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Chemistry: Oxidation-Reduction Reactions


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About this Lesson

  • Type: Video Tutorial
  • Length: 10:49
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 116 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Reactions in Aqueous Solutions (10 lessons, $14.85)
Chem: Solutions: Precipitation, Acid-Base, & Redox (3 lessons, $4.95)

This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

About this Author

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Founded in 1997, Thinkwell has succeeded in creating "next-generation" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technology-based textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

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wonderful experience
~ dorothy4


wonderful experience
~ dorothy4


I grew up in California, and about ten years ago moved to Virginia. And one of the first things I noticed was that on the roads there weren't the kind of classic, vintage automobiles that I saw so commonly in California. First I thought that Virginians just had no class, but I realized that there was an entirely different reason. And that is that cars on the East Coast have a much more difficult time because of rust and the salt that's used on the roads. The rust would eat up the cars and so the cars would need to be replaced. And they just wouldn't last as long. So rusting cars and rusting iron, in general, is one of the many, many examples of what we call a redox reaction, a reaction in which there's a reduction and an oxidation simultaneously.
So let's take a couple of minutes now and carefully define exactly what we mean as chemists by a redox reaction. To do this let's start out with a demonstration. I have here a solution of silver nitrate, which we've learned in previous lectures is silver ions and nitrate ions hydrated in solution. And I want to take a piece of copper metal and just simply put it into the silver solution. And let's just watch what happens. What you'll see starting to happen here are two things. One is that the copper is being coated by almost a black or metallic coating. That coating is actually silver metal. At the same time, as we watch this continue, we're going to see a darkening of the solution, and the solution eventually will turn a pretty blue color. In fact, let's make believe that we're forwarding in time here. I'm going to simply replace this by an experiment that I started before we started to roll the cameras. This is what it would look like 10 minutes, 15 minutes from now. You'll notice that there's a large amount on the bottom of a very shiny metallic material, almost crystalline like. I can see that the light is catching these crystals of silver metal. And notice that the solution is now a beautiful blue color. That pretty blue indicates the formation of a Cu^2+ ion. So let's look at the molecular level now on what we're actually looking at.
What happened in this reaction is that two silver cations reacted with copper in the metallic form to give us metallic silver and copper now in the 2+ oxidation stage, or the 2+ form. If we look at what happened to silver, silver went from a 1+ cation to a neutral material. So think about that. In order for it to lose charge, it had to gain electrons. Because remember the charge of the electron is a minus 1 charge. So silver gained additional electrons, and we call that process reduction.
In contrast, if we look at what happened to the copper, the copper went from a neutral species to a 2+ cation. And so the copper actually had to have lost 2 electrons, once again because of the negative charge on the electron in order to give us the overall 2+ copper product.
We can break this reaction down further and consider individually the story of each one of our reactants. We have, on the one hand, silver going from Ag^+ to Ag^0. And in fact, in order for that equation to be balanced, we needed 2 equivalents of silver from Ag^+ to Ag^0. So silver plus 2 electrons takes us to Ag^0. Notice that the charges balance on this. This is our reduction step. The silver we say is being reduced. And our oxidation step is Cu^0 going to Cu^2+, plus 2 electrons. We refer to these as half reactions. Neither one of them makes any physical sense, but it's a convenient tool for us to divide the chemical reaction in half, and consider what is happening to each one of the partners in this electron trading process, if you will.
Although the copper is being oxidized, think about the role it is playing. It is actually acting as the reducing agent in that it is giving away its electrons. That might strike you as odd. Why would it be reducing if it's giving away its electronics? But think about what's happening. Let's go back to the reduction. Silver in this case is being reduced, although it is gaining electrons. It's being reduced though. We use that term reduction because the charge of the electron is negative. So what's actually being reduced here is not the number of electrons, but the charge of the overall species going from plus to neutral. So silver in this case is being reduced in that it is accepting electrons. It's accepting the negative charges, but it is acting as the oxidizing agent. It is the thing responsible for oxidizing the copper. That again can be a confusing juxtaposition of terms here. The reduction step involves silver, but silver at the same time is acting as the oxidant, because it is removing the electron, whereas copper is participating in an oxidation. And it's doing so by acting as the reducing agent, or the reductant. It's the thing giving away electrons.
So let's go to a second example, and this will give us a chance to get a little more practice becoming familiar with these terms. I'm going to now look at the product of the following two half reactions, magnesium metal going to Mg^2+, and hydrogen ion, H^+, going to hydrogen gas. So what I'm going to be doing now in order to do this reaction is I have a solution of hydrochloric acid. We know from past units hydrochloric acid is H^+Cl^- in solution. And I'm going to add to the hydrochloric acid a strip of magnesium ribbon, just metallic magnesium. Now right away, as we zoom in on this flask, you can see the bubbles coming off of the magnesium. That's liberating hydrogen gas. Meanwhile, the magnesium is being consumed. What's that you say? How do we know that that's really hydrogen gas? Watch. We know one thing about hydrogen gas. We know that it is reactive in air, that it ignites in air. Let's make sure that this really is hydrogen. Gives you a startle, especially if you're right here. Let me tell you. That sound you heard was the rushing in and rushing out of gas as we got this ignition of the hydrogen and the oxygen in the atmosphere. I'll leave that there.
In just a moment we'll see that all of the magnesium is going to be completely consumed by the acid. Meanwhile, let's review. The magnesium in this reaction is being oxidized. It's participating in the oxidation step in that it is losing electrons. Remember, once again, oxidation is the losing of electrons. Meanwhile, the hydrogen, in this case the one acting as the oxidant, is in fact being reduced, going from a +1 to a neutral state. We can combine these two half reactions to give us the overall reaction of what's going on here. And notice I've included in this case the spectator ions, so these are actually going to be part of the participating reactions. So the overall reaction that we did was magnesium plus hydrochloric acid goes to Mg^2+. We still have the chloride remaining. And notice now that cancels the charge of the magnesium, and then the hydrogen gas. We also could write this in a more simplified form, removing the two chlorides on either side of the equation, the spectator ions in other words, and just talk again about the business of the reaction, which is MgH^+ goes to Mg^2+ plus hydrogen gas, once again the oxidation step and the reduction step that we've identified before.
Finally, coming full circle, let's talk again about rusting cars, and more generally the reaction of oxygen with iron. Indeed the term oxidation originally came from the notion that oxygen in our atmosphere was capable of removing electrons, was capable as acting as an oxidant. So let's look at how that happens when we consider the reaction again of O[2] and iron. Notice first of all the fate of the iron. It goes from a neutral form, from elemental iron, initially to iron in the 2+ oxidation state. It's having electrons removed. Where are those electrons going? To oxygen of course, oxygen is going from a neutral state, and it's accepting electrons. Each oxygen atom is gaining two electrons that it took from the iron. In the oxygen in the final product, which is water, the oxygen now formally is oxygen 2-. We don't write it as oxygen 2-, because 0^2- is a very unstable ion, but it's accompanied by two hydrogens. And they neutralize that charge on the oxygen. Nonetheless, formally, we now consider this as an oxygen in the 2- oxidation state. And we'll talk more about oxidation states very soon.
So what have we learned? We've learned that in addition to precipitation reactions and acid-base reactions in aqueous solution, a very common reaction for aqueous solution is the redox reaction. And we can characterize a redox reaction as a reaction in which the oxidation state changes, in other words, reactions where electrons are transferred from one species to another.
Reaction in Acqueous Solutions
Reactions Involving Solutions
Oxidation-Reduction Page [2 of 2]

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