A redox is a chemical reaction in which one of the reactants is reduced and the other oxidizes. Reduction and oxidation are processes that refer to the transfer of electrons between elements or compounds and are designated by the oxidation state. An atom oxidizes as its oxidation number increases and decreases as this value decreases. Redox reactions are critical to basic life functions, such as photosynthesis and respiration. More steps are required to balance a redox than with normal chemical equations. The most important aspect is to determine if redox actually occurs.
Steps
Part 1 of 3: Identifying a Redox Reaction
Step 1. Learn the rules for assigning the oxidation state
The oxidation state (or number) of a species (each element of the equation) is equal to the number of electrons that can be acquired, given away or shared with another element during the chemical bonding process. There are seven rules that allow you to determine the oxidation state of an element. They must be followed in the order presented below. If two of them are in contrast, use the first to assign the oxidation number (abbreviated "n.o.").
- Rule # 1: A single atom, by itself, has a n.o. of 0. For example: Au, n.o. = 0. Also Cl2 has an n.o. of 0 if it is not combined with another element.
- Rule # 2: the total oxidation number of all atoms of a neutral species is 0, but in an ion it is equal to the ionic charge. The "no. of the molecule must be equal to 0, but that of any single element can be different from zero. For example, H.2Or has a n.o. of 0, but each hydrogen atom has an n.o. of +1, while that of oxygen -2. The ion Ca2+ has an oxidation state of +2.
- Rule # 3: For compounds, group 1 metals have a n.o. of +2, while those of group 2 of +2.
- Rule # 4: The oxidation state of fluorine in a compound is -1.
- Rule # 5: The oxidation state of hydrogen in a compound is +1.
- Rule # 6: The oxidation number of oxygen in a compound is -2.
- Rule # 7: In a compound with two elements where at least one is a metal, the elements of group 15 have a n.o. of -3, those of group 16 of -2, those of group 17 of -1.
Step 2. Divide the reaction into two half reactions
Even if the half reactions are only hypothetical, they help you to easily understand if a redox is in progress. To create them, take the first reagent and write it as a half reaction with the product that includes the element in the reagent. Then take the second reagent and write it as a half reaction with the product that includes that element.
-
For example: Fe + V2OR3 - Fe2OR3 + VO can be divided into the following two half-reactions:
- Fe - Fe2OR3
- V.2OR3 - VO
-
If there is only one reagent and two products, create a half reaction with the reagent and the first product, then another with the reagent and the second product. When combining the two reactions at the end of the operation, do not forget to recombine the reagents. You can follow the same principle if there are two reagents and only one product: create two half reactions with each reagent and the same product.
- ClO- - Cl- + ClO3-
- Semireaction 1: ClO- - Cl-
- Semireaction 2: ClO- - ClO3-
Balance Redox Reactions Step 3 Step 3. Assign the oxidation state to each element of the equation
Using the seven rules mentioned above, determine the n.o. of all sorts of the chemical equation you have to solve. Even if a compound is neutral, its constituent elements have an oxidation number other than zero. Remember to follow the rules in order.
- Here are the n.o. of the first half reaction of our previous example: for the single Fe atom 0 (rule # 1), for Fe in Fe2 +3 (rule # 2 and # 6) and for O in O3 -2 (rule # 6).
- For the second half-reaction: for V in V2 +3 (rule # 2 and # 6), for O in O3 -2 (rule # 6). For V it is +2 (rule # 2), while for O -2 (rule # 6).
Balance Redox Reactions Step 4 Step 4. Determine if one species is oxidized and the other is reduced
By looking at the oxidation number of all species in the half-reaction, you determine if one oxidizes (its n.o. increases) and the other decreases (its n.o. decreases).
- In our example, the first half reaction is an oxidation, because Fe starts with a n.o. equal to 0 and reaches +3. The second half reaction is a reduction, because V starts with a n.o. of +6 and reaches +2.
- As one species oxidizes and the other reduces, the reaction is redox.
