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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry>Unit 4

Lesson 5: Oxidation–reduction (redox) reactions

# Balancing redox equations

AP.Chem:
TRA‑2 (EU)
,
TRA‑2.C (LO)
,
TRA‑2.C.1 (EK)

## Introduction

A rusted buoy on a rocky beach.
Rust is formed by the redox reaction of iron and oxygen gas in the presence of moisture. Image credit: "Badentarbat Bay: Corroded Buoy on the Beach" by DeFacto on Wikimedia Commons, CC BY-SA 4.0.
Oxidation–reduction or redox reactions are reactions that involve the transfer of electrons between chemical species (check out this article on redox reactions if you want a refresher!). The equations for oxidation-reduction reactions must be balanced for both mass and charge, which can make them challenging to balance by inspection alone. In this article, we’ll learn about the half-reaction method of balancing, a helpful procedure for balancing the equations of redox reactions occurring in aqueous solution.

## The half-reaction method of balancing redox equations

To balance a redox equation using the half-reaction method, the equation is first divided into two half-reactions, one representing oxidation and one representing reduction. The equations for the half-reactions are then balanced for mass and charge and, if necessary, adjusted so that the number of electrons transferred in each equation is the same. Finally, the half-reaction equations are added together, giving the balanced overall equation for the reaction.
Let’s see how this procedure works for a simpler redox reaction. For example, consider the reaction between the $\ce{Co^3+}$ ion and nickel metal:
$\ce{Co^3+}(aq) + \ce{Ni}(s) \rightarrow \ce{Co^2+}(aq) + \ce{Ni^2+}(aq)$
Is this equation balanced? It appears to be balanced with respect to mass since there is one C, o atom and one N, i atom on each side of the equation. However, it is not balanced for charge: the net charge on the left side of the equation is 3, plus, while the net charge on the right side is 4, plus. To help us balance the equation for charge, we’ll use the half-reaction method.
To start, let’s split the equation into separate oxidation and reduction half-reactions:
Oxidation half-reaction: The oxidation half-reaction shows the reactants and products participating in the oxidation process. Since N, i metal is being oxidized to $\ce{Ni^2+}$ in this reaction, we might start by writing out that process:
$\text{Oxidation:}\; \ce{Ni}(s) \rightarrow \ce{Ni^2+}(aq)$
However, this isn’t the complete oxidation half-reaction! Like the overall equation, our half-reaction is balanced for mass but not charge. We can balance it for charge by adding two electrons to the right side of the equation so that the net charge on each side is 0:
$\text{Oxidation:}\; \ce{Ni}(s) \rightarrow \ce{Ni^2+}(aq) + \blueD{2\,e^-}$
Now that the oxidation half-reaction is balanced, it tells us that two electrons are produced for every atom of nickel oxidized. But where do those electrons go? We can follow their trail to the reduction half-reaction.
Reduction half-reaction: The reduction half-reaction shows the reactants and products participating in the reduction step. In this case, our equation should show $\ce{Co^3+}$ being reduced to $\ce{Co^2+}$. It should also include an electron on the left side of the equation for charge balance:
$\text{Reduction:}\; \ce{Co^3+}(aq) + \blueD{e^-} \rightarrow \ce{Co^2+}(aq)$
The balanced reduction half-reaction tells us that one electron is consumed for every $\ce{Co^3+}$ ion reduced. Importantly, the electrons for this process come from the oxidation half-reaction.
Next, we’ll want to add the balanced half-reactions together to get the balanced overall equation. First, however, we need to make sure that the electrons will cancel out when we combine the half-reactions (we can't have stray electrons floating around!). Right now, the oxidation half-reaction involves the transfer of two electrons, while the reduction half-reaction involves the transfer of only one electron. So, we need to multiply the reduction half-reaction by 2:
\begin{aligned} &2[\ce{Co^3+}(aq) + e^- \rightarrow \ce{Co^2+}(aq)] \\\\ &\ce{2Co^3+}(aq) + 2\,e^- \rightarrow \ce{2Co^2+}(aq) \end{aligned}
Now, we can add the two half-reactions together, cancelling out the electrons on both sides:
\begin{aligned} &\ce{Ni}(s) \rightarrow \ce{Ni^2+}(aq) + \blueD{\cancel{2\,e^-}} \\\\ &\ce{2Co^3+}(aq) + \blueD{\cancel{2\,e^-}} \rightarrow \ce{2Co^2+}(aq) \\\\[-0.90em] &\overline{\phantom{\ce{Ni}(s) + \ce{2Co^3+}(aq) \rightarrow \ce{Ni^2+}(aq) + \ce{2Co^2+}(aq)}} \\[-0.65em] &\ce{Ni}(s) + \ce{2Co^3+}(aq) \rightarrow \ce{Ni^2+}(aq) + \ce{2Co^2+}(aq) \end{aligned}
The resulting equation has equal numbers of each type of atom on both sides of the equation (1 N, i and 2 C, o) as well as the same net charge on each side (6, plus). Together, this means that the equation is balanced for both mass and charge!

