An oxidereduction, reductionoxidation or simply redox reaction is a chemical reaction in which one of the reactants is reduced while the other is oxidized. Oxide reduction refers to the transfer of electrons between elements or compounds and is designated according to the oxidation state. An atom is oxidized when its oxidation number increases and is reduced when its oxidation number decreases. Redox reactions are essential for basic life functions, such as photosynthesis and respiration. Balancing a redox reaction requires more steps than balancing a regular chemical equation. The most important step is to identify if the reaction that is taking place is really a redox reaction.
Steps
Part 1 of 3: Identify a Redox Reaction
Step 1. Learn the rules for assigning an assignment status
The oxidation state of a species (each element in the equation) is a number equivalent to the number of electrons that can be gained, lost, or shared with another element during the chemical bonding process. There are seven rules that will allow you to determine the oxidation state of an element. You must follow them in the order described. If two of the rules conflict, use the first rule to assign the oxidation state (ODE).
 Rule # 1: An individual atom, by itself, has an ODE of 0. For example: Au, ODE = 0. CI_{2} it also has an ODE of 0, as long as it is not combined with another element.
 Rule # 2: the total ODE of all atoms in a neutral species is always 0, but in an ion it is equal to the charge of that ion. The ODE of the molecule must equal 0, but the ODE of each element in the molecule may not be zero. For example, H_{2}Or it has an ODE of 0, each hydrogen atom has an ODE of +1, while the oxygen atom has an ODE of 2. The Ca ion^{2+} it has an oxidation state of +2.
 Rule # 3: In compounds, Group 1 metals have an ODE of +1 and Group 2 metals have an ODE 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 state of oxygen in a compound is 2.
 Rule # 7: in compounds of two elements where at least one is a metal, the elements of group 15 have an ODE of 3; those in group 16 have an ODE of 2; and those in group 17 have an ODE of 1.
Step 2. Divide the reaction into two halfreactions
Although the halfreactions are hypothetically reactions, dividing the equation you can easily determine whether or not a redox reaction occurs. To do this, take the first reagent and write it as a halfreaction with the product that includes the reagent element. Then take the second reagent and write it as a halfreaction with the product that includes that element.

For example, Fe + V_{2}OR_{3}  Faith_{2}OR_{3} + VO is divided into the following two halfreactions.
 Faith  Faith_{2}OR_{3}
 V_{2}OR_{3}  VO

If there is only one reactant and two products, create a half reaction with the reactant and the first product and a half reaction with the reactant and the second product. When you are finally going to combine the halfreactions, don't forget to combine the reagents again. You can do the same if there are two reagents and only one product: use each reagent with the same product in the halfreactions.
 ClO^{}  Cl^{} + ClO_{3}^{}
 Halfreaction 1: ClO^{}  Cl^{}
 Halfreaction 2: ClO^{}  ClO_{3}^{}
Step 3. Assign each element in the equation its oxidation state
Using the seven oxidation state assignment rules, determine the oxidation state of each species from the given chemical equation. While a compound may be neutral, the elements that make up for that compound will have a charged oxidation state. Remember to follow the rules in the established order.
 In the first halfreaction of the previous example: the ODE of the Fe atom that is alone is 0 (rule # 1), the ODE of the Fe in Fe_{2} is +3 (rules 2 and 6) and the ODE of O in O_{3} is 2 (rule # 6).
 In the second halfreaction: the ODE of the V in V_{2} is +3 (rules 2 and 6) while the ODE of O in O_{3} is 2 (rule # 6). The ODE of V is +2 (rule # 2), while that of O is 2 (rule # 6).
Step 4. Determine if one species is oxidized while the other is reduced
Observing the oxidation states of each species in the halfreaction, determine if one of the species is oxidized (increases its oxidation state), while the other is reduced (decreases its oxidation state).
 In the example, the first halfreaction is oxidized because Fe starts with an ODE of 0 and increases to 3. The second halfreaction is reduced because V starts with an ODE of +6 and decreases to +2.
 Since one of the species is oxidized and the other is reduced, then the equation is an oxidereduction reaction.
Part 2 of 3: Balancing a Redox Reaction in a Neutral or Acidic Solution
Step 1. Divide the reaction into two halfreactions
You will already have your equation divided into two halfreactions since you had to do it in the previous step to determine whether or not a redox reaction occurred. If they have already confirmed that it is a redox reaction, then the first step will be to divide the reaction into two halfreactions. To do this, take the first reactant and write it as a half equation with the product that includes that element in the reactant. Then take the second reagent and write it as a halfreaction with the product that includes that element.

