REDOX REACTIONS AND BALANCING EQUATIONS



REDOX REACTIONS AND BALANCING EQUATIONS

This is an important topic. Consider the mitochondria in the cell being the site of cellular respiration and the production of ATP which is a redox process. Corrosion, "rusting", is a redox reaction and a big problem in industry. A battery is a redox reaction. Redox is the transfer of electrons while acid/base chemistry involved the transfer of protons. Precipitation, unlike these two types of reactions, does not involve a transfer as such but the combination of existing ions in solution such that an insoluble compound is formed.

I. OXIDATION NUMBERS

• BIG GOAL IN REDOX . . . Keeping up with where the electrons go!!!!

In redox keeping an account of the number of electrons and where they go is very important. So we need oxidation numbers. This is a way of assigning numbers to keep up with electrons and that we need a system.

My system is one of many and NOT like the one in most texts. Also please see that oxidation numbers are a lot like "charges" we have discussed in the past, although not totally the same thing. Definition: Oxidation numbers are assigned per "atom" of an element in a compound or polyatomic ion. I really do not like to use the term "atom" here since as soon as there is a reaction, atoms become ions or bond covalently to form molecules.

My system depends on using a PRIORITY list that is as follows:

Element/Group Oxidation Number

F –1 always

Group 1A +1 always

Group 2A +2 always

H +1 usually

O –2 usually

What is this list and how is it used? This list is to be used to assign oxidation numbers to all elements in any compound or polyatomic ion. First, some important facts.

1. All free pure elements have an oxidation number of 0 since they are only made of neutral atoms.

2. For all molecules and formulas for ionic compounds, the sum of the oxidation numbers in them equals zero. Remember, this is the basis for knowing when you have a formula correct for an ionic compound.

3. For polyatomic ions, the sum of the oxidation numbers equals the charge you have learned (or are supposed to have learned!!!!) on the polyatomic ion.

4. The "charge" of a single atom that has become an ion, ex. Na+, is its oxidation number. As in the case of the sodium ion here, the oxidation number is +1.

Consider several examples to show how to assign oxidation numbers.

CrO42– The question would be, what is the oxidation number of each element in this ion?

Now consult the list, Start at the top of the list since priority lists are that way, the most important item is at the top!! Go down the list until you find something on the list that is also in the compound or ion with which you are dealing. Well, there is nothing on this list to help until you get down to O. The oxidation number of O is –2. That is the ONE of the O's!!! See the definition of oxidation numbers. Now there are 4 O's in this ion, so there is a total of –8 for the sum of the oxidation numbers of all 4 O's. Well, what must the Cr's oxidation number be? The oxidation numbers must add to be the "charge" on the total ion, –2. Thus the Cr is +6 since +6 and –8 = –2.

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What about Cr2O72–? Again, there is nothing to help from the list until O. Each O is –2, there are 7 of them so 7 ∞ 2 = –14. Now what about the Cr? The Cr's must equal +12 in order for the total of the –14 and +12 to equal the charge on the original polyatomic ion of –1. HOWEVER, this +12 is for 2 Cr's so the oxidation number of each Cr is +12/2 = +6. You must remember that the oxidation number is for one "atom" of an element.

What about OF2? OK. F is at the top of the list and gets –1 immediately. There are 2 F's so there is a total of –2 for them. Thus the O must be +2 so that the total of the oxidation numbers equals zero this time. Why zero? There is NO charge on this molecule. NOTE, O is not –2 this time as it usually is because of the priority position of F at the top of the list. There is more chemistry to the explanation of why this is true that we will encounter later.

What about NaH? Note, Na is above H on the list so the Na get +1 first. Thus the H must be –1 to add to be zero. This is not the H on the list. That H is +1. This is a form of H known as hydride. Note, it was the second part of the compound and in combination with a metal. Thus the ending of this is "-ide".