## Balancing redox equations in acidic or basic solution

We just used the half-reaction method to balance a simple redox equation. However, many of the redox reactions that occur in aqueous solution are more complicated than the example shown above. In these cases, we may need to add $\ce{H2O}$ molecules and either $\ce{H+}$ ions (for reactions occurring in acidic solution) or $\ce{OH-}$ ions (for reactions occurring in basic solution) to fully balance the equation. Example 1 below shows this process for a reaction taking place in acidic solution, while Example 2 shows this process for a reaction taking place in basic solution.

## Example 1: Balancing redox equations in acidic solution

Balance the equation for the reaction of copper metal with the nitrate ion in acidic solution.
$\ce{Cu}(s) + \ce{NO3-}(aq) \rightarrow \ce{Cu^2+}(aq) + \ce{NO2}(g)$
To balance the equation, let’s follow the half-reaction method we just learned. Since this reaction is occurring in acidic solution, we can use $\ce{H+}$ ions and $\ce{H2O}$ molecules to help balance the equation.

### Step 1: Divide the equation into half-reactions

Let’s start by separating the unbalanced equation into two half-reactions:
\begin{aligned} &\text{Oxidation:}\; \ce{Cu}(s) \rightarrow \ce{Cu^2+}(aq) \\\\ &\text{Reduction:}\; \ce{NO3-}(aq) \rightarrow \ce{NO2}(g) \end{aligned}
Note that neither half-reaction is fully balanced! We’re going to do that in the next step.

### Step 2: Balance each half-reaction for mass and charge

The oxidation half-reaction is already balanced for mass, so we just need to balance it for charge. We can do so by adding two electrons to the right side of the equation, making the net charge 0 on both sides:
$\text{Oxidation:}\; \ce{Cu}(s) \rightarrow \ce{Cu^2+}(aq) + \blueD{2\,e^-}$
What about the reduction half-reaction? This equation is unbalanced in terms of both mass and charge. Let’s balance it for mass first: we know that the N atoms are already balanced (there is one on each side of the equation). However, the O atoms are not. We can balance the O atoms by adding one $\ce{H2O}$ molecule to the right side of the equation:
$\ce{NO3-}(aq) \rightarrow \ce{NO2}(g) + \blueD{\ce{H2O}(l)}$
There are now two unbalanced H atoms on the right side of the equation. Since the reaction is in acidic solution, we can balance these atoms by adding $\ce{H+}$ ions to the left side:
$\ce{NO3-}(aq) + \blueD{\ce{2H+}}(aq) \rightarrow \ce{NO2}(g) + \ce{H2O}(l)$
Next, let’s balance the equation for charge. To do so, we must add one electron to the left side of the equation so that the net charge on each side is 0:
$\text{Reduction:}\; \ce{NO3-}(aq) + \ce{2H+}(aq) + \blueD{e^-} \rightarrow \ce{NO2}(g) + \ce{H2O}(l)$

