General Chemistry (The contents will be updated periodically)
Q4: How can you balance a chemical equation?
A chemical change, or a chemical reaction, can be represented by a chemical equation. In many cases, chemical equations must be balanced so that the Law of Conservation of Mass is followed. Here, "balance" really means adding coefficients in front of reactants and products to make the same element having the same numbers on both sides of the equation. So in a balanced equation, if you have 5 O atoms on the left side, then you should also have 5 O atoms on the right side; otherwise, the equation is not balanced. It should be noted that the O atoms refer to the total number of O atoms, not the O atoms from a single molecule.
In general, when you balance a chemical equation, remember to (1) start from an atom that is converted from one single form to another; (2) balance polyatomic anions as a group if the group does not fall apart after the reaction. In the following sections, I will show you how to balance several typical reactions, especially redox reactions.
a. Combusion Reactions
These are typically referring to reactions of a substance with oxygen (O2). For example, butane (C3H8) can react with O2 to form CO2 and H2O as shown below:
C3H8 + O2 → CO2 + H2O
Clearly, this reaction is not balanced because the number of atoms of each type is not equal on both sides, for example, you have 3 C atoms on the left side but only one C on the right side. Since C only converts to one substance (CO2), so you should start to balance C first, not O, which converts to two substances. Because there are 3 C atoms on the left, you can add 3 in front of CO2 to balance C atoms as shown below:
C3H8 + O2 → 3 CO2 + H2O
Now C atoms are balanced. Let's look at H atom. Since there are 8 H atoms on the left side, you can add 4 in front of H2O to make it balanced as shown below:
C3H8 + O2 → 3 CO2 + 4 H2O
At this stage, the only atom left unbalanced is O. Since there are 10 O atoms (= 3 x 2 + 4 x 1) on the right side, so if you add 5 in front of O2 on the left side, then O atoms will be balanced as shown below:
C3H8 + 5 O2 → 3 CO2 + 4 H2O (balanced equation)
At this stage, you have 3 C atoms, 8 H atoms, and 10 O atoms on both sides. This equation is now balanced.
b. Decomposition Reactions
A decomposition reaction is a reaction that a reactant decomposes to one or more substances, such as the decomposition of washing soda, Na2CO3:
Na2CO3 → Na2O + CO2 (balanced equaton)
As a matter of fact, this equation is automatically balanced, no need to do anything.
Another example is the decomposition of potassium chlorate KClO3. This compound is relatively stable at room temperature; however, at a higher temperature, it will decompose to KCl and O2 as shown below:
KClO3 → KCl + O2
As written, K and Cl are already balanced, but not the O atom. Since you have an odd number of O on the left side and an even number of O on the right side, you can add 2 in front of KClO3 and 3 in front of O2 to balance the O atoms as shown below:
2 KClO3 → KCl + 3 O2
Now K and Cl are not balanced anymore. Since you have 2 K and 2 Cl on the left side, you can add 2 in front of KCl on the right side to make this equation balanced as shown below.
2 KClO3 → 2 KCl + 3 O2 (balanced equation)
c. Replacement Reactions
Replacement reactions are those that an element from one reactant is replaced by another. One example is the reaction between HCl (aq) and Zn (s) as shown below:
HCl (aq) + Zn (s) → ZnCl2 + H2
Since Zn is balanced as written, we need to look at H and Cl atoms. On the left side you have one H and one Cl; whereas on the right side you have 2 H atoms and 2 Cl atoms, so adding 2 in front of HCl will make this equation balanced as shown below
2 HCl (aq) + Zn (s) → ZnCl2 + H2 (balanced equation)
d. Precipitation Reactions
A precipitation reaction can be considered a double replacement reaction. Sometimes, the reaction is balanced automatically and no further action is required, but most times they aren't, so balancing is needed. For example, the reaction between Al2(SO4)3 and BaCl2 as shown below:
Al2(SO4)3 + BaCl2 → BaSO4 + AlCl3
Ba2+ is balanced, but not the Cl-, Al3+ and SO42-. You can start with Al atom. Since there are two Al atoms on the left side, you can add 2 in front of AlCl3 on the right side:
Al2(SO4)3 + BaCl2 → BaSO4 + 2 AlCl3
Now the total number of Cl atoms on the right side is 6, so you can add 3 in front of BaCl2:
Al2(SO4)3(aq)+ 3 BaCl2 (aq) → BaSO4 (a) + 2 AlCl3 (aq)
As a result of this action, there are 3 Ba atoms on the left side, you can add 3 in front of BaSO4 to make the equation balanced as shown below:
Al2(SO4)3 (aq) + 3 BaCl2 (aq) → 3 BaSO4 (s) + 2 AlCl3 (aq) (balanced euqtion)
e. Acid-base Neutralization Reactions
