There are many kinds of chemical reactions; Some create explosions, others emit light, and some set off a bunch of other reactions. The type of chemical reaction we will be focusing on today is the redox reaction.
- This article is about redox reactions
- We will learn about oxidation states and how they are important to redox reactions
- Next, we will learn how to use the half-reaction method to balance reactions
- Lastly, we will learn about how photosynthesis is a prime example of a redox reaction
Redox reaction definition
Let's start by looking at the definition of a redox reaction.
A redox reaction involves the transfer of electrons from one species to another. In a redox reaction, the oxidation state of two species changes, with one being reduced (decreased) and one being oxidized (increased).
The oxidation state (also called oxidation number) corresponds to the number of electrons a species will give or receive during bonding. There are several rules to determine the oxidation state of a species, these are:
Single elements and single element compounds have an oxidation state of 0
The sum of the oxidation states in a neutral compound is equal to 0
The sum of the oxidation states in a charged compound is equal to that charge
In a compound, the more electronegative species (the species that is more likely to accept an electron) will have a negative oxidation state, while the less electronegative species (the species that is more likely to donate an electron) will have a positive oxidation state.
For ions, the oxidation state is equal to the charge
Some elements have a usual oxidation state. Transition metals tend to be way more varying.
| Element | Typical Ox. State | Exceptions |
| Hydrogen | + 1 | Metal hydrides (-1) |
| Group 1 Metals | +1 | None |
| Group 2 Metals | +2 | None |
| Oxygen | -2 | Peroxides and F2O |
| Fluorine | -1 | None |
| Chlorine | -1 | Compounds with O or F |
Now that we know the basics of oxidation states, let's work on a problem.
What is the oxidation state of the metal cation in these compounds?
AgNO3, ZnCl2, Co2O3
1. Let's start with AgNO3. NO3 is an ion with a charge of -1, so Ag has an oxidation state of +1.
2. For ZnCl2, we know that Cl usually has an oxidation state of -1. This is the case here, since there is no O or F in this compound. Since we have 2 mols of Cl to balance out Zn, then the ox. state of Zn is +2.
3. Lastly, for Co2O3, O3 is an ion with a charge of -2. Since we need 2 mols of Co to balance, then the oxidation state of Co is +1.
While we can calculate the oxidation number for every element in a compound, it is easier to calculate the number for just the cation (positively charged species) and anion (negatively charged species) as a whole. Ions like NO3 and SO4 will not be the thing that is being oxidized/reduced, so calculating the oxidation number for each element isn't needed.
Redox reaction example
Now that we know how to calculate oxidation states, let's look at a redox reaction example.
In this reaction, which species is being oxidized, and which is being reduced?
$$Zn + CuSO_4 \rightarrow Cu + ZnSO_4$$
1. Let's start by identifying the oxidation states on the left side. Zn is a lone neutral element, so its oxidation state is 0.
2. SO4 is an ion with a charge of -2, so Cu must have a charge of +2 since this is a neutral compound.
3. Now for the right side. Cu is now alone, so it's oxidation state is 0.
4. Zn is now bonded to SO4, so it's oxidation state is +2.
So, our changes in oxidation state are Cu (+2 → 0) and Zn (0 → +2). Since copper's oxidation state has decreased, it is being reduced. That means that zinc is being oxidized since its oxidation state has increased.
The accepting species is the oxidizing agent, while the species that is donating the electrons is the reducing agent. The wording may be a bit confusing at first, but it can be helpful that the "agent" is the species doing that action and not having that action done to it. Balancing redox reactions
An important component of redox reactions is that they must be balanced by both charge and mass. In the previous example, every species was neutral, but that isn't the case for every redox reaction. We use the half-reaction method to balance these equations, where one or more of the species is charged.
The half-reaction method of balancing is used to balance redox reactions. A redox reaction is split into the oxidation half-reaction and reduction half-reaction. These half-reactions are individually balanced and then recombined into the full balanced equation.