Part 2 of 3: Balancing an Redox into an Acid or Neutral Solution
Balance Redox Reactions Step 5 Step 1. Divide the reaction into two half reactions
You should have done this in the previous steps to determine if it is a redox. If, on the other hand, you have not done so, because in the text of the exercise it is expressly stated that it is a redox, the first step is to divide the equation into two halves. To do this, take the first reagent and write it as a half reaction with the product that includes the element in the reagent. Then take the second reagent and write it as a half reaction with the product that includes that element.
-
For example: Fe + V2OR3 - Fe2OR3 + VO can be divided into the following two half-reactions:
- Fe - Fe2OR3
- V.2OR3 - VO
-
If there is only one reagent and two products, create a half reaction with the reagent and the first product and another with the reagent and the second product. When combining the two reactions at the end of the operation, do not forget to recombine the reagents. You can follow the same principle if there are two reagents and only one product: create two half reactions with each reagent and the same product.
- ClO- - Cl- + ClO3-
- Semireaction 1: ClO- - Cl-
- Semireaction 2: ClO- - ClO3-
Balance Redox Reactions Step 6 Step 2. Balance all elements in the equation except hydrogen and oxygen
Once you have established that you are dealing with redox, it is time to balance it. It begins by balancing all the elements in each half-reaction other than hydrogen (H) and oxygen (O). Below you will find a practical example.
-
Semireaction 1:
- Fe - Fe2OR3
- There is one Fe atom on the left side and two on the right, so multiply the left side by 2 to balance.
- 2Fe - Fe2OR3
-
Semireaction 2:
- V.2OR3 - VO
- There are 2 atoms of V on the left side and one on the right, so multiply the right side by 2 to balance.
- V.2OR3 - 2VO
Balance Redox Reactions Step 7 Step 3. Balance the oxygen atoms by adding H.2Or to the opposite side of the reaction.
Determine the number of oxygen atoms on either side of the equation. Balance this by adding water molecules to the side with fewer oxygen atoms until the two sides are equal.
-
Semireaction 1:
- 2Fe - Fe2OR3
- On the right side there are three O atoms and zero on the left. Add 3 molecules of H2Or on the left side to balance.
- 2Fe + 3H2O - Fe2OR3
-
Semireaction 2:
- V.2OR3 - 2VO
- There are 3 O atoms on the left side and two on the right side. Add a molecule of H.2Or on the right side to balance.
- V.2OR3 - 2VO + H2OR
Balance Redox Reactions Step 8 Step 4. Balance the hydrogen atoms by adding H.+ to the opposite side of the equation.
As you did for oxygen atoms, determine the number of hydrogen atoms on either side of the equation, then balance them by adding H atoms+ from the side that has less hydrogen, until they are the same.
-
Semireaction 1:
- 2Fe + 3H2O - Fe2OR3
- There are 6 H atoms on the left side and zero on the right side. Add 6 H+ to the right side to balance.
- 2Fe + 3H2O - Fe2OR3 + 6H+
-
Semireaction 2:
- V.2OR3 - 2VO + H2OR
- There are two H atoms on the right side and none on the left. Add 2 H+ left side to balance.
- V.2OR3 + 2H+ - 2VO + H2OR
Balance Redox Reactions Step 9 Step 5. Equalize the charges by adding electrons from the side of the equation that requires them
Once the hydrogen and oxygen atoms are balanced, one side of the equation will have a greater positive charge than the other. Add enough electrons to the positive side of the equation to bring the charge back to zero.
- Electrons are almost always added from the side with the H atoms+.
-
Semireaction 1:
- 2Fe + 3H2O - Fe2OR3 + 6H+
- The charge on the left side of the equation is 0, while the right side has a charge of +6, due to hydrogen ions. Add 6 electrons on the right side to balance.
- 2Fe + 3H2O - Fe2OR3 + 6H+ + 6e-
-
Semireaction 2:
- V.2OR3 + 2H+ - 2VO + H2OR
- The charge on the left side of the equation is +2, while on the right side it is zero. Add 2 electrons to the left side to bring the charge back to zero.
- V.2OR3 + 2H+ + 2e- - 2VO + H2OR
Balance Redox Reactions Step 10 Step 6. Multiply each half-reaction by a scale factor, so that the electrons are even in both half-reactions
The electrons in the parts of the equation must be equal, so that they cancel out when the half-reactions are added together. Multiply the reaction by the lowest common denominator of the electrons to make them equal.