The reaction between an acid and a base is relatively easy to balance. Since the nature of this type of reactions is that it is a reaction between H+ and OH- with a 1:1 molar ratio, you can always start by adding an appropriate coefficient before an acid or a base to balance the equation. For example, the reaction between NaOH and H2SO4 as shown below:
NaOH (aq) + H2SO4 → Na2SO4 + H2O
Because every H2SO4 provides 2 H+ for the reaction, you can just add 2 in front of NaOH (aq), and consequently add 2 in front of H2O to make the reaction balanced as shown below:
2 NaOH (aq) + H2SO4(aq) → Na2SO4 (aq) + 2 H2O (l) (balanced equation)
f. Redox Reactions
A redox reaction involves electron transfer from one substance to another. Some equations can be balanced by observation such as the reaction between O2 and H2 to form H2O as shown below:
H2 + O2 → H2O
Because there are two O atoms on the left side, you can just add 2 in front of H2O on the right side, then you add 2 in front of H2 to balance H atoms. The balanced equation will be:
2 H2 (g) + O2 (g) → 2 H2O (l) (balanced equation)
Alternatively, you can use the half-reaction method to balance redox reactions. The general steps include (a) redox couples; (b) write and balance reduction and oxidation half-reactions; (c) multiply the smallest whole number to reduction and oxidation reactions to ensure they both have the same number of electrons. (d) add two half-reactions together to cancel the same substances to get the balanced redox reaction. I will use the following examples to demonstrate in detail how to follow this procedure to balance a redox reaction.
Example 1: Al (s) + Cu2+ (aq) → Al3+ (aq) + Cu (s)
In this reaction, redox couples are
Al (s) →Al3+ (Oxidation reaction)
Cu2+→Cu (s) (Reduction reaction)
Next, let's try to balance each half-reaction. We need to make sure mass and charge are balanced. For the oxidation reaction, the mass is already balanced, but not the charge. We need to add 3 electrons to the right side to balance the charge. So the balance oxidation reaction will be
Al (s) →Al3+ + 3 e- (balanced oxidation reaction)
Following the same procedure, the balanced reduction reaction will be
Cu2+ + 2 e-→Cu (s) (balanced reduction reaction)
Keep in mind that 2 electrons were added to the left side to make both sides have a total charge of zero.
Because a net non-zero charge cannot appear in the final balanced equation, we need to multiply a whole number to the reduction reaction and another to the oxidation reaction to make both have the same number of electrons. In this case, we multiply 2 to the oxidation reaction and 3 to the reduction reaction to get the following half-reactions:
2 Al (s) →2 Al3+ + 6 e- (balanced oxidation reaction)
3 Cu2+ + 6 e-→ 3 Cu (s) (balanced reduction reaction)
Now we can add these two equations together: left on the left side and right on the right side:
2 Al (s) + 3 Cu2+ + 6 e-→2 Al3+ + 6 e- + 3 Cu (s)
After cancelling 6 e-, you have the final balanced redox reacton:
2 Al (s) + 3 Cu2+ →2 Al3+ + 3 Cu (s) (balanced redox equation)
Example 2: KMnO4 + H2SO4 + FeSO4→K2SO4 + MnSO4 + Fe2(SO4)3 + H2O
The redox couples for this reaction are
KMnO4 → MnSO4 (Reduction Reaction)
FeSO4 → Fe2(SO4)3 (Oxidation Reaction)
For the reduction reaction, Since Mn is balanced, we can add K+ on the right side to make sure K atoms are balanced:
(1) KMnO4 → MnSO4 + K+
To make S atom balanced, we add H2SO4 because it is the only source of S atom:
(2) KMnO4 + H2SO4→ MnSO4 + K+
As a result, we add 4 H2O on the right to balance O:
(3) KMnO4 + H2SO4→ MnSO4 + K+ + 4 H2O
We also add 4 in front of H2SO4 to balance H atoms:
(4) KMnO4 + 4 H2SO4→ MnSO4 + K+ + 4 H2O
Consequently, we add 3 other SO42- to the right to make it balanced (we did not add 4 in front of MnSO4 to balance SO4-):
(5) KMnO4 + 4 H2SO4→ MnSO4 + K+ + 4 H2O + 3 SO42-
Now the mass is balanced, we add 5 electrons in the left to make the charge balanced:
(6) KMnO4 + 4 H2SO4 + 5 e-→ MnSO4 + K+ + 4 H2O + 3 SO42- (balanced reduction reaction)
Now let's look at the oxidation reaction. We first add 2 in front of FeSO4 and one SO42-tothe left to get the following:
(7) 2 FeSO4 + SO42-→ Fe2(SO4)3
Now the mass is balanced and we add 2 e- to the right to make the charge balanced:
(8) 2 FeSO4 + SO42-→ Fe2(SO4)3 + 2 e- (balanced oxidation reaction)
Next step is to multiply 2 to the equation (6) and 5 to the equation (8), and add both together to get the following equation:
(9) 2 KMnO4 + 8 H2SO4 + 10 e- + 10 FeSO4 + 5 SO42-→ 2 MnSO4 + 2 K+ + 8 H2O + 6 SO42- + 5 Fe2(SO4)3 + 10 e-
After cancelling 10 e- and 5 SO42-, we get the following:
(10) 2 KMnO4 + 8 H2SO4 + 10 FeSO4 + → 2 MnSO4 + 2 K+ + 8 H2O + SO42- + 5 Fe2(SO4)3
After you combine 2 K+ and SO42- to K2SO4, the final balanced equation is:
2 KMnO4 + 8 H2SO4 + 10 FeSO4 + → 2 MnSO4 + K2SO4 + 8 H2O + 5 Fe2(SO4)3
You can count the numbers of each atom on both sides to confirm this equation is balanced.