Let's break down how to use this method,
Redox half-reactions
To write these half-reactions, we need to follow these steps:1. Write the unbalanced reaction in its ionic form2. Determine the oxidation state of every species in the reaction3. Determine which species is being oxidized, and which is being reduced4. Split the reaction into the two half-reactions5. Use electrons to balance the charge for each half-reaction.6. Multiply the necessary half-reactions so that the number of electrons cancels7. Combine each half-reaction and cancel like terms
Let's start with an example. Balance the redox reaction using the skeleton equation below:
$$Fe^{3+} + Sn^{2+} \rightarrow Fe^{2+} + Sn^{4+}$$
We already are given the reaction in its ionic form, so we start by determining the oxidation state for each species.
This step is pretty simple since all of our elements are ions, their oxidation numbers are just their charges.
Now we need to split this reaction into its half-reactions. Let's start with Fe.
$$Fe^{3+} \rightarrow Fe^{2+}$$
This half-reaction is the reduced half-reaction, since Fe's oxidation number is being reduced. Species being reduced gain an electron, so we want to balance our charge by adding an electron to the left side
$$Fe^{3+} + e^- \rightarrow Fe^{2+}$$
Now for our oxidation half-reaction.
$$Sn^{2+} \rightarrow Sn^{4+}$$
Species being oxidized lose an electron(s), so we balance by adding 2 electrons to the right side.
$$Sn^{2+} \rightarrow Sn^{4+} + 2e^-$$
Before we combine these reactions, we need to make sure that the electrons will cancel. We can multiply the reduction reaction by 2 so that our electrons will cancel.
$$2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$$
Now we can combine our reactions to get the final, balanced reaction
$$2Fe^{3+} + Sn^{2+} \rightarrow 2Fe^{2+} + Sn^{4+}$$
The final reaction shouldn't have any electrons listed as reactants or products, so make sure that you cancel them out. Also, the reduction half-reaction will always have its electrons on the left and the oxidation half-reaction will have them on the right.
Redox reactions in acidic and basic solutions
When a redox reaction is taking place in an acidic or basic solution, we need to alter the steps just a bit. These steps take place after step 4 in our half-reaction method.
- For acidic solutions: Use H2O to balance oxygen atoms and use H+ to balance hydrogen atoms.
- For basic solutions: Use H2O to balance oxygen atoms and use H+ to balance hydrogen atoms. Add OH- to neutralize any H+ atoms and convert them into H2O molecules. (this step is at the end)
Let's start with an example in an acidic solution.
Write the balanced redox reaction (in acidic solution) using the skeleton formula below:
$$MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}$$
We start by determining each species' oxidation state:
Fe is simple since its oxidation state is just its charge, so it's oxidation state goes from +2 to +3. This means that Fe is being oxidized. The oxidation state of oxygen is -2, and we have 4 mols of it. This means that Mn must have an ox. state of +7 since the total charge of the compound is -1. Mn is being reduced since its oxidation state goes from +7 to +2.
Now we need to draw our half-reactions. Let's start with Mn
$$MnO_4^- \rightarrow Mn^{2+}$$
First, we must balance our oxygen by adding 4 mols of H2O to the right side
$$MnO_4^- \rightarrow Mn^{2+} + 4H_2O$$
Next, we balance our hydrogen by adding 8 mols of H+ to the left side
$$MnO_4^- + 8H^+ \rightarrow Mn^2+ + 4H_2O$$
Mn goes from +7 to +2, meaning it gains 5 electrons, so we need to add those 5 electrons to the equation
$$MnO_4^- +5e^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O$$
Now that this half-reaction is balanced, we can move on to the oxidation half-reaction
$$Fe^{2+} \rightarrow Fe^3+$$
Fe loses an electron in this reaction, so we can add that electron to the right side
$$Fe^{2+} \rightarrow Fe^{3+} + e^-$$
When we combine these reactions, we want the electrons to cancel, so we have to multiply the whole equation by 5.
$$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$$
Lastly, we can combine these two reactions to get the final redox reaction.
$$MnO_4^- + 8H^+ +5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$$
Reactions in acidic solution will always have H+ present, and reactions in basic solution will always have OH-. Now onto a basic solution problem.