### Step 3: Equalize the number of electrons transferred

Since two electrons are lost in the oxidation half-reaction and one electron is gained in the reduction half-reaction, we need to multiply the reduction half-reaction by 2:
\begin{aligned} &2[\ce{NO3-}(aq) + \ce{2H+}(aq) + e^- \rightarrow \ce{NO2}(g) + \ce{H2O}(l)] \\\\ &\ce{2NO3-}(aq) + \ce{4H+}(aq) + 2\,e^- \rightarrow \ce{2NO2}(g) + \ce{2H2O}(l) \end{aligned}

### Step 4: Add the half-reactions together

Combining the two half-reactions and cancelling out the electrons, we get
\begin{aligned} &\ce{Cu}(s) \rightarrow \ce{Cu^2+}(aq) + \blueD{\cancel{2\,e^-}} \\\\ &\ce{2NO3-}(aq) + \ce{4H+}(aq) + \blueD{\cancel{2\,e^-}} \rightarrow \ce{2NO2}(g) + \ce{2H2O}(l) \\\\[-0.90em] &\overline{\phantom{\ce{Cu}(s) + \ce{2NO3-}(aq) + \ce{4H+}(aq) \rightarrow \ce{Cu^2+}(aq) + \ce{2NO2}(g) + \ce{2H2O}(l)}} \\[-0.65em] &\ce{Cu}(s) + \ce{2NO3-}(aq) + \ce{4H+}(aq) \rightarrow \ce{Cu^2+}(aq) + \ce{2NO2}(g) + \ce{2H2O}(l) \end{aligned}
And we’re done! Let’s check our work: there are equal numbers of each type of atom on both sides of the equation (1 C, u, 2 N, 6 O, and 4 H) and the net charge is the same on each side (2, plus), so the equation is balanced!

## Example 2: Balancing redox equations in basic solution

Balance the equation for the reaction of permanganate and iodide ions in basic solution.
$\ce{MnO4-}(aq) + \ce{I-}(aq) \rightarrow \ce{MnO2}(s) + \ce{I2}(aq)$
Again, let’s use the half-reaction method to balance this equation. This time, however, we can only use $\ce{OH-}$ ions and $\ce{H2O}$ molecules and to help balance the equation since the reaction is taking place in basic solution.

### Step 1: Divide the equation into half-reactions

In this reaction, the iodide ion is oxidized and the permanganate ion is reduced:
\begin{aligned} &\text{Oxidation:}\; \ce{I-}(aq) \rightarrow \ce{I2}(aq) \\\\ &\ce{Reduction:}\; \ce{MnO4-}(aq) \rightarrow \ce{MnO2}(s) \end{aligned}

### Step 2: Balance each half-reaction for mass and charge

Let’s start with the oxidation half-reaction, which needs to be balanced for both mass and charge. First, we place the coefficient 2 in front of $\ce{I-}$ to achieve mass balance:
$\blueD{2}\,\ce{I-}(aq) \rightarrow \ce{I2}(aq)$
Then, we add two electrons to the right side of the equation to achieve charge balance:
$\text{Oxidation:}\; \ce{2I-}(aq) \rightarrow \ce{I2}(aq) + \blueD{2\,e^-}$
Next, let’s turn to the reduction half-reaction, which also needs to be balanced for both mass and charge. We’ll start with mass: since there is already one M, n atom on each side of the equation, we only need to balance the O atoms. We could do so by adding $\ce{OH-}$ ions and $\ce{H2O}$ molecules to either side of the equation in a trial and error approach, but this method can be tricky and time-consuming. Instead, let's first balance the half-reaction as if it occurs in acidic solution:
$\ce{MnO4-}(aq) + \blueD{\ce{4H+}(aq)} \rightarrow \ce{MnO2}(s) + \blueD{\ce{2H2O}(l)}$
Then, to account for the fact that the half-reaction actually takes place in basic solution, let's add $\ce{OH-}$ to both sides of the equation to neutralize the $\ce{H+}$:
\begin{aligned} &\ce{MnO4-}(aq) + \underbrace{\ce{4H+}(aq) + \blueD{\ce{4OH-}(aq)}} \rightarrow \ce{MnO2}(s) + \ce{2H2O}(l) + \blueD{\ce{4OH-}(aq)} \\[0.75em] &\kern9.50em\ce{4H2O}(l) \\\\ &\ce{MnO4-}(aq) + \ce{2H2O}(l) \rightarrow \ce{MnO2}(s) + \ce{4OH-}(aq) \end{aligned}
Notice that we combined the $\ce{H+}$ and $\ce{OH-}$ ions on the left side of the equation to form new $\ce{H2O}$ molecules, and then we eliminated the $\ce{H2O}$ molecules that appeared on both sides of the equation.
Finally, let’s balance the half-reaction for charge. To do so, we’ll add three electrons to the left side of the equation, making the net charge on each side 4, minus:
$\text{Reduction:}\; \ce{MnO4-}(aq) + \ce{2H2O}(l) + \blueD{3\,e^-} \rightarrow \ce{MnO2}(s) + \ce{4OH-}(aq)$

