, let and represent two halogens. The sodium salts of these halides would be and . In solution the would produce ions and would produce ions. If were the more active halogen, then placing in a solution containing ions would result in a reaction in which received electrons from ions. (Working with both elemental halogen gases and the halogen ions can be confusing, so in the procedure, the halogen gases are referred to as elemental halogens.)
In water solution, none of the halogen gases or ions are darkly colored enough for an observer to determine exactly what is present in the final solution. The halogen gases, however, do have distinctive colors when dissolved in mineral oil. A technique called solvent extraction is used to separate and identify the products of the reactions in this part of the lab.
Your teacher may have already prepared reference solutions of the halogen gases in mineral oil so that you can see the distinctive colors of , , and when dissolved in mineral oil. You will also want to know if the halide ions produce any color with mineral oil. If the reference solutions are available, you may skip procedure steps 1, 2, 3 and 4. If the reference solutions are not available, you will produce them in steps 1, 2, 3, and 4.
Mineral oil is a non-polar solvent. Halide ions are much more soluble in water (polar) than non-polar solvents. The elemental halogens, , , and , are non-polar molecules and therefore, are much more soluble in non-polar solvents. When water solutions containing both halide ions and elemental halogens are shaken with mineral oil, the elemental halogens present will concentrate in the mineral oil and the ions will concentration in the water. When the water and mineral oil separate (immiscible) the substances dissolved in them are separated. This process is called solvent extraction.
Apparatus and Materials (per lab group) for Part II
- Chlorine water, in water,
- Bromine water, in water,
- Iodine water, in water,
- Mineral oil,
- Sodium chloride solution, , ,
- Sodium bromide solution, , ,
- Potassium iodide solution, , ,
- Beryl type pipets, 7
- Test tubes, , 12
- Stoppers for test tubes, 12
- Test tube rack
Procedure for Part II
- As a reference, test to see what color develops when each halogen (elemental) is dissolved in mineral oil. Place one dropper full of chlorine water, one of bromine water, and one of iodine water, into three separate test tubes.
- Add one dropper full of mineral oil to each test tube, stopper the tube, and shake it for .
- Let the mineral oil rise to the top and record the color that each halogen shows when dissolved in mineral oil. Record your observations in the Part II Data Table.
- Test to see if halide ions give a color to mineral oil by repeating steps 1, 2, and 3 with solutions of , , and with mineral oil. Record your observations.
- Set up six test tubes in a test tube rack. Label the tubes 1 through 6.
- You will now react each elemental halogen with the OTHER two halide ion solutions to determine if the ions reduce the halogens. Place one dropper full of into test tube 1 and one dropper full of into test tube 2.
- Add one dropper full of chlorine water to each of test tubes 1 and 2. Stopper and shake each.
- Add one dropper full of mineral oil to each of test tubes 1 and 2. Stopper and shake each.
- When the mineral oil layer has separated, determine its color and decide whether has reaction has occurred. If the color of chlorine appears in the mineral oil, then no reaction has occurred. If either the bromine or the iodine color appears in the mineral oil layer, then there was a reaction.
- Record both the color and the reaction results for in the Data Table.
- Repeat the test using bromine water. Add one dropper full of to test tube 3 and one dropper full of to test tube 4. Add one dropper full of bromine water to each of test tubes 3 and 4. Stopper and shake each.
- Add one dropper full of mineral oil to each of test tubes 3 and 4. Stopper and shake each.
- When the mineral oil has separated, determine its color and whether or not a reaction occurred. It the color of bromine appears, no reaction has occurred. If either chlorine or iodine color appears, then there was a reaction. Record both color and reaction results in the Data Table.
- Repeat the test using iodine water. Add one dropper full of to test tube 5 and one dropper full of to test tube 6. Add one dropper full of iodine water to each of test tubes 5 and 6. Stopper and shake each.
- Add one dropper full of mineral oil to each of test tubes 5 and 6. Stopper and shake each.
- Record the color of the mineral oil layer and the reaction results for in the Data Table.
Data for Part II
Post-Lab Questions for Part II
- Why is it assumed that the halide ions will not dissolve in mineral oil?
- Write net ionic equations for all the reactions that occurred with the halogens.
- List the reduction half-reactions of the halogens in order of decreasing activity.
Demonstrations for Chapter 23
The Electrolysis of Water
Brief description of demonstration
A solution of sodium sulfate is electrolyzed using either a Hoffman apparatus. With the Hoffman apparatus, the gas collected in the separate tubes has a ratio of 2:1, and the colors in each tube change to yellow and blue. The gas is later collected into test tubes, and appropriate test for hydrogen and oxygen are done.
Apparatus and Materials
sodium sulfate solution,
bromothymol blue solution
- stirring rod
- disposable pipette
- DC power supply capable of delivering at
- Hoffman apparatus
- Graphite or platinum electrodes
sodium hydroxide solution,
- Alligator clip leads for power supply
- 2 test tunes,
- wooden splint and matches
- Ring stand
- Utility clamp
Add the sodium sulfate solution to a large beaker or other clear container. Add of bromothymol blue solution and stir. The solution should be green. If the solution is blue, add a few drops of and stir until the solution turns green. If it is yellow, add a few drops of solution and stir until the solution turns green. This is an indication of a neutral solution. Close the stopcocks on the Hoffman apparatus and fill the bulb with the green solution. Once it is full, open the stopcock of one of the arms slightly, and the arm will fill with solution. Close the stopcock once it is full. Repeat this procedure with the other arm. Connect the wire leads to the electrodes on the Hoffman apparatus, and then connect the leads to the power supply. Turn on the power supply, and adjust the EMF to about . Gas will collect into each arm, hydrogen from the negative side (the cathode) and oxygen at the positive side (the anode). The volume of hydrogen will be twice that of the volume of the oxygen, a nice proof of Avogadro’s Law and the expression of the empirical formula of water. The solution at the anode turns yellow, and the solution at the cathode turns blue, due to the formation of and ions, respectively. After the hydrogen arm is about of the way full, turn off the power supply and disconnect the alligator clips. Perform the standard tests for hydrogen and oxygen by collecting test tubes of each by opening the stopcock of each tube, dispensing each gas into separate test tubes.
Even though this demonstration is run at low voltage, the amperes are high, so there is an electrical shock hazard. The acid and base solutions can cause chemical burns. Use caution. Do not mix the hydrogen and oxygen, as it can cause an explosion hazard.
Pour the solutions down the sink and flush the down with 100 times excess of water.