What about FeCl3? Oops!!! Nothing on the list in this! What to do???? You have seen compounds like this before this year. You should know this is iron(III) chloride. WHY? Use the negative part of the compound to determine the positive part. What does this mean? You know that Cl in a compound is –1 from the chart. Now, there are 3 Cl's for a total of –3. Thus the Fe must be +3 to equal 0.

REMEMBER . . . WHEN THERE IS NOTHING ON THE LIST IN WHAT YOU ARE WORKING WITH, USE THE NEGATIVE PART TO DETERMINE THE POSITIVE PART'S OXIDATION NUMBER!!!!

ALWAYS GIVE THE ELECTRONS TO THE MOST ELECTRONEGATIVE ELEMENT.

I could go on and on. Sometimes oxidation numbers are 0 and sometimes even fractions.

II. NOW WHAT TO DO WITH THESE NUMBERS??? One use is to answer this question.

Is the following equation for a redox reaction?

CH4 + O2 --( CO2 + H2O

This reaction involving the addition of O2 are combustion reactions. You must know the products for the combustion of a hydrocarbon or a carbohydrate are always CO2 and H2O. Consider again CH4 + O2 --( CO2 + H2O. It is not even necessary for me to balance this to answer the question. You must now use the technique to assign oxidation numbers to every element in this equation.

A quick check should reveal the following to be true if you use the technique as explained.

The H is +1 and the C is –4 in CH4 while the O2 is 0 since it is an element. The C is +4 and the O is –2 in the CO2 while the H is +1 and the O is –2 in the water. SO WHAT??? This clearly shows some of the elements' oxidation numbers changed from the reactant side (left) to the product side (right). That means some electrons were transferred between them and indeed this is a REDOX reaction. The test is to see if any oxidation numbers change. If they do, the reaction is redox, plain and simple as that!!!!

Now let's examine the specifics of this reaction.

The O is O2 has an O oxidation number while the O's on the right are –2. What does this mean? Each O in O2 had to GAIN 2 electrons in order for the O to become –2 during the reaction. Gaining electrons is called REDUCTION and always results in something becoming less positive since electrons are negative.

The C goes from –4 on the left to +4 on the right of the arrow. What has happened??? The only way to understand this is to think about what you are moving in this reaction. ONLY negative electrons. Thus to go from –4 to +4, think about a number line. You must move 8 places or in terms of the chemistry, lose 8 electrons. Thus you are losing 8 negatives and become more positive. Losing electrons is called OXIDATION. Remember, oxidation and reduction occur simultaneously!!!

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• Sometimes the answer to the question is not obvious at first.

This was a question from a typical chemistry exam on redox.

12. Which of the following equations represent oxidation-reduction reactions?

1. CH4 + O2 --( 2H2O + CO2

2. Zn + 2HCl --( ZnCl2 + H2

3. 2Na + 2H2O --( 2NaOH + H2

4. MnO2 + 4HCl --( Cl2 + 2H2O + MnCl2

a. 1 only b. 2 and 3 only c. 1 and 4 only d. 1, 3, and 4 only

e. All are oxidation-reduction reactions.

In this case, you must be able to assign oxidation numbers to each element in each compound. These numbers are a means of indicating the loss and gain of electrons. If one element experiences a change in this number from the reactant side to the product side, then at least one other element will also have a change. The important thing to know is that is all you need to determine to KNOW that the equation represents a redox reaction.

• A helpful hint: any time an element exists on one side of the arrow by itself but is combined with other elements on the other side of the arrow, that element is undergoing either reduction or oxidation, and the reaction is redox.

You should determine that 1, 2, and 3 are all redox reactions.

NOTE . . . Acid/base reactions are NOT redox and neither are precipitation reactions.

III. REDOX REACTIONS IN AQUEOUS SOLUTIONS

Now we will study the specific type of redox in solution since that is the emphasis of this chapter.