Example 3: MnO4- (aq) + C2O4-2 (aq)→ Mn2+ + CO2 (g) (acidic condition)
First, let's try to figure out the oxidation and reduction half-reactions. For the Mn atom, the oxidation state decreases from +7 to +2, so the reduction half-reaction will be :
(1) MnO4- → Mn2+
whereas C atom changes its oxidation state from +3 to +4, so the oxidation half-reaction will be :
(2) C2O4-2 → CO2
For the reduction reaction (1), we can add H2O on the right and H+ on the left to balance O and H atoms due to the fact that the reaction was carried out in an acidic condition in water. Keep in mind that it is pretty common to use H+/H2O to balance a reaction if the reaction is carried out under an acidic condition in an aqueous solution. For the reduction reaction, you simply add 2 in front of CO2 and 2 e- on the right side to make it balanced. The followings are two balanced half-reactions:
(3) MnO4- + 8 H+ + 5 e- →Mn2+ + 4 H2O
(4) C2O42- → 2 CO2 + 2 e-
The next step is to add two half-reactions together; however, before you do that, you need to mutiply 2 to the reaction (3) and 5 to the reaction (4). As a result, you will have
(5) 2 MnO4- + 16 H+ + 10 e- → 2 Mn2+ + 8 H2O
(6) 5 C2O42- → 10 CO2 + 10 e-
Now you can add two reactions together. After canceling 10 e-, you get the following balanced equation:
2 MnO4- (aq) + 16 H+ + 5 C2O42- (aq)→ 2 Mn2+ + 10 CO2 (g) + 8 H2O
Example 4: CN- (aq) + MnO4- (aq)→ CNO- + MnO2 (s) (basic condition)
Following the general procedure, you can easily identify oxidation and reduction pairs as shown below:
(1) Oxidation Reaction: CN- → CNO- and
(2) Reduction reaction: MnO4 → MnO2
Let's look at the oxidation reaction first. C and N atoms are balanced, but not the O. Since the reaction is carried out under a basic condition, we can add OH- to the left side of the reaction to balance the O and use H2O to balance the H atom:
(3) CN- + 2 OH-→ CNO- + H2O + 2 e-
For the reduction reaction, Mn is already balanced, but not the O atom. Again, because the reaction is carried out under a basic condition, we can add OH- to the right side of the reaction to balance the O and use H2O to balance the H atom:
(4) 3 e- + MnO4- + 2 H2O → MnO2 + 4 OH-
Since reaction 3 has 2 e- and reaction 4 has 3 e-, we will multiply 3 to the equation (3) and 2 to the equation (4) to get the following: - + 6 HO-→ 3 CNO- + 3 H2O + 6 e-
(6) 6 e- + 2 MnO4- + 4 H2O → 2 MnO2 + 8 OH-
Now we can add two reactions together. After cancelling H2O and OH-, the balanced equation will be:
3 CN- (aq) + H2O (l) + 2 MnO4- (aq) → 3 CNO- (aq) + 2 MnO2 (s) + 2 OH- (aq)
Keep in mind that when you use H+/H2O to balance the H and O, you need to add H+ on the side that has more O atoms, whereas when you use OH-/H2O to balance the H and O, you need to add OH- to the side that has less O atom. For example, for the following reaction:
H2 + O2 → H2O (acidic condition)
Reduction couple will be:
O2 (g) → H2O
Left side has more O atom, so H+ will be added to the left side:
O2 (g) + 4 H+ + 4 e-→ 2 H2O
Oxidation couple will be:
H2 (g) → H2O
Right side has more O, so H+ should be added on the right side:
2 H2 (g) + H2O → H2O + 4 H+ + 4 e-
Adding two together gives the balanced equation as below:
2 H2 (g) + O2 (g) → 2 H2O (l) (balanced equation)
