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All equations and equation-like statements, whether in words or symbols, are numbered, simply for ease of referring to them. The numbering is of the form [A-B] , where A is the question number and B is the equation number within the section for that question. For "equations" (statements) that are not balanced, NB follows the equation number.

1. a. In words: iron metal + molecular oxygen `-->` iron(III) oxide. [1-1 NB]

In symbols, using correct formulas for each chemical, and showing phases:

Fe(s) + O_{2}(g) `-->` Fe_{2}O_{3}(s) [1-2 NB]

Balanced: 4 Fe(s) + 3 O_{2}(g) `-->` 2 Fe_{2}O_{3}(s) [1-3]

If you had trouble with anything above, it is probably due to material from earlier in the course. So far, there is nothing here that is specific to redox.

b. In order to write the half cells, we need to analyze the reaction, and see which atom gained electrons and which one lost electrons. This requires some understanding of the chemicals. In this case, we start with elemental iron and end up with iron in its 3+ form. That is, we go from Fe^{0} to Fe^{3+}; the Fe has lost 3 electrons. The oxygen goes from its elemental form to its common ion, 2-; the O gains 2 electrons (per atom!). In making those statements, you recognize that the elemental forms are neutral, with oxidation number zero. The compound iron(III) oxide, contains Fe in the 3+ state (the roman numeral tells you that) and O in its common 2- form.

Now write what we figured out above as formal half cell equations, adding electrons as needed to make them balanced:

Fe^{0} `-->` Fe^{3+} + 3 e^{-} [1-4]

O_{2}^{0} + 4 e^{-} `-->` 2 O ^{2-} [1-5]

[or O^{0} + 2 e^{-} `-->` O ^{2-}. [1-6] It doesn't really matter whether you write the O part in terms of one O atom or one O_{2} molecule -- so long as you do it correctly, and keep track. But O_{2} is the correct chemical, so I'll use it from the start.]

We have described the two half cells. We need to combine these, so that the electrons come out even -- cancel out. That is, the number of electrons given off (lost) by the Fe must equal the number taken (gained) by the O. At this point, one equation involves 3 electrons and one involves 4. To make the electrons come out the same, we multiply the first equation by 4 and the second by 3. We get:

4 Fe^{0} `-->` 4 Fe^{3+} + 12 e^{-} [1-7]

3 O_{2}^{0} + 12 e^{-} `-->` 6 O ^{2-} [1-8]

Each equation is itself balanced, and the number of free electrons is the same in the two equations. Now, add them together:

4 Fe^{0} + 3 O_{2}^{0} + 12 e^{-} `-->` 4 Fe^{3+} + 12 e^{-} + 6 O ^{2-} [1-9]

As desired, the number of electrons is the same on each side, so cancel them out. We get:

4 Fe^{0} + 3 O_{2}^{0} `-->` 4 Fe^{3+} + 6 O ^{2-} [1-10]

Check that the equation is balanced. Once we write balanced half cells, each step in the work should give a balanced equation. The equation here is balanced, and in fact is pretty good chemistry. It shows the two elements getting together, with electrons being transferred from the Fe to the O. But at this point the equation shows ions, and we know our product is the solid ionic compound. Look at those ions, and you will see that there are just enough ions on the right side, so that we have...

4 Fe^{0} + 3 O_{2}^{0} `-->` 2 Fe_{2}O_{3} [1-11]

Finally, tidy it up... We don't usually show the charge zero in final equations, and we need phases...

4 Fe (s) + 3 O_{2} (g) `-->` 2 Fe_{2}O_{3} (s) [1-12]

This is the same as we got in part a. That is, equations 1-12 and 1-3 are the same.

* In this case, the method of part a is simpler than the method of part b. The purpose of trying part b here is so you can see how it works, in a relatively simple case. In many cases, you will be unable to balance a redox equation by the simple method a; knowing how to use method b will be important. *

2. a. In words: hydrogen peroxide `-->` water + molecular oxygen [2-1 NB]

In symbols, using correct formulas for each chemical, and showing phases:

H_{2}O_{2} (aq) `-->` H_{2}O (l) + O_{2}(g) [2-2 NB]

Balanced: H_{2}O_{2} (aq) `-->` H_{2}O (l) + 1/2 O_{2}(g). [2-3]

That has a fraction in it, so multiply through the entire equation by 2:

2 H_{2}O_{2} (aq) `-->` 2 H_{2}O (l) + O_{2}(g) [2-4]

b. Which atoms gain or lose electrons? The key is recognizing that H_{2}O_{2} is a special case. Peroxides contain O in the somewhat unusual oxidation state of -1. The two atoms of O^{-1} in the H_{2}O_{2} become O^{2-} in water and O^{0} in molecular oxygen. That is, one O gains electrons and one O loses electrons.

Writing those statements as half cell reactions...

O^{-} + e^{-} `-->` O^{2-} [2-5]

O^{-} `-->` O^{0} + e^{-} [2-6]

There is one electron in each of those half cells, so we can simply add those two equations together and the electrons will cancel out:

2 O^{-} `-->` O^{2-} + O^{0} [2-7]

Now just tidy up... 2 O^{-} is really H_{2}O_{2}, and so forth. Add phases, and we get the same equation as we got in part a.