### Step 3: Equalize the number of electrons transferred

To equalize the number of electrons transferred in the two half-reactions, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 (resulting in each half-reaction containing six electrons):
\begin{aligned} &3[\ce{2I-}(aq) \rightarrow \ce{I2}(aq) + 2\,e^-] \\\\ &\ce{6I-}(aq) \rightarrow \ce{3I2}(aq) + 6\,e^- \\\\ &2[\ce{MnO4-}(aq) + \ce{2H2O}(l) + 3\,e^- \rightarrow \ce{MnO2}(s) + \ce{4OH-}(aq)] \\\\ &\ce{2MnO4-}(aq) + \ce{4H2O}(l) + 6\,e^- \rightarrow \ce{2MnO2}(s) + \ce{8OH-}(aq) \end{aligned}

### Step 4: Add the half-reactions together

Finally, let’s add the two half-reactions together, making sure to cancel out the electrons in each equation:
\begin{aligned} &\ce{2MnO4-}(aq) + \ce{4H2O}(l) + \blueD{\cancel{6\,e^-}} \rightarrow \ce{2MnO2}(s) + \ce{8OH-}(aq) \\\\ &\ce{6I-}(aq) \rightarrow \ce{3I2}(aq) + \blueD{\cancel{6\,e^-}} \\\\[-0.90em] &\overline{\phantom{\ce{2MnO4-}(aq) + \ce{4H2O}(l) + \ce{6I-}(aq) \rightarrow \ce{2MnO2}(s) + \ce{8OH-} + \ce{3I2}(aq)}} \\[-0.65em] &\ce{2MnO4-}(aq) + \ce{6I-}(aq) + \ce{4H2O}(l) \rightarrow \ce{2MnO2}(s) + \ce{3I2}(aq) + \ce{8OH-}(aq) \end{aligned}
Checking our work, we see that there are 2 M, n, 12 O, 8 H, and 6 I atoms as well as a net charge of 8, minus on both sides of the equation. So, the equation is balanced!

## Summary

We can use the half-reaction method to balance the equations of redox reactions occurring in aqueous solution. In this method, a redox equation is separated into two half-reactions, one involving oxidation and one involving reduction. Each half-reaction is balanced for mass and charge, and then the two equations are recombined with appropriate coefficients so that the electrons cancel. To balance more complex redox equations, it is sometimes necessary to add $\ce{H+}$ ions and $\ce{H2O}$ molecules (if the reaction occurs in acidic solution) or $\ce{OH-}$ ions and $\ce{H2O}$ molecules (if the reaction occurs in basic solution) to the equation.

