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Chemistry: Redox Reactions: Half-Reaction Method


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About this Lesson

  • Type: Video Tutorial
  • Length: 11:32
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 124 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Final Exam Test Prep and Review (49 lessons, $64.35)
Chemistry: Looking In-Depth at Redox Reactions (5 lessons, $7.92)

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 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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Recent Reviews

Several errors
~ sovietcanuckistan

It seems to me he has several errors in his work. There's a missing + sign at one point, (when he's adding the half-reactions) though that's not a big deal. A much bigger deal is that he doesn't seem to properly balance his chromium in the complex acid balancing equation. It's ironic he says "you have to double check", but he apparently doesn't.

short and informative
~ dorothy4

short and to the point - most excellent

Several errors
~ sovietcanuckistan

It seems to me he has several errors in his work. There's a missing + sign at one point, (when he's adding the half-reactions) though that's not a big deal. A much bigger deal is that he doesn't seem to properly balance his chromium in the complex acid balancing equation. It's ironic he says "you have to double check", but he apparently doesn't.

short and informative
~ dorothy4

short and to the point - most excellent

Now we're going to explore an alternative approach to balancing redox reactions, and in fact, we'll use the same two examples of chemical reactions that we used in the previous tutorial so that you can compare these two methods. Now, let's remind ourselves, first of all, that oxidation is losing electrons, reduction is gaining electrons, the most important part of this whole process, because our approach this time will be to identify the individual pieces of the reaction where the oxidation and reduction are occurring, to break that into two separate equations or half-reactions, and to balance those half-reactions separately. Then at the end we'll combine them together.
So let's see how this works. Again, let's just go ahead so we can kind of keep track of where we are--the steps we're going to go through, we're going to write the half-reactions, we're going to assign oxidation numbers or oxidation states, we'll balance the redox atoms, then we go ahead and balance oxygen with water, hydrogen, or hydroxide. We'll balance electrons for the two reactions, and then finally after we put in the electrons we'll add the half-reactions together.
So for our first problem, again, the more trivial example of copper metal reacting with silver, being oxidized by silver to give us copper 2+ and silver metal. The first step is simply to write half-reactions down. In this equation silver is involved and, in fact, we see silver plus going to silver elemental, and copper is going from elemental copper to copper 2+. Now we assign the oxidation numbers. Remember that for elemental silver the oxidation number is zero in it's elemental form and likewise copper will also be a zero.
Our next step is to balance each one of the half-reactions for charge. So silver plus, plus an electron goes to silver, and copper goes to copper 2+ plus two electrons. Now, remember we know that because its oxidation state is zero. Balance the electrons for the two reactions. So in this case what we're going to do is since this requires one electron, this requires two electrons, we need to multiply this half-reaction by two. So we have two times that half-reaction--silver plus going to silver--and only one times copper goes to copper 2+ and two electrons. Now, notice that in doing this we have equal numbers of electrons in the reduction step and in the oxidation step. At the end of this we'll combine these. We don't have any oxygens or hydrogens to balance in this case, so we'll just combine these two sets together and we end up with copper solid, 2 silver plus goes to copper 2+ and silver as a solid.
So a trivial example, but notice the difference. We kept the two pieces of the reaction, the oxidation and the reduction, separate until the very last step when we combined them after adjusting so that the electrons used were equal to the electrons gained.
Here comes this more challenging problem. We want to balance the following reaction. We've seen this before. It's dichromate and chlorite going to chromic and Cl[2]O[7]. Step one is we want to write the individual half-reactions. So in this case we know that chromium is going from a form of dichromate to chromic. Now again, notice that there are lots of things that aren't balanced here. All I'm looking at are the redox changes--chromium as a form of dichromate goes to chromic. Likewise, we go from chlorite to Cl[2]O[7]. Dichloroheptoxide. Now, assign oxidation numbers. Recall that each one of these chromium atoms has a 6+ oxidation state and each one of those is going to a chromium 3. Notice we haven't balanced anything yet. We're just looking at oxidation states. Here we have chlorite going to Cl[2]O[7], and again, we're going from a 3+ oxidation state to a 7+ oxidation state. Again, that's 7+ for each one of those chlorines.