3. In words: ferrous ion + dichromate ion `-->` ferric ion + chromium(III) ion [3-1 NB]

Fe^{2+} + Cr_{2}O_{7}^{2-} `-->` Fe^{3+} + Cr^{3+} [3-2 NB]

That equation seems to describe the given information. However, it must be incomplete; there is no O on one side. As we go on, we must, somehow, deal with that.

There is more than one way to proceed. In fact, I'm going to show two ways to do this here. At the outset, they are not correct; in fact, both do the same things -- but in a different order. I'll comment on them some more after presenting the two methods. For clarity, they are shown here with different indents.

Method 1

Write half-cells:

Fe^{2+}-->Fe^{3+}+ e^{-}. [3-3] That's straightforward.

Cr^{6+}+ 3 e^{-}-->Cr^{3+}. [3-4] In writing this, we focus simply on the Cr. Its oxidation state is 6+ in the dichromate ion.We can combine the Fe and Cr half cells in the ratio of 3 Fe to 1 Cr. That is, multiply the Fe half cell equation by 3, and then add the Fe and Cr half cell equations together. That gives:

3 Fe^{2+}+ Cr^{6+}-->3 Fe^{3+}+ Cr^{3+}[3-5]That equation is balanced: it has the same atoms and the same charge on both sides. In a sense, we have solved the redox part of this. However, the chemistry really isn't right. Cr

^{6+}is not the correct chemical -- even though it does represent the key atom in its correct oxidation state. So, we now need to "clean" this up -- to put it in better chemical form. But none of the steps that follow are redox.We're going to replace that Cr

^{6+}by the more appropriate Cr_{2}O_{7}^{2-}. To prepare for that let's multiple the previous equation by 2, so we have 2 Cr. That gives:

6 Fe^{2+}+ 2 Cr^{6+}-->6 Fe^{3+}+ 2 Cr^{3+}[3-6]Of course, Cr

^{6+}is not the actual chemical species that is involved; the relevant chemical is the dichromate ion, which contains Cr in the 6+ oxidation state. Therefore, we now replace the 2 Cr^{6+}by Cr_{2}O_{7}^{2-}. But that introduces 7 oxygens on the left. What do we do? As a start, let's just show them on the right. They are ordinary oxygens in compounds, with a 2- oxidation state; we'll show them, for the moment, simply as O^{2-}. That gives:

6 Fe^{2+}+ Cr_{2}O_{7}^{2-}-->6 Fe^{3+}+ 2 Cr^{3+}+ 7 O^{2-}[3-7]That equation, too, is balanced: it has the same atoms and the same charge on both sides. However, the chemistry still isn't right. Writing free oxide ions may help us see the steps, but it is not a likely product. What to do? The clue comes from the statement of the original question: the reaction occurs in acidic aqueous solution. Water and H

^{+}ions are around; we are free to use them as we need to. Let's add some H^{+}ions -- enough to get rid of those oxide ions. 14 will do. Adding 14 H^{+}ions -- to both sides -- gives the final result. With phases, it is:

6 Fe^{2+}(aq) + Cr_{2}O_{7}^{2-}(aq) + 14 H^{+}(aq)-->6 Fe^{3+}(aq) + 2 Cr^{3+}(aq) + 7 H_{2}O (l) [3-8]Method 2

At the start of answering this question, I noted that there are alternative ways of doing this. Let's look at another possible approach. Here I will use an approach commonly found in chem textbooks. As we start this alternative, let's note that it ends up doing exactly the same things we did above; it just does them in a different order.

We start, as usual, by describing the two half cells. For Fe, it is simple, and the equation above stands:

Fe^{2+}-->Fe^{3+}+ e^{-}. [3-3]However, for the Cr half-cell, the more complex one, we take a different approach. We start by writing the basic description that is given:

Cr_{2}O_{7}^{2-}-->Cr^{3+}[3-9 NB]

Trivially, we see that there must be 2 Cr on the right:

Cr_{2}O_{7}^{2-}-->2 Cr^{3+}[3-10 NB]That has excess O on the left. Balance the O, by adding O to the right -- in an appropriate chemical form, H

_{2}O.

Cr_{2}O_{7}^{2-}-->2 Cr^{3+}+ 7 H_{2}O [3-11 NB]That has excess H on the right. Balance the H, by adding H to the left -- in an appropriate chemical form, H

^{+}.

Cr_{2}O_{7}^{2-}+ 14 H^{+}-->2 Cr^{3+}+ 7 H_{2}O [3-12 NB]Now balance the charge by adding electrons as needed:

Cr_{2}O_{7}^{2-}+ 14 H^{+}+6 e^{-}-->2 Cr^{3+}+ 7 H_{2}O [3-13]

This equation is fully balanced -- for each atom and for charge. It is a final balanced equation for the Cr half cell.We now have equations for two half-cell reactions ( [3-3] & [3-13] ). Combine them as usual, and clean up as needed; the result will be the same final result we got earlier, equation [3-8].

Comment on the two methods shown here for balancing this reaction... The most important comment is that the two methods are entirely equivalent. In fact, you'll find that we did almost exactly the same things in both cases, but we did them in a different order.

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