For example, Fe + V_{2}OR_{3}  Faith_{2}OR_{3} + VO is divided into the following two halfreactions:
 Faith  Faith_{2}OR_{3}
 V_{2}OR_{3}  VO

If there is only one reactant and two products, create a halfreaction with the reactant and the first product and a halfreaction with the reactant and the second product. When you are finally going to combine halfreactions, don't forget to combine the reagents again. You can do the same if there are two reagents and only one product: use each reagent with the same product in the halfreactions.
 ClO^{}  Cl^{} + ClO_{3}^{}
 Halfreaction 1: ClO^{}  Cl^{}
 Halfreaction 2: ClO^{}  ClO_{3}^{}
Step 2. Balance all elements of the equation except hydrogen and oxygen
Once you've determined that a redox reaction occurs, it's time to balance it. Begin by balancing all the elements in each halfreaction other than hydrogen (H) and oxygen (O). Do it by following these steps:

Halfreaction 1:
 Faith  Faith_{2}OR_{3}
 There is 1 Fe atom on the left side and 2 on the right. Multiply the left one by 2 to balance.
 2Fe  Faith_{2}OR_{3}

Halfreaction 2:
 V_{2}OR_{3}  VO
 There are 2 V atoms on the left side and 1 on the right. Multiply the right by 2 to balance.
 V_{2}OR_{3}  2VO
Step 3. Balance the oxygen atoms by adding H_{2}Or on the opposite side of the reaction.
Determine the number of oxygen atoms on each side of the equation. Balance the equation by adding water molecules on the side with the fewest oxygen atoms until both sides are equal.

Halfreaction 1:
 2Fe  Faith_{2}OR_{3}
 There are 3 O atoms on the right side and none on the left. Add 3 molecules of H_{2}Or on the left side to balance.
 2Fe + 3H_{2}O  Faith_{2}OR_{3}

Halfreaction 2:
 V_{2}OR_{3}  2VO
 There are 3 O atoms on the left side and two on the right. Add 1 molecule of H_{2}Or on the right side to balance.
 V_{2}OR_{3}  2VO + H_{2}OR
Step 4. Balance the hydrogen atoms by adding H^{+} on the opposite side of the equation.
Just as you did with the oxygen atoms, now determine the hydrogen atoms for each side of the equation. Then balance it by adding H atoms^{+} on the side with the fewest atoms until both sides are equal.

Halfreaction 1:
 2Fe + 3H_{2}O  Faith_{2}OR_{3}
 There are 6 atoms on the left side and none on the right. Add 6 H^{+} on the right side to balance.
 2Fe + 3H_{2}O  Faith_{2}OR_{3} + 6H^{+}

Halfreaction 2:
 V_{2}OR_{3}  2VO + H_{2}OR
 There are 2 H atoms on the right side and none on the left. Add 2 H^{+} on the left side to balance.
 V_{2}OR_{3} + 2H^{+}  2VO + H_{2}OR
Step 5. Balance the charges by adding electrons on the corresponding side of the equation
Once you've balanced the hydrogen and oxygen, one side of the equation will be more positive than the other. Add enough electrons to the side of the equation that is more positive to make the charge equal to zero.
 Electrons are almost always added on the side that has the H atoms^{+}.