Take some water and pour it into a beaker. Add NaNO2 crystals and stir. They dissolved easily. Then add Kl crystals, stir and they too dissolve easily. Nothing happens at this point. Any idea why no precipitate formed? All the ions are still soluble in water even when they mix with each other.

Then when I add a few drops of 3.OM H2SO4 and presto. A reaction takes place quite notably. Brown gas is given off with a strong smell and black solid I2 begins to collect all over the top of the original solution. What has happened? Some redox reaction will not occur in solution until the pH is changed. Thus the addition of the strong acid prompted the reaction to take place. We now need a balanced equation for this reaction. Let's see how to balance redox in aqueous solution with the reaction from the demo.

NO2– + I– --( NO(g) + I2(s)

This is the net equation for this reaction. You will see countless numbers of these. You will usually never know that other ions such as the Na+ and the K+ were ever around. You should realize that the ions on the left of the arrow in this did not just appear from thin air. However, in all honesty you will not have to concern yourself with that detail. How do you go about balancing this net ionic equation?

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Step One

Take the equation and split it into 2 "logical" half reactions.

NO2– --( NO

I– --( I2

What is "logical" about these

The substance on the left with “N” in it makes the one on the right with “N” in it.

It would make NO SENSE to have I– --( NO!!!!

Step Two

Balance all elements by using coefficients EXCEPT hydrogen and oxygen.Don’t even worry about the O and H yet.

NO2– --( NO

2I– --( I2 The only thing I need to do is put the 2 in front of the I–.

Step Three

Now balance the O's by adding H2O's wherever needed. This mean add the waters to the right or to the left. Just add them on the side that needs more O’s.

NO2– --( NO + H2O Notice, adding 1 water balances the O's in this half.

2I– --( I2 Obviously no O's to even deal with so no waters needed!!

But now you have introduced H's in the first half reaction. No worry. Step 4!

Step Four

Balance H's by adding H+'s wherever needed. Again this means add to the left or the right as needed.

2H+ + NO2 --( NO + H2O Note, adding the 2 H+'s on the left took care of the H's.

2I– --( I2 Again, no H's to worry about.

The steps up to this point balance all the elements BUT NOT THE CHARGE!!!!

What do you do to balance the charge? What charge you may be asking? Notice each half reaction, and you will see the charge is not the SAME on both sides of the arrow.

NOW WE MUST DO STEP 5!!!

Step Five

2H+ + NO2– --( NO + H2O

On the left of the arrow is a total of –1 charge when you consider the 2 H+'s and the –1 on the nitrite. Treat these charges just like adding positive and negative numbers from algebra I. It does not matter where the charges are, just add them. Notice on the right there are no charges. The right has a total charge of 0. So now what, you may ask?

These are REDOX reactions which means we are going to be moving electrons from one substance to another. We can now add electrons in order to balance the charge.

BIG POINT!!!! You will add the electrons to the side of the reaction that is more positive.

How many electrons you add depends on the total charge with which you are working.

e– + 2H+ + NO2– --(NO + H2O

By adding 1 electron on the left, the total charge there is now 0. Notice, I added the electron to the more positive side and only enough to make the two sides equal.

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Now what about the next half reaction?

2I– --( I2 + 2e–

Why add 2 electrons to the right? The left of the arrow has –2 total charge on the 2 iodides. There was no charge on the right, neutral iodine. Thus the right side is more positive. Adding 2 electrons makes the right –2 just like the left. Always remember, add only enough electrons to make the more positive side the same as the other side which is more negative to begin with.

Step Six

So now what do you do with this set of half reaction equations? You have balanced the elements and the charges in each half. Now you want to put the halves together to result in the total balanced equation. BUT, there is one principle that must be met. The electrons lost by one half must be equal to the electrons gained by the other half. NOTICE, electrons lost are one the right of the arrow as products while electrons gained are one the left of the arrow.

What can we do to make these electrons equal? Treat this just like an algebra situation where you can multiply in this case to make the electrons equal so you can then add the equations and the electrons will "cancel out"!!! I do NOT like this statement but I understand that is what you will be thinking.

