Balancing Redox Reactions
Supporting original reading. Follow the AP topic pathway for reviewed scope, worked reasoning and targeted practice.
Chemistry I · original Chapter 7 · combined Chapter 7
- Recap of Oxidation and Reduction
- Oxidizing/Reducing Agents
- Balancing Oxidation-Reduction Reactions
Section 2
Balancing Redox Reactions
So now that you have a general understanding of oxidation and reduction as well as some background in assigning some states, we can look into more complex reactions concerning redox.
Just to recap, we learned in the previous section that compounds can be oxidized or reduced and is associated with the transfer of electrons. We also learned that we can assign oxidation states to atoms in molecules to account for electrons lost and gained. And we also learned a handy mnemonic device that we can use if we forget the definitions of oxidation-reduction: LEO goes....GER. If you forgot what this meant already, just review the previous section.
Now we can move on to some additional terms that you may encounter when dealing with these types of reactions. Lets use the combustion reaction again to help us with these definitions.
Image 1
In CH₄+2O₂→CO₂+2H₂O, carbon changes from −4 to +4 and loses eight electrons in oxidation-number bookkeeping. Oxygen changes from 0 to −2 per atom; four oxygen atoms gain those eight electrons in total. CH₄ is the reducing agent because it is oxidized while reducing oxygen. O₂ is the oxidizing agent because it is reduced while oxidizing the carbon-containing fuel.
Interactive 2.1 : General Summary of Oxidation/Reduction
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Agents?
So since A was oxidized and caused reduction in molecule B, then A is the reducing agent. B is thus the oxidizing agent.
Electron Transfer
Electrons are being transferred from molecule A to B, this must mean that molecule A is being oxidized (losing electrons) and B is being reduced (gaining electrons)
Alright, so the first step in this approach is to just figure out the oxidation states for everything.
For CH₃OH, hydrogen is +1 and oxygen is −2. A neutral molecule then requires x+4(+1)+(−2)=0, so its carbon has oxidation number −2. The species here is methanol. Ethanol, C₂H₅OH, has two carbon atoms with different individual oxidation numbers and must not silently be substituted into this one-carbon calculation.
O2 has an oxidation state of 0 since it’s in its elemental form.
For CO2, O has a -2 oxidation state, so C must have a oxidation state of +4 to remain neutral.
Balancing Oxidation-Reduction Reactions
So far, we have been just giving you guys the balanced reactions and from there, you determine the oxidation states. But it is equally as important as to determine a balanced equations using just the oxidation states. In chemistry literature, this method is usually known as the oxidation states method. Basically, it’s just using the oxidation states of the reaction to determine the overall balanced equation. Let’s look at an example how this is done.
Example 7.2.1
Let’s consider the combustion of methanol, CH₃OH, using the unbalanced skeleton CH₃OH+O₂→CO₂+H₂O. This deliberately corrects the original ethanol label so that the following one-carbon formula and electron count agree.
Balance this equation using the oxidation states method.
For H2O, O has a -2 oxidation state and H has +1, so therefore it is balanced.
Methanol’s carbon changes from −2 to +4, a loss of six electrons per carbon atom. An oxygen atom from O₂ changes from 0 to −2, a gain of two electrons, so three oxygen atoms receive six electrons. Accounting for all atoms gives 2CH₃OH+3O₂→2CO₂+4H₂O. Check carbon, hydrogen and oxygen counts after using the electron ratio; redox accounting alone does not replace atom balance.
The half is there to “cancel” the 2 subscript on the oxygen to give us a solid number of 3 oxygen atoms. The main thing you have to understand from this is that the oxidation number decrease (2) has to equal to oxidation number increase (6), which is where the balancing comes in (having to have 3 oxygen atoms)
Now the rest of the balancing is just book keeping and making sure that the numbers all match up:
Multiplying by 2 to get rid of the fraction gives us the final balanced equation:
Now I’m sure that this was a little confusing, so let’s look at another example to maybe clear things up. If you are having trouble, have no fear, the more practice you get, the better you’ll get at this. Trust me, it wasn’t the easiest for us to understand either.
Example Problem 7.2.2
Use the oxidation state method to balance the following reaction:
Alright, so at first glance this is a little intimidating because the compounds seem larger, but again, its not that bad once we break it down.
Step 1:
First see if you can easily balance this reaction through inspection. This should always be your first step, why makes thing harder than it should be? Problem is that you can’t balance this that easily. Sorry, if you could do that, then there would be no point in teaching you this method.
Step 2:
Find the oxidation states for each atom in the equation. Breaking it down atom by atom let’s look at the oxidation states for just the reactants:
For HNO3:
- O: has a -2 oxidation state. There are 3 of those so it totals -6
- H: There is only 1, so it has a +1 oxidation state
- N: Since HNO3 is neutral, N has to have a oxidation state of +5
For H3AsO3:
- H: +1 and since there are 3, it totals at +3
- O: -2 and since there are 3, it totals at -6
- As: In order to achieve neutrality, it must have a +3 oxidation state
Now, let’s look at the oxidation states for the products:
For NO:
- O: it has a -2 oxy. state
- N: To be neutral, N must have an oxidation state of +2
For H3AsO4:
- H: Totaling +3 (+1 for each H atom)
- O: Totaling -8 (-2 for each O atom)
- As: in order to be neutral, it must have an oxidation state of +5
For H2O:
- O: -2 oxidation state
- H: +1 for each atom, meaning +2 total
Step 3:
Now that we know the oxidation states, lets have a more detailed look at the changes in oxidation state for each atom:
Now since oxygen and hydrogen keep the same oxidation state in the reaction, they are not considered in contributing to the balancing of the reaction.
Step 4
Now we have to remember that the decrease of oxidation state has to equal the increase in oxidation state meaning that there must be 3 As atoms for every 2 N atoms to achieve an overall balance.
Step 5
Since we know this, we can write the reaction for the reactant side:
Step 6
Use inspection method to balance the rest of the equation, which should be easy enough to yield your final, balanced equation:
Now that you’ve had practice with this method, it’ll be important to know that there is another method you can use as well, which will be discussed in the next section.
Glossary and deeper reading · 6 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Oxidation
Oxidation is electron loss, represented by an increase in oxidation number. It occurs together with reduction so that electrons are conserved in the overall reaction. For example, a zinc atom becoming a zinc ion with charge +2 undergoes oxidation.
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Reduction
Reduction is electron gain, represented by a decrease in oxidation number. It is paired with oxidation in an electron-transfer reaction. For example, a copper ion with charge +2 gaining two electrons to form copper metal undergoes reduction.
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Reducing agent
A reducing agent causes another species to gain electrons. The agent supplies those electrons and is itself oxidized. For example, zinc acts as a reducing agent when it transfers electrons to copper ions.
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Redox reaction
A redox reaction combines oxidation of one species with reduction of another. Electron loss and gain must balance in the net equation along with atoms and total charge. For example, zinc reducing copper ions involves zinc oxidation and copper-ion reduction.
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Oxidation number
An oxidation number is a formal electron-accounting value assigned using specified rules. Changes in oxidation number help identify oxidation and reduction but are not always actual atomic charges. For example, oxygen usually has oxidation number −2 in water while hydrogen has +1.
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Oxidizing agent
An oxidizing agent causes another species to lose electrons. The agent accepts those electrons and is itself reduced. For example, a copper ion acts as an oxidizing agent when it accepts electrons from zinc metal.