|
Abstract Solar cells are popping up on rooftops everywhere these days and are a model for clean, renewable energy. Did you ever look at those solar panels and wonder how we can get electricity produced by solar cells when the sun is not shining? It is a great question because solar panels do not produce electricity when it is dark outside. One strategy to overcome this challenge is to store the energy produced by solar cells during the day in the form of a fuel that can be used at a later time. In this project, you will explore a cutting-edge method for storing renewable energy by breaking up water molecules into hydrogen and oxygen. The hydrogen and oxygen are fuels that can be burned in devices such as fuel cells to produce clean electricity when it is dark!Objective Examine water's usefulness as a renewable energy source by observing how efficient a cobalt-based catalyst can be at helping to form molecular oxygen.Introduction The sun is setting on a brisk fall evening, and all of a sudden your TV turns off right in the middle of your favorite show! Without the sun, the solar cells on your roof cannot generate electricity to run your appliances. That is not a good scenario! Many forms of renewable energy, such as solar energy and wind energy, are not available at all times of the day and night. These renewable energies are intermittent sources of energy. We, as a society, however, need energy to be available at all times. For this reason, renewable energy sources like solar energy pose particular challenges to engineers if they are to be used by power plants that generate electrical energy for your house, or for your car in the place of gasoline. Currently, power plants that supply energy to your house run on coal, natural gas, or gasoline as a fuel source and can do so 24 hours a day, 7 days a week. However, these fuels are carbon-based, meaning that they can generate pollutants. Additionally, fossil fuels are unsustainable, meaning that they will eventually run out. Using clean, renewable energy would be a solution to these problems. Sunlight is a form of renewable energy that is virtually limitless: more solar energy strikes the earth each hour than the entire world uses in a year! However, for solar energy to become practical, we need a cheap and efficient method of storing the solar energy for when the sun does not shine. One method to achieve efficient storage of solar energy is in chemical bonds. Specifically, what scientists and engineers seek to do is use the energy from the sun to rearrange low energy bonds to form high energy bonds. The high energy bonds can be used later to deliver the energy back to us when we need it. This concept is comparable to when you charge a reusable battery. You get electricity from the wall socket, store it in the battery, and when you need to run your phone or computer, the energy is available. In fact, society is already familiar with this concept of energy storage in chemical bonds in the form of fossil fuels (like gasoline and natural gas). By burning these fuels, energy contained within those high-energy bonds is released (and used by humans) along with carbon dioxide (CO A primary goal for renewable energy research is to develop fuel storage methods that are scalable, sustainable, and do not used carbon-based fuels. The ideal energy cycle would be a fully renewable one where the fuels can be utilized (burned to extract the energy from the high-energy bonds) and the waste products (low energy bonds) can be captured and reused to form the same fuel again. One attractive approach for renewable energy storage is to use solar energy to drive the rearrangement of water (low-energy bonds) into molecular hydrogen gas and oxygen gas (high-energy bonds). This transformation, often called water splitting, provides attractive alternatives to hydrocarbon fuels because combustion of hydrogen fuel (which is really the "burning" of hydrogen fuel, H Achieving efficient "splitting" of water requires a catalyst that assists oxygen-oxygen bond formation between two oxygen atoms derived from water. A catalyst is a material or molecule that increases the rate of a reaction between other starting materials, but is not used up in the reaction. Since the catalyst increases the reaction rate, less energy input is required to produce the product. This means that with the same amount of energy available, the product can be produced at a faster rate—it is more energy efficient. Equation 1:
Consider the reaction in Equation 1 above. With a catalyst present, we do not need as much energy to make C as we would have needed if the catalyst was absent, and with a catalyst present, we can make C much more quickly. One example to help explain this is to consider that A is a taxi driver and B is a person at the airport. The goal of the "reaction" is to get A and B together so that the taxi driver can take the passenger to a desired location called C. Without a catalyst, both the taxi driver and the person would wander around the airport, requiring a large amount of energy (through walking or driving) and time to bump into each other. However, if there is a taxi stand, acting as the catalyst, the taxi driver and the person would be able to find each faster with less energy wasted. One fuel-forming reaction that interests scientists and engineers is the splitting, or breaking up, of water to form molecular hydrogen and oxygen using light from the sun. This chemical reaction is shown below in Equation 2. The molecular hydrogen and oxygen can then be stored separately, and later brought together as fuel in what is called a fuel cell. Equation 2:
Sun-light does not directly act on water to cause it to split into these elemental components. Hence catalysts are needed to effect the overall transformation. In Nature, the water-splitting reaction is accomplished by the process of photosynthesis inside of the leaf. Outside of the leaf, solar fuels other than hydrogen may be produced with the protons and electrons extracted from water, including the reduction of carbon dioxide to methanol. However, all water-splitting schemes require oxygen (O2) production. This turns out to be a very inefficient step and a barrier to using artificial photosynthesis (the splitting of water into H2 and O2) as a renewable energy. Thus the focus of many research programs involves the design and development of catalysts that assist water oxidation to molecular oxygen (O2).