- Half-reaction 1 contains 6 electrons, while half-reaction 2 contains 2. Multiplying half-reaction 2 by 3, it will have 6 electrons, the same number as the first.
-
Semireaction 1:
2Fe + 3H2O - Fe2OR3 + 6H+ + 6e-
-
Semireaction 2:
- V.2OR3 + 2H+ + 2e- - 2VO + H2OR
- Multiplication by 3: 3V2OR3 + 6H+ + 6e- - 6VO + 3H2OR
Balance Redox Reactions Step 11 Step 7. Combine the two half reactions
Write all reactants on the left side of the equation and all products on the right side. You will notice that there are equal terms on one side and the other, such as H2O, H+ and it's-. You can delete them and only the balanced equation will remain.
- 2Fe + 3H2O + 3V2OR3 + 6H+ + 6e- - Fe2OR3 + 6H+ + 6e- + 6VO + 3H2OR
- The electrons on both sides of the equation cancel each other out, arriving at: 2Fe + 3H2O + 3V2OR3 + 6H+ - Fe2OR3 + 6H+ + 6VO + 3H2OR
- There are 3 molecules of H.2O and 6 H ions+ on both sides of the equation, so delete those as well to get the final balanced equation: 2Fe + 3V2OR3 - Fe2OR3 + 6VO
Balance Redox Reactions Step 12 Step 8. Check that the sides of the equation have the same charge
When you finish balancing, make sure the charge is the same on both sides of the equation.
- For the right side of the equation: the n.o. of Fe is 0. In V2OR3 the "no. of V is +3 and of O is -2. Multiplying by the number of atoms of each element we get V = +3 x 2 = 6, O = -2 x 3 = -6. The charge is canceled.
- For the left side of the equation: in Fe2OR3 the "no. of Fe is +3 and of O is -2. Multiplying by the number of atoms of each element gives Fe = +3 x 2 = +6, O = -2 x 3 = -6. The charge is canceled. In VO the n.o. for V it is +2, while for O it is -2. The charge is also canceled on this side.
- Since the sum of all charges is zero, our equation is correctly balanced.
Part 3 of 3: Balancing a Redox in a Basic Solution
Balance Redox Reactions Step 13 Step 1. Divide the reaction into two half reactions
To balance an equation in a basic solution just follow the steps described above, adding one last operation at the end. Again, the equation should already be split to determine if it is a redox. If, on the other hand, you have not done so, because in the text of the exercise it is expressly stated that it is a redox, the first step is to divide the equation into two halves. To do this, take the first reagent and write it as a half reaction with the product that includes the element in the reagent. Then take the second reagent and write it as a half reaction with the product that includes that element.
-
For example, consider the following reaction, to be balanced in a basic solution: Ag + Zn2+ - Ag2O + Zn. It can be divided into the following half reactions:
- Ag - Ag2OR
- Zn2+ - Zn
Balance Redox Reactions Step 14 Step 2. Balance all elements in the equation except hydrogen and oxygen
Once you have established that you are dealing with redox, it is time to balance it. It begins by balancing all the elements in each half-reaction other than hydrogen (H) and oxygen (O). Below you will find a practical example.
-
Semireaction 1:
- Ag - Ag2OR
- There is an Ag atom on the left side and 2 on the right, so multiply the right side by 2 to balance.
- 2Ag - Ag2OR
-
Semireaction 2:
- Zn2+ - Zn
- There is a Zn atom on the left side and 1 on the right side, so the equation is already balanced.
Balance Redox Reactions Step 15 Step 3. Balance the oxygen atoms by adding H.2Or to the opposite side of the reaction.
Determine the number of oxygen atoms on either side of the equation. Balance the equation by adding water molecules to the side with fewer oxygen atoms until the two sides are equal.
-
Semireaction 1:
- 2Ag - Ag2OR
- There are no O atoms on the left side and there is one on the right side. Add a molecule of H.2Or to the left side to balance.