Halfreaction 1:
 2Fe + 3H_{2}O  Faith_{2}OR_{3} + 6H^{+}
 The charge on the left side of the equation is 0, while the charge on the right side has a 6+ charge due to hydrogen ions. Add 6 electrons to the right side to balance.
 2Fe + 3H_{2}O  Faith_{2}OR_{3} + 6H^{+} + 6e^{}

Halfreaction 2:
 V_{2}OR_{3} + 2H^{+}  2VO + H_{2}OR
 The charge on the left side of the equation is 2+, while the charge on the right side is 0. Add 2 electrons to the left side to make the charge equal to zero.
 V_{2}OR_{3} + 2H^{+} + 2e^{}  2VO + H_{2}OR
Step 6. Multiply each halfreaction by a scale factor so that the electrons are equal in the two halfreactions
You must equalize the electrons on each side of the equation, so that adding the two halfreactions will cause the electrons to cancel each other out. Multiply the reaction by the least common multiple of both electrons to make them equal.
 Halfreaction 1 has 6 electrons while halfreaction 2 has 2 electrons. If you multiply the halfreaction 2 by 3, it will have 6 electrons and it will be equal to the first halfreaction.

Halfreaction 1:
 2Fe + 3H_{2}O  Faith_{2}OR_{3} + 6H^{+} + 6e^{}

Halfreaction 2:
 V_{2}OR_{3} + 2H^{+} + 2e^{}  2VO + H_{2}OR
 Multiply by 3: 3V_{2}OR_{3} + 6H^{+} + 6e^{}  6VO + 3H_{2}OR
Step 7. Combine the two halfreactions
Write all the reactants on the left side of the equation and all the products on the right side of the equation. You will notice that there are equivalent terms on both sides of the equation, like H_{2}O, H^{+} and e^{}. Cancel the equivalent terms and the remaining terms will form a balanced equation.
 2Fe + 3H_{2}Or + 3V_{2}OR_{3} + 6H^{+} + 6e^{}  Faith_{2}OR_{3} + 6H^{+} + 6e^{} + 6VO + 3H_{2}OR
 The electrons on both sides of the equation will cancel out and the result will be: 2Fe + 3H_{2}Or + 3V_{2}OR_{3} + 6H^{+}  Faith_{2}OR_{3} + 6H^{+} + 6VO + 3H_{2}OR
 There are 3 H ions_{2}O and 6^{+} on each side of the equation that also cancel, resulting in the following final balanced equation: 2Fe + 3V_{2}OR_{3}  Faith_{2}OR_{3} + 6VO
Step 8. Check that each side of the equation has the same charge
Once you've finished balancing, check the equation to make sure the loads are balanced on both sides of the equation. The charges on each side must be equal.
 On the right side of the equation: the ODE of Fe is 0. In V_{2}OR_{3}, the ODE of V is +3 and that of O is 2. Multiplying by the number of atoms of each element, you will get V = +3 x2 = 6 and O = 2 x 3 = 6. The charges are canceled.
 On the left side of the equation: in Fe_{2}OR_{3}, the ODE of Fe is +3 and that of O is 2. Multiply by the number of atoms of each element, Fe = +3 x 2 = +6 and O = 2 x 3 = 6. The charges are canceled. In VO, the ODE of V is +2, while that of O is 2. Charges are also canceled on this side.
 Because all charges are equal to zero, the equation is correctly balanced.
Part 3 of 3: Balancing a Redox Reaction in a Basic Solution
Step 1. Divide the reaction into two halfreactions
To balance in a basic solution you have to follow the same steps as described above, only with an extra step at the end. Again, the equation should already be divided into two halfreactions since you should have divided it in the previous step to determine if a redox reaction occurred. If they have already confirmed that it is a redox reaction, then the first step will be to divide it into two halfreactions. To do this, take the first reagent and write it as a halfreaction with the product that includes the reagent element. Then take the second reagent and write it as a halfreaction with the product that includes that element.