A catalyst formed from cobalt ions (Co) in a phosphate (Pi) buffered solution (Co-Pi) has recently been discovered, in Dr. Nocera's lab at MIT, that is capable of water oxidation to O2 at low overpotentials and high rates. The catalyst is comprised of inexpensive, earth-abundant elements, and its formation is robust under a host of conditions. See the video below for more details about the discovery of the Co-Pi catalyst. In this science project, you will construct an experimental set-up that will allow you to synthesize this cutting-edge cobalt-based catalyst. You will perform measurements to determine how much the catalyst reduces the energy needed for the water-splitting reaction to form oxygen. Does it sound too challenging? If you watch the video below you’ll see that setting up the experiment really isn’t that hard. Especially if you follow all the steps in the Experimental Procedure! Once you get the hang of it, you can use your set-up to try to discover your own new and improved catalysts! Maybe you'll find the one that we'll all be using someday to power the planet.
Terms, Concepts, and Questions to Start Background Research
Questions
Bibliography This resource will give you more information about catalysts:
More information about renewable energy can be found at these websites:
More information about water splitting and the original discovery of the Co-Pi catalyst can be found using these resources:
For more information about setting up a circuit on a breadboard, or electronics terms, use this primer:
Materials and Equipment
Disclaimer: Science Buddies occasionally provides information (such as part numbers, supplier names, and supplier weblinks) to assist our users in locating specialty items for individual projects. The information is provided solely as a convenience to our users. We do our best to make sure that part numbers and descriptions are accurate when first listed. However, since part numbers do change as items are obsoleted or improved, please send us an email if you run across any parts that are no longer available. We also do our best to make sure that any listed supplier provides prompt, courteous service. Science Buddies receives no consideration, financial or otherwise, from suppliers for these listings. (The sole exception is any Amazon.com or Barnes&Noble.com link.) If you have any comments (positive or negative) related to purchases you've made for science fair projects from recommendations on our site, please let us know. Write to us at scibuddy@sciencebuddies.org. Shop for Supplies at Science Buddies Online Store
|
||||||||||||||||||||||||||||||||||||||||||||||||
| Note Before Beginning: This science fair project requires you to hook up one or more devices in an electrical circuit. Basic help can be found in the Electronics Primer. However, if you don't have experience in putting together electrical circuits you may find it helpful to have someone who can answer questions and help you troubleshoot if your project isn't working. A science teacher or parent may be a good resource. If you need to find another mentor, try asking a local electrician, electrical engineer, or person whose hobbies involve building things like model airplanes, trains, or cars. You may also need to work your way up to this project by starting with an electronics project that has a lower level of difficulty. |
![]() ![]() |
| Figure 1. The photo on the left (1a) shows how the various components in the circuit should be wired onto the breadboard. The completed circuit, including breadboard and multimeter leads, is shown in the photo on the right (1b). |
![]() ![]() |
![]() ![]() |
| Figure 2. The photos above show one possible method for suspending the nickel electrodes in the jar to make the electrochemical cell. |
![]() |
| Figure 3. The completed galvanostatic electrochemical cell is pictured above on the top (3a) and represented as a schematic on the bottom (3b). |
| Technical Note #1 | |||||||||
|
By following steps 1-8 above you have constructed a simplified galvanostat. This means that the electrochemical cell passes the same current through the cell at all times, and the voltage read-out varies based on the efficiency or property of the electrodes. The four 9-volt batteries generate a maximum of 36 (9v per battery x 4 batteries in series =36v), you typically will get less than 36 volts depending on how fresh the batteries are. The circuit is completed by two other resistors attached in series. One resistor is the electrochemical cell itself (the nickel electrodes in the phosphate buffer), and the other resistor is a 10,000 Ohm resistor you specifically place in series. This 10,000 Ohm resistor is critical to stabilize the electrochemical cell and ensure that a constant current is passed at all times. The reason for this is that most of the voltage (~30v) drops across the 10,000 Ohm resistor, and approximately 1.5-3v are dropped across the electrochemical cell. It is important to drop most of the voltage through the resistor because this will set the current that passes through the rest of circuit. Small variations in the electrochemical cell will have little effect because the 10,000 Ohm resistor is the dominant factor. Using Equation 3 below, and assuming that approximately 30 volts are dropped over the 10,000 Ohm resistor, the current can be calculated to be 3mA (30v / 10,000 Ohms = 0.003 A = 3mA). This calculation indicates there are 3mA of current flowing through the electrochemical cell. Equation 3:
| |||||||||
| Technical Note #2 | ||||||||||||||
|
The voltage readout you measure in step 2, above, is the voltage required by the electrochemical cell to maintain a constant current of 3mA (see Technical Note #1 to learn how the current was calculated). This voltage is the sum of the energy required to drive the water-splitting reaction Equation 4:
If the reaction was 100% efficient, it would require only 1.23v to maintain the 3mA of current passing through the cell. The voltmeter/multimeter would thus read 1.23v. As an example, let's imagine that the initial voltmeter/multimeter reading (in step 2 above) for your electrochemical cell was 2.46v. According to Equation 4, this would mean that without the catalyst, the reaction was 50% efficient (1.23v / 2.46v = 0.50 = 50%). If voltage readings after the addition of the catalyst dropped to 2.12v, then the efficiency would rise to 58% (1.23v / 2.12v = 0.58 = 58%). | ||||||||||||||
![]() |
| Figure 4. As shown in the picture above, only a small amount of cobalt nitrate should be added to the buffer at a time. |
![]() ![]() |
| Figure 5. In the beginning, as shown in photo on the left, there are no bubbles on the nickel electrodes. As the reaction proceeds, the gases formed can be seen as tiny bubbles covering the electrodes as shown in the photo on the right. |
| Technical Note #3 |
| Above, we saw that the efficiency of the reaction is determined by the voltage drop across the electrochemical cell. Higher voltages lead to lower efficiency. But what about the speed at which we produce hydrogen and oxygen? Above, we were running our cell at 3mA, and this current is directly proportional to the rate of hydrogen and oxygen production. What if we slow this rate to 1mA, 0.1mA, or 0.01mA? How would that affect the voltage? Knowing the relationship between the current (rate) and the voltage (energy input) is essential to designing water splitting systems that are practical. As described in Variation 1, change the resistor so that the current is varied between 3mA and 30mA. Plot the cell voltage as a function of the log of the current passed in the cell. Do you get a straight line? What is the slope? Repeat this experiment with catalysts formed for different metal salts (see Variation 2). Plot all of your results on the same graph to determine which catalyst performs the best. Remember that lower voltages and higher currents equal better catalysts. |
Variations
Credits
Elizabeth R. Young, Ph.D., MIT
Yogesh Surendranath, MIT
Thomas Teets, MIT
Edited by Sandra Slutz, Ph.D., Science Buddies
Last edit date: 2011-11-04 12:00:00
If you like this project, you might enjoy exploring related careers.
![]() |
Chemist Everything in the environment, whether naturally occurring or of human design, is composed of chemicals. Chemists search for and use new knowledge about chemicals to develop new processes or products. |
![]() |
Chemical Engineer Chemical engineers solve the problems that affect our everyday lives by applying the principles of chemistry. If you enjoy working in a chemistry laboratory and are interested in developing useful products for people, then a career as a chemical engineer might be in your future. | |
![]() |
Electrical & Electronics Engineer Just as a potter forms clay, or a steel worker molds molten steel, electrical and electronics engineers gather and shape electricity and use it to make products that transmit power or transmit information. Electrical and electronics engineers may specialize in one of the millions of products that make or use electricity, like cell phones, electric motors, microwaves, medical instruments, airline navigation system, or handheld games. |
![]() |
Fuel Cell Engineer Most of the world's energy comes from fossil fuels. However, the amount of fossil fuels is finite, and many people are concerned about where our energy will come from in the future. We can turn to alternative, renewable sources of fuel, such as our sun (solar energy) and the winds (wind energy). But what happens when the sun doesn't shine or the winds don't blow? Would we be stuck? Well, that is where the fuel cell comes in. A fuel cell is an electrochemical device that generates electricity through a reaction between a fuel, like hydrogen, and an oxidant, like oxygen. This reaction produces few greenhouse gas emissions other than water or water vapor. The job of the fuel cell engineer is to design new fuel cell technology that improves the reliability, functionality, and efficiency of the fuel cell. Do you like the idea of using your math and science skills to work on mankind's future energy needs? Then start "fueling your future" and read more about this career. | |
|
Join Science Buddies
Become a Science Buddies member! It's free! As a member you will be the first to receive our new and innovative project ideas, news about upcoming science competitions, science fair tips, and information on other science related initiatives. |