Balance the redox reaction (in basic solution) given the skeleton equation:
$$Zn + NO_3^- \rightarrow Zn^{2+} + NO_2$$
The oxidation states for each species are the same as their charge.
Now onto the half-reactions.
$$Zn \rightarrow Zn^{2+}$$
Zn is being oxidized, since its oxidation number is increasing. This also means we need to add our 2 electrons to the right side.
$$Zn \rightarrow Zn^2+ + 2e^-$$
Now for our reduction reaction,
$$NO_3^- \rightarrow NO_2$$
We can add one H
2O molecule to our right side to balance O,$$NO_3^- \rightarrow NO_2 + H_2O$$
Next, we add 2H+ to the left to balance H
$$NO_3^- + 2H^+ \rightarrow NO_2 + H_2O$$
Since this is a reduction, we add an electron to the left side
$$NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O$$
For our electrons to cancel, we multiply this reaction by 2
$$2NO_3^- + 4H^+ + 2e^- \rightarrow 2NO_2 + 2H_2O$$
Now we can combine our reactions
$$2NO_3 + 4H^+ + Zn \rightarrow Zn^{2+} + 2NO_2 + 2H_2O$$
Now for the fun part. We need to add 4OH- to both sides, so we can make water to cancel our hydrogen molecules
$$2NO_3 + 4H^+ + Zn + 4OH^- \rightarrow Zn^{2+} + 2NO_2 + 2H_2O + 4OH^- $$
We combine the H+ and OH- to make water
$$2NO_3 + Zn + 4H_2O \rightarrow Zn^{2+} + 2NO_2 + 2H_2O + 4OH^- $$
Lastly, we can subtract the 2 water molecules on the right from the left. So, our final reaction is:
$$2NO_3 + Zn + 2H_2O \rightarrow Zn^{2+} + 2NO_2 + 4OH^- $$
For basic reactions, there should not be any H+ atoms in the final reaction.
Photosynthesis redox reaction
Probably the most famous of the redox reactions is photosynthesis.
Fig.1-During photosynthesis, carbon dioxide and sunlight are converted into glucose and oxygen gas.
There are two main reactions in photosynthesis: a light-dependent and a light-independent reaction. Both of these reactions are redox reactions. Let's look at the light-dependent reaction.
Which species are oxidized and reduced in this reaction?
$$2H_2O + 2NADP^+ 3ADP + 3P + light \rightarrow 2NADPH + 2H^+ + 3ATP + O_2$$
First things first, we have a bunch of things that, while they are a part of the reaction as a whole, aren't directly a part of the redox. We can simplify this reaction to be:
$$2H_2O + 2NADP^+ \rightarrow 2NADPH + 2H^+ + O_2$$
Now on to giving each species its oxidation number. H has an oxdation state of +1 on both sides, O starts with -2, then changes to 0. NADP is a bit trickier. Instead of calculating the oxidation state, we can look at the half-reaction.
$$NADP^+ \rightarrow NADPH$$$$NADP^+ + H^+ \rightarrow NADPH$$
$$NADP^+ + H^+ + 2e^- \rightarrow NADPH$$
We see that 2e- are added on the left side to balance the reaction. Since NADP+ is gaining electrons, it is being reduced. Oxygen goes from an oxidation state of -2 to 0, so it is being oxidized.
Redox Reactions - Key takeaways
- A redox reaction involves the transfer of electrons from one species to another. In a redox reaction, the oxidation state of two species changes, with one being reduced (decreased) and one being oxidized (increased).
- The oxidation state (also called oxidation number) corresponds to the number of electrons a species will give or receive during bonding.
- By calculating oxidation states, we can determine which species is being reduced and which species is being oxidized
- The species that is accepting the electrons is the oxidizing agent, while the species that is donating the electrons is the reducing agent.
- We use the half-reaction method to balance redox reactions
- We use H+ and H2O to balance acidic redox reactions, and those species plus OH- for basic ones
- Photosynthesis is an important example of a redox reaction
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