Now, we balance the redox atoms. This takes into account the fact that there are two chromiums involved here, so we need a 2 here. Likewise, we have two chlorines here, so we need to have the coefficient of 2 here such that we have a balancing of the redox atoms. Next we want to balance each half-reaction for charge. So in this case we have a 2- here and a 3+ here. We've got again a difference of 6+ going to 3+ as far as our oxidation states, and we have two chromiums, again, going through that change, so that's a total of 6 electrons that we're having to add in order to reduce the chromium from this form to this form.
Down here we now need 8 electrons in order to do the oxidation from chlorite to Cl[2]O[7]. Now again, remember, we go from 3+ to 7+. That's 4 electrons per chlorine, or a total of 8 electrons in this half-reaction. Now we're ready. We want to balance the oxygen atoms in half-reactions. So we're going to use water and hydrogen atoms, so we have now, in this case for the upper one we have 14 hydrogens on this side. In fact, let me back up a step. First of all, we have 7 oxygens on the left so we're going to need 7 water molecules so that we balance oxygen. Then we balance the hydrogen, so 14 H+'s to account for the 14 hydrogens that we introduced with the water. Now we do the other half-reaction. Notice we're keeping our half-reactions separate. So in this case, in order to balance oxygen we need 3 water molecules on this side. That requires 6 hydrogens on the right side to balance the hydrogens.
Now, at this stage everything should match for the individual half-reactions at least. Everything should be balanced. And this is a convenient place to stop and check yourself. Everything for each one of those two half-reactions should be balanced independently. But what we haven't done yet is make sure that the electrons used equals the electrons gained.
So the next step is to balance electrons for the two half-reactions. That's going to require multiplying this half-reaction by 4 so that we have a total of 24 electrons, and the lower reaction by 3, so again, we have a total of 24 electrons that are given up. So we need 24 electrons here. We get them from this half-reaction. So you get the strategy again. We're still keeping them as separate half-reactions, the oxidation and the reduction step.
Finally we bring it all together. We add everything together and we end up with, in this case, a balanced equation. Now notice that we've got hydrogens on each side at this stage and we're going to have waters also on each side, and so we need to cancel, and so this reduces down to this final equation where we just have 38 hydrogens on one side. So now as a check everything should balance. Charge should balance; atoms on both sides should balance, so you want to doubly check yourselves.
As I mentioned last time, we have the possibility of having this reaction occur in basic solution instead of acidic solution. So we have a couple of approaches now. One answer is like we did last time to simply take this final equation and add enough hydroxide to neutralize the H+. So we're going to add 38 hydroxides to both sides, and in that case we'll have water on this side and hydroxides on the other side, and we'd have a balanced equation. Or we can go back to the original half-cell reaction but use hydroxide in water to balance oxygen atoms. So let's go ahead and see what that would look like.
So we're going to back up here, back to where we have the half-reactions, except I'm going to use hydroxide in water now to balance. Now, remember, I need two hydroxides on one side and water on the other side to balance out the total number of oxygens. So, in other words, I'm looking to see where I need to add more oxygen and to that side I add the two hydroxides and put a water on the other side. That makes sure my hydrogen is balanced. So this is what that half-reaction would look like. We do the same thing down here so that our oxygens balance out.
Now, at this stage we have balanced half-reactions, balanced in base, not in acid, so we now need once again to multiply by the appropriate numbers of electrons, because each one of these equations is balanced by itself, but we need to make sure that the electrons used equals the electrons gained. So once again we'll multiply by 4 and by 3 to get a total of 24 electrons that are consumed, 24 electrons that are given off--and remember, this is a reduction step, this is an oxidation step. Don't forget that. We add them all together, we end up with this colossal mess, which reduces down to this final equation.
And again, just remember, when we balanced it in acid we had 38 H+'s. Now notice we've got 38 hydroxides, so that neutralization method that we talked about being an alternative gives us the same answer. And again, you should just check yourself on that. Make sure that that makes sense and that you can do this as well.
Also, if you go back one tutorial to the oxidation state method, you'll find that the same answer applies. There is, in fact, one unique answer to this where all charges balance and all atoms balance, so you can't miss it as long as you just check yourself that you have complete balance.
Oxidation-Reduction Reactions
Looking In-Depth at Redox Reactions
Balancing Redox Reactions Using the Half-Reaction Method Page [1 of 2]

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