Introduction to Redox Reactions

Redox reactions are a very important part of the AP® Chemistry exam, and they can be quite challenging. I know that I personally spent more time on understanding redox reaction than on any other concept on the test. I’ll try to make it a little easier for you than it was for me.

So, first things first. What is a redox reaction? A redox reaction is simply a reaction in which one element is reduced (gains electrons) and another is oxidized (loses electrons). Here’s a handy mnemonic to help you remember that: OIL RIG (oxidation is losing (electrons), reduction is gaining (electrons)). A good way to tell if a reaction is a redox reaction is if an element changes oxidation states (for example, C{ u }^{ + } becomes Cu) or if an element acquires an oxygen atom or two (hence the name oxidation).

Enough about redox reactions themselves. How do you balance a redox reaction? It’s actually fairly simple; there are just a set of steps you need to take and rules you need to follow.

When balancing a redox reaction, you need to know the half-reactions of the reactants. As implied by the name, two half reactions can be added to get a total net reaction. For example, this equation:

C{ u }^{ + }(aq)+Fe(s)\rightarrow F{ e }^{ 3+ }(aq)+Cu(s)

…splits into these two half-reactions:

Fe(s)\rightarrow F{ e }^{ 3+ }(aq)

C{ u }^{ + }(aq)\rightarrow Cu(s)

Although the potentials of the half-reactions must be found for certain redox reaction calculations, balancing a redox reaction requires none of that.

Balancing a Redox Reaction

Balancing a redox reaction is actually a fairly simple problem because you’re given the direction of the reaction. You don’t have to figure out what’s happening, just how it’s happening. We’ll demonstrate this in the neutral solution example. There are also specific rules for aqueous reactions in acidic or basic conditions in addition to the rules for neutral conditions.

The Half-Equation Method is used to balance redox reactions. This method separates the equation into two half-equations, one for oxidation and another for reduction. Here are the steps involved in the half-equation method for a redox reaction in a neutral solution:

1. Calculate the charges on each side. Balance them by adding electrons ({ e }^{ - }) to the more positive side. One handy tip: { e }^{ - } and { H }^{ + } are usually found on the same side.

2. The { e }^{ - } on each side should be equal. If they are not, multiply one or the other of the half-reactions by an integer to make them equal. If this does not make sense, just continue reading; the examples should make it clear.

3. Add the half-equations together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.

4. Check the equation to ensure that it is balanced.

Those are the basic steps for how to balance a redox reaction in a neutral solution. Now, let’s take a look at a basic example of balancing a redox equation in a neutral solution before we move on to the more complex situation in acidic and basic solutions.

Balance this redox equation:

C{ u }^{ + }(aq)+Fe(s)\rightarrow F{ e }^{ 3+ }(aq)+Cu(s)

First, we split the equation into two half-reactions. The substance getting reduced will have electrons in its reactants, and the substance getting oxidized will have electrons in its products.

Fe(s)\rightarrow F{ e }^{ 3+ }(aq)

C{ u }^{ + }(aq)\rightarrow Cu(s)

In this particular case, we can see that Fe loses electrons (becoming more positive), while Cu gains an electron (becoming less positive). As you can see, this is all clearly laid out in the information given to us: Fe becomes more positive, and Cu becomes less positive. Isn’t that nice? So, the half-reactions are:

Fe(s)\rightarrow F{ e }^{ 3+ }(aq)+3{ e }^{ - }

C{ u }^{ + }(aq)+{ e }^{ - }\rightarrow Cu(s)

Now, if we were to put these two reactions together, we’d have a net gain of two electrons. That won’t work, so we multiply the bottom equation by 3 to balance the top equation’s electrons.