- H.2O + 2Ag - Ag2OR
-
Semireaction 2:
- Zn2+ - Zn
- There are no O atoms on either side of the equation, which is therefore already balanced.
Balance Redox Reactions Step 16 Step 4. Balance the hydrogen atoms by adding H.+ to the opposite side of the equation.
As you did for oxygen atoms, determine the number of hydrogen atoms on either side of the equation, then balance them by adding H atoms+ from the side that has less hydrogen, until they are the same.
-
Semireaction 1:
- H.2O + 2Ag - Ag2OR
- There are 2 H atoms on the left side and none on the right side. Add 2 H ions+ to the right side to balance.
- H.2O + 2Ag - Ag2O + 2H+
-
Semireaction 2:
- Zn2+ - Zn
- There are no H atoms on either side of the equation, which is therefore already balanced.
Balance Redox Reactions Step 17 Step 5. Equalize the charges by adding electrons from the side of the equation that requires them
Once the hydrogen and oxygen atoms are balanced, one side of the equation will have a greater positive charge than the other. Add enough electrons to the positive side of the equation to bring the charge back to zero.
- Electrons are almost always added from the side with the H atoms+.
-
Semireaction 1:
- H.2O + 2Ag - Ag2O + 2H+
- The charge on the left side of the equation is 0, while on the right side it is +2 due to hydrogen ions. Add two electrons to the right side to balance.
- H.2O + 2Ag - Ag2O + 2H+ + 2e-
-
Semireaction 2:
- Zn2+ - Zn
- The charge on the left side of the equation is +2, while on the right side it is zero. Add 2 electrons to the left side to bring the charge to zero.
- Zn2+ + 2e- - Zn
Balance Redox Reactions Step 18 Step 6. Multiply each half-reaction by a scale factor, so that the electrons are even in both half-reactions
The electrons in the parts of the equation must be equal, so that they cancel out when the half-reactions are added together. Multiply the reaction by the lowest common denominator of the electrons to make them equal.
In our example, both sides are already balanced, with two electrons on each side
Balance Redox Reactions Step 19 Step 7. Combine the two half reactions
Write all reactants on the left side of the equation and all products on the right side. You will notice that there are equal terms on one side and the other, such as H2O, H+ and it's-. You can delete them and only the balanced equation will remain.
- H.2O + 2Ag + Zn2+ + 2e- - Ag2O + Zn + 2H+ + 2e-
- The electrons on the sides of the equation cancel each other out, giving: H.2O + 2Ag + Zn2+ - Ag2O + Zn + 2H+
Balance Redox Reactions Step 20 Step 8. Balance positive hydrogen ions with negative hydroxyl ions
Since you want to balance the equation in a basic solution, you need to cancel out the hydrogen ions. Add an equal value of OH ions- in order to balance those H+. Make sure you add the same number of OH ions- on both sides of the equation.
- H.2O + 2Ag + Zn2+ - Ag2O + Zn + 2H+
- There are two H ions+ on the right side of the equation. Add two OH ions- on both sides.
- H.2O + 2Ag + Zn2+ + 2OH- - Ag2O + Zn + 2H+ + 2OH-
- H.+ and OH- combine to form a water molecule (H.2O), giving H2O + 2Ag + Zn2+ + 2OH- - Ag2O + Zn + 2H2OR
- You can delete a water molecule on the right side, obtaining the final balanced equation: 2Ag + Zn2+ + 2OH- - Ag2O + Zn + H2OR
Balance Redox Reactions Step 21 Step 9. Check that both sides of the equation have zero charge
After balancing is done, make sure the charge (equal to the oxidation number) is the same on both sides of the equation.
- For the left side of the equation: Ag has an n.o. of 0. The Zn ion2+ has a n.o. by +2. Each OH ion- has an n.o. of -1, which multiplied by two gives a total of -2. The +2 of Zn and the -2 of the OH ions- cancel each other out.
- For the right side: in Ag2O, Ag has an n.o. by +1, while O is -2. Multiplying by the number of atoms we obtain Ag = +1 x 2 = +2, the -2 of O vanishes. Zn has a n.o. of 0, as well as the water molecule.
- Since all charges result in zero, the equation is correctly balanced.