## Want to join the conversation?

• Why is the reduction half reaction is written like this:
2H​+​​(aq)+2e​−​​→H​2​​(g)
Isn't it:
H​+​​(aq)+2e​−​​→H​2​​(g)

Since the base equation contains:
H​+​​(aq) -> H​2​​(g)

I do not understand this...
• H​+​​(aq)+2e​−​​→H​2​​(g) balances neither the charge nor the mass.
Left side: Charge of -1; one H
Right side: 0 charge; 2 H atoms (bonded as H2)
For balancing Redox reactions, it is necessary to first balance the main atoms (through adjusting stoichiometric coefficients), then the charges (through electron transfer and as per conditions - acidic/neutral/basic).
• In the example of combustion reaction, the oxidation number of C in C8H18 comes out to be -9/4. How is this possible? Electrons cannot be transferred in fractions right?
• It may be better to consider that there are two different carbon oxidation numbers rather than the fractional one.

The two on each end are bonded to 3 hydrogens so are -3, the six in the middle are bonded to 2 hydrogens so are -2
• It says above for Disproportionate Reaction: "If we analyze the oxidation numbers for chlorine, we see that the reactant ClO− ​is being oxidized to ClO​3​− (where the oxidation number increases from +1 to +5)."

I can't see what goes from +1 to + 5. Is it Cl? If so, how so?
• Cl is +1 in ClO^- and goes to +5 in ClO3^-

As oxygen is more electronegative than chlorine (in the Pauling scale), we assign it its preferred oxidation number which is -2.

In a molecule the oxidation numbers of all atoms sum to the charge.

So in ClO^- the oxidation numbers need to sum to -1. If oxygen is -2 then chlorine needs to be +1

-1 = Cl + O
-1 = x + -2
x = +1

For ClO3^- it's the same idea, but now we have 3 oxygens.

-1 = Cl + 3O
-1 = x + -6
x = +5
• 1.)Im not sure how to recognize if the oxidation number is -2 or +2.....
2.)For example, in CO2, is 2 the oxidation number or the number of atoms or the valency??
Pls answer in detail...Im really confused.
(1 vote)
• The oxidation number is the number of valence electrons an atom is assumed to have when the electrons are counted according to certain arbitrary rules.
Two important rules are;
1. The oxidation number of O in its compounds is almost always -2.
2. The charge on a molecule or ion is equal to the sum of the oxidation numbers of its atoms.
Thus the oxidation number of an O atom in CO₂ is -2.
The two O atoms have a total oxidation number of -4.
Since CO₂ is neutral, the oxidation number of C is +4.
• In the Practice question above, it does not specify what type of redox reaction it is. How do I know what type it is and what to put in as my answer? P.S. I got it wrong.
• Compare the reaction to the types explained in the article --
1) Is there molecular oxygen (O2) involved? NOPE not oxidation
2) Has a single reactant undergone both oxidation and reduction? NOPE not disproportionation
3) Are two elements trading places within a compound?
Doesn't look like it, but what if we look at the complete equation (rather than the net ionic equation we are given).
• there must be a counter ion that came with the "H+" ... let's pretend that we added HCl ...
This gives us the reaction:
Al(s) + HCl(aq) → AlCl3(aq) + H2(g)
Now we can see that it fits the definition for a single replacement reaction!

To answer the question you don't need to know what type of reaction it is, you just need to make sure that you've balanced electrons (charges) as well as atoms.

The method shown above is to separate the oxidation and reduction reactions into half reactions ... did you try that?
• What do you mean by saying "we can check to see if any reactants and products appear on both side"? Do you have an example?
• In this context, I think what he means is that if any ion is in an identical form on the left and right side of the equation then they cancel out and can be crossed out. In other words, identify any spectator ions.
• Where does the water from the combustion example go? (car engine example) Steam?
• Yes, you are correct in the form of vapour it comes out of the engine and enters the air.
• For Fe2O3, I know that the oxidation mu,her for oxygen is always -2 but then it says that the oxidation state of iron is +3. How does that make the compound neutral? Wouldn't the charge be +1 and not neutral?
• You aren't taking in to account that there are 2 iron atoms and 3 oxygen atoms in one Fe2O3 molecule.

3 oxygens with oxidation numbers of -2 each, -2 * 3 = -6
Fe2O3 is a neutral molecule so we know the oxidation numbers all must sum to 0
So the two irons must have oxidation numbers that equal +6, +6 / 2 = +3