Fe(s)\rightarrow F{ e }^{ 3+ }(aq)+3{ e }^{ - }3Cu^{ + }(aq)+3e^{ - }\rightarrow 3Cu(s)

Now we can add them together:

Fe(s)+3C{ u }^{ + }(aq)+3{ e }^{ - }\rightarrow F{ e }^{ 3+ }(aq)+3Cu(s)+3{ e }^{ - }

The electrons cancel each other, and we are left with a balanced reaction:

Fe(s)+3C{ u }^{ + }(aq)\rightarrow F{ e }^{ 3+ }(aq)+3Cu(s)

Sounds good? Sounds simple? Good, because there’s more ground to be covered here.

Balancing Redox Equations in Acidic and Basic Solutions

The neutral steps are the same, but there are several new steps to be noted.

1. Balance all the elements in the equation other than O and H.

2. Balance the oxygen atoms by adding water ({ H }_{ 2 }O) molecules to the other side of the equation. This works because water has one O.

3. Balance the hydrogen atoms (including those added in Dtep 2 to balance the oxygen atom) by adding { H }^{ + } ions to the opposite side of the equation.

4. Calculate the charges on each side. Balance them by adding electrons ({ e }^{ - }) to the more positive side. One handy tip: { e }^{ - } and { H }^{ + } are usually found on the same side.

5. The { e }^{ - } on each side should be equal. If they are not, multiply one or the other of the half-reactions by an integer to make them equal. If this does not make sense, just continue reading; the examples should make it clear.

6. Add the half-equations together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.

7. (Only for reactions in basic solutions) Add a number of { OH }^{ - } to both sides equal to the number of { H }^{ + } in the reaction, and combine the { H }^{ + } and { OH }^{ - } together to make { H }_{ 2 }. If that doesn’t make sense, just continue to the example.

8. Check the equation to ensure that it is balanced.

Got it? Let’s try an example. This one is in an acidic solution.

C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)+HN{ O }_{ 2 }(aq)\rightarrow C{ r }^{ 3+ }(aq)+N{ O }_{ 3 }^{ - }(aq)

Separate it into two half-reactions:

C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)\rightarrow C{ r }^{ 3+ }(aq)

HN{ O }_{ 2 }(aq)\rightarrow N{ O }_{ 3 }^-(aq)

Balance all elements other than H or O. In this case, we only need to balance Cr:

C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)\rightarrow 2C{ r }^{ 3+ }(aq)

HN{ O }_{ 2 }(aq)\rightarrow N{ O }_{ 3 }^-(aq)

Balance all Os with { H }_{ 2 }O on the other side:

C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)\rightarrow 2C{ r }^{ 3+ }(aq)+7{ H }_{ 2 }O(l)

HN{ O }_{ 2 }(aq)+{ H }_{ 2 }O(l)\rightarrow N{ O }_{ 3 }^{ - }(aq)

Balance all hydrogens by adding { H }^{ + } to the other side:

C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)+14{ H }^{ + }(aq)\rightarrow 2C{ r }^{ 3+ }(aq)+7{ H }_{ 2 }O(l)

HN{ O }_{ 2 }(aq)+{ H }_{ 2 }O(l)\rightarrow N{ O }_{ 3 }^{ - }(aq)+3{ H }^{ + }(aq)

Balance the charges on both sides. For the first half-reaction, the charge on the left is +12 and the charge on the right is +6, so we need six electrons on the left side. For the second half-reaction, the charge on the left is 0 and the charge on the right is +2, so we need two electrons on the right side:

C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)+14{ H }^{ + }(aq)+6{ e }^{ - }\rightarrow 2C{ r }^{ 3+ }(aq)+7{ H }_{ 2 }O(l)

HN{ O }_{ 2 }(aq)+{ H }_{ 2 }O(l)\rightarrow N{ O }_{ 3 }^{ - }(aq)+3{ H }^{ + }(aq)+2{ e }^{ - }

Now, there are too many electrons on the first half-reaction, so we need to multiply the second half-reaction by 3 to compensate:

C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)+14{ H }^{ + }(aq)+6{ e }^{ - }\rightarrow 2C{ r }^{ 3+ }(aq)+7{ H }_{ 2 }O(l)