2X (e– + 2H+ + NO2 --(NO + H2O) By multiplying this entire half by two, there will be 2 electrons on the left of this reaction and there were already 2 on the right of the second reaction.

2I– --( I2 + 2e– Now you can add the two halves together since the electrons are equal. Nature does this automatically while all we are doing is writing something on paper to represent this. I am glad nature's ability to carry out chemistry does not depend on our ability to understand it on paper. The result is the following BALANCED NET IONIC EQUATION!!!

2I– + 4H+ + 2NO2– --( 2NO + 2H2O + I2

You will notice the electrons do not appear in the final equation. They have "canceled" each other out mathematically, BUT in reality this is related to the chemistry as follows. The 2 electrons lost by the iodide half reaction are the same 2 electrons gained in the other half reaction involving the nitrite.

This equation is now completely balanced and could be used to do any stoichiometry problems, etc., like any other balanced equation.

Now what if the reaction occurs in a basic solution. How do you balance these?

You can use the same steps as I have already taught you to use to balance an acid redox. YOU MUST LEARN THESE STEPS. After doing the first 6 steps, you simply convert the H+'s to OH–'s.

Example: The following reaction occurs in a basic solution.

Cr(OH)3 + ClO3– --(CrO42– + Cl–

First, the two logical half reactions.

Please note that I have already added H2O and H+ to balance O's and H's. I did not need any other coefficients at first since all other elements were already balanced. I have indicated what I added in bold print. I have also added the electrons needed to balance the charge for each half reaction.

H2O + Cr(OH)3 --( CrO42– + 5H+ + 3e- Why 3 electrons here? Total charge = 0 as on the left.

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6e– + 6H+ + ClO3– --( Cl– + 3H2O Why 6 electrons on the left? Total charge is –1 on both sides.

Remember!!! ADD electrons to the side that is more positive to begin with and only add as many as needed to make that side equal to the other. Special note: The subscripts in polyatomic ions such as the 3 in the ClO3– have nothing to do with the total charge of the reaction. Only the –1 on the chlorate ion itself is involved.

Now we must make the electrons equal which requires that we multiply the first half reaction by 2 and thus we will have the following: 2H2O + 2Cr(OH)3 --( 2CrO42– + 1OH+ + 6e–

Now we can add the two half reactions and will have the following.

6H+ + ClO3– + 2H2O + 2Cr(OH)3 --( Cl– + 3H2O + 2CrO42– + 1OH+

We have some problems. First, there are waters and hydrogen ions on both sides of the equation. To make things easier in the rest of this work, we will need to do an algebra trick here and "collect" all the waters and hydrogen ions on one side. Since everything in a balanced equation is always left positive, we will collect such that this is the case. This will mean 1 water left on the right along with 4H+'s.

ClO3– + 2H2O + 2Cr(OH)3 --( Cl– + H2O + 2CrO42– + 4H+

I trust you understand how this is like algebra. I would have "subtracted 2 H2O's and 6 H+'s from each side."

Obviously this is not basic yet since the H+'s are clearly in the equation.

To accomplish this conversion, we will add 4 OH–'s to each side. Why? Well we have 4 H+'s on the right so we are essentially "neutralizing them on paper". At the same time we are adding the 4 OH–'s to both sides as in any algebra or conservation principle. What you do to one side you must do to the other is NOT new to you.

4OH– + ClO3– + 2Cr(OH)3--( Cl– + H2O + 2CrO42– + 4H+ + 4OH–

Now we will make an assumption that is clearly logical, the 4H+ + 4OH– on the right is essentially 4 H2O's if you add them together. DO SO!!! Now there are 5 waters on the right when you collect all the waters as we have been discussing.

4OH– + ClO3– + 2H2O + 2Cr(OH)3 --( Cl– + 5H2O + 2CrO42–

This represents a basic equation as you can clearly see the OH–'s in the equation. Please note that I did everything the same as if this had been acidic until I added the OH–'s. You add as many OH–'s as you have H+'s on both sides of the equation. THEN you make water out of the H+'s and OH–'s that are on the same side of the arrow and collect all the waters together.

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OXIDIZING AND REDUCING AGENTS

One common confusion about redox is how to identify the oxidizing and reducing agents in the reaction. First, the two agents are always reactants in the original equation. They are not the H+'s and H2O's added to balance H's and O's. Consider the original example I did for the acidic redox problem. These are the original half reactions before I even worried about the electrons being equal.

e– + 2H+ + NO2– --( NO + H2O

Note that here the electron is on the left of the equation. This electron is literally being "added" in this reaction as a reactant. This half reaction is the reduction half since this is what adding electrons is called.

l– --( I2 + 2e–

Notice here the electrons are on the right of the equation. This means the electrons are being produced or "lost from the reaction" as the case is. When electrons are lost, this is the oxidation half reaction.

Thus you must always have electrons on opposite sides of the two half reactions since always both oxidation and reduction must occur at the same time.

Now what about identifying the agents? The fact is this.

The reactant that is in the half reaction with the electrons on the right is the reducing agent even though that is the oxidation half reaction.

Conversely, the reactant in the half reaction where the electrons are on the left of the equation is the oxidizing agent. The half reaction that is losing electrons is forcing the other reactant to gain them, or to be reduced. Thus the agent is the reducing agent that loses the electrons. Turn this around and the reactant that is gaining the electrons is forcing the other reactant to lose electrons, or to be oxidized. I confess that this is bothersome but nonetheless you must learn it some way.

Another consideration is with oxidation numbers. If an element becomes more positive from the reactant side to the product side, then the compound or ion that contains this element is the reducing agent since it is losing electrons if it is more positive. The opposite is true if an element becomes more negative. I actually prefer this method. Just remember, the agents are always on the left of the arrow!!! They are the original reactants.

Disproportionation

There is an interesting type of redox reaction where you will have only 1 reactant which is forming at least two products or perhaps vice versa. Consider the following examples.

P4(s) --( PH3(g) + HPO32–(aq)

IO3–(aq) + I–(aq)--( I3–(aq)

Both of these were performed in an acidic solution. Now what do you do?

You will still write two half reactions, but starting with the same reactant in the case of the P4 reaction and in the case of the I3– reaction you will have the same product for both half reactions. You would then proceed to follow the same steps as you have already used. Note, the P4 is both the oxidizing and reducing agent!!

Example: P4(s) --( PH3(g) + HPO32–(aq)

The two half reactions are as follows: P4(s) --( PH3(g) and P4(s) --( HPO32–(aq)

IO3–(aq) + I–(aq)--( I3–(aq)

The two half reactions are as follows:

Now balance as all other redox reactions. IO3–(aq) --( I3–(aq) and I–(aq) --( I3–(aq)

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BIG NOTE: All the above steps were done to balance this equation for a redox reaction that occurs in an ACIDIC solution.

I ALSO STRESS THE FOLLOWING: There are definitely other techniques for balancing redox equations than this one. I assure you different professors at different schools and in different classes at the same school will use different techniques. You need to consider which is easiest for you to understand. I seem to have had more success with this one for most of my students. NO ONE technique is better chemistry than any other.

We will now do a step that bothers me in that this is NOT the way the chemistry is actually done in the reaction in a beaker. If you want a redox reaction to occur in a basic solution, you add a base. We have balanced this up to this point as if this is an "acidic" redox. Now we will do something on paper that works to convert this to the balanced equation in "basic" solution. I want you to clearly understand this bothers me because this is just a technique that works and makes this easier!!! I repeat, this is not what happens in the real reaction. This is just a convenient “short cut” to switch from acidic to basic solutions.

Important Point . . . Only do the steps you need in this procedure!!!!

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