3HN{ O }_{ 2 }(aq)+3{ H }_{ 2 }O(l)\rightarrow 3N{ O }_{ 3 }^{ - }(aq)+9{ H }^{ + }(aq)+6{ e }^{ - }

Now we just add them together and cancel the electrons and any common terms (in this case, 9{ H }^{ + } and 3{ H }_{ 2 }O):

C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)+5{ H }^{ + }(aq)+3HN{ O }_{ 2 }(aq)\rightarrow 2C{ r }^{ 3+ }(aq)+4{ H }_{ 2 }O(l)+3N{ O }_{ 3 }^{ - }(aq)

There you have it, a redox reaction in an acidic solution! Now let’s do one in a basic solution.

Ag(s)+Z{ n }^{ 2+ }(aq)\rightarrow A{ g }_{ 2 }O(aq)+Zn(s)

Separate it into two half-reactions:

Ag(s)\rightarrow A{ g }_{ 2 }O(aq)

Z{ n }^{ 2+ }(aq)\rightarrow Zn(s)

Balance all elements other than H or O. In this case, we only need to balance Ag:

2Ag(s)\rightarrow A{ g }_{ 2 }O(aq)

Z{ n }^{ 2+ }(aq)\rightarrow Zn(s)

Balance all O‘s with { H }_{ 2 }O on the other side:

2Ag(s)+{ H }_{ 2 }O(l)\rightarrow A{ g }_{ 2 }O(aq)

Z{ n }^{ 2+ }(aq)\rightarrow Zn(s)

Balance all hydrogens by adding { H }^{ + } to the other side:

2Ag(s)+{ H }_{ 2 }O(l)\rightarrow A{ g }_{ 2 }O(aq)+2{ H }^{ + }(aq)

Z{ n }^{ 2+ }(aq)\rightarrow Zn(s)

Balance the charges on both sides. For the first half-reaction, the charge on the left is 0 and the charge on the right is +2, so we need two electrons on the right side. For the second half-reaction, the charge on the left is +2 and the charge on the right is 0, so we need two electrons on the left side:

2Ag(s)+{ H }_{ 2 }O(l)\rightarrow A{ g }_{ 2 }O(aq)+2{ H }^{ + }(aq)+2{ e }^{ - }

Z{ n }^{ 2+ }(aq)+2{ e }^{ - }\rightarrow Zn(s)

Next we would balance the electrons, but the electrons are already balanced. So, we just add the two half-reactions together:

2Ag(s)+{ H }_{ 2 }O(l)+Z{ n }^{ 2+ }(aq)+2{ e }^{ - }\rightarrow A{ g }_{ 2 }O(aq)+2{ H }^{ + }(aq)+Zn(s)+2{ e }^{ - }

The electrons cancel, and there are no other common terms to cancel:

2Ag(s)+Z{ n }^{ 2+ }(aq)+{ H }_{ 2 }O(l)\rightarrow A{ g }_{ 2 }O(aq)+Zn(s)+2{ H }^{ + }(aq)

We can’t have { H }^{ + } in a basic solution, so we add { OH }^{ - } to both sides to turn the { H }^{ + } into water:

2Ag(s)+Z{ n }^{ 2+ }(aq)+{ H }_{ 2 }O(l)+2{ OH }^{ - }(aq)\rightarrow A{ g }_{ 2 }O(aq)+Zn(s)+2{ H }_{ 2 }O(l)

Cancel one { H }_{ 2 }O:

2Ag(s)+Z{ n }^{ 2+ }(aq)+2{ OH }^{ - }(aq)\rightarrow A{ g }_{ 2 }O(aq)+Zn(s)+{ H }_{ 2 }O(l)

That makes sense? Just remember, in a basic solution, you can’t have any { H }^{ + } left over, while in an acidic solution, you can’t have any { OH }^{ - } left over.

That’s how you balance a redox reaction! Whew. That was a lot of information. If you have any questions, let us know in the comments. Good luck with your AP® Chemistry exam; I hope this crash course helps you!

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