For example, balance the following reaction in a basic solution: Ag + Zn^{2+}  Ag_{2}O + Zn. This reaction is divided into the following halfreactions:
 Ag  Ag_{2}OR
 Zn^{2+}  Zn
Step 2. Balance all elements of the equation except hydrogen and oxygen
Once you've determined that a redox reaction occurs, it's time to balance it. Begin by balancing all the elements in each halfreaction other than hydrogen (H) and oxygen (O). Do it by following these steps:

Halfreaction 1:
 Ag  Ag_{2}OR
 There is an Ag atom on the left side 2 on the right side. Multiply the left one by 2 to balance.
 2Ag  Ag_{2}OR

Halfreaction 2:
 Zn^{2+}  Zn
 There is 1 Zn atom on the left side and 1 on the right, therefore the equation is already balanced.
Step 3. Balance the oxygen atoms by adding H_{2}Or on the opposite side of the reaction.
Determine the number of oxygen atoms on each side of the equation. Balance the equation by adding water molecules on the side with the fewest oxygen atoms until both sides are equal.

Halfreaction 1:
 2Ag  Ag_{2}OR
 There is no O atom on the left side and one on the right. Add a molecule of H_{2}Or on the left side to balance.
 H_{2}O + 2Ag  Ag_{2}OR

Halfreaction 2:
 Zn^{2+}  Zn
 There are no O atoms on either side, therefore the equation is already balanced.
Step 4. Balance the hydrogen atoms by adding H^{+} on the opposite side of the equation.
Just as you did with the oxygen atoms, now determine the hydrogen atoms for each side of the equation. Then balance it by adding H atoms^{+} on the side with the fewest atoms until both sides are equal.

Halfreaction 1:
 H_{2}O + 2Ag  Ag_{2}OR
 There are 2 H atoms on the left side and none on the right. Add 2 H^{+} on the right side to balance.
 H_{2}O + 2Ag  Ag_{2}O + 2H^{+}

Halfreaction 2:
 Zn^{2+}  Zn
 There are no H atoms on either side, therefore the equation is already balanced.
Step 5. Balance the charges by adding electrons on the corresponding side of the equation
Once you've balanced the hydrogen and oxygen, one side of the equation will be more positive than the other. Add enough electrons to the side of the equation that is more positive to make the charge equal to zero.
 Electrons are almost always added on the side that has the H atoms^{+}.

Halfreaction 1:
 H_{2}O + 2Ag  Ag_{2}O + 2H^{+}
 The charge on the left side of the equation is 0, while the right side has a charge of 2+ due to hydrogen ions. Add 2 electrons to the right side to balance.
 H_{2}O + 2Ag  Ag_{2}O + 2H^{+} + 2e^{}

Halfreaction 2:
 Zn^{2+}  Zn
 The charge on the left side of the equation is 2+ while the charge on the right side is 0. Add 2 electrons to the left side to make the charge equal to zero.
 Zn^{2+} + 2e^{}  Zn
Step 6. Multiply each halfreaction by a scale factor so that the electrons are equal in the two halfreactions
You must equalize the electrons on each side of the equation, so that adding the two halfreactions will cause the electrons to cancel each other out. Multiply the reaction by the least common multiple of both electrons to make them equal.
In the example, both sides are already balanced with 2 electrons on each side
Step 7. Combine the two halfreactions
Write all the reactants on the left side of the equation and all the products on the right side of the equation. You will notice that there are equivalent terms on both sides of the equation, like H_{2}O, H^{+} and e^{}. Cancel the equivalent terms and the remaining terms will make up a balanced equation.
 H_{2}O + 2Ag + Zn^{2+} + 2e^{}  Ag_{2}O + Zn + 2H^{+} + 2e^{}
 The electrons on both sides of the equation cancel out and the equation will look like this: H_{2}O + 2Ag + Zn^{2+}  Ag_{2}O + Zn + 2H^{+}
Step 8. Balance the positive hydrogen ions with negative hydroxyl ions
Since you must balance in a basic solution, now you have to cancel the hydrogen ions. Add the same amount of OH ions^{} to balance the H ions^{+}. When you go to add the OH ions^{}Remember to add the same amount to both sides of the equation.
 H_{2}O + 2Ag + Zn^{2+}  Ag_{2}O + Zn + 2H^{+}
 On the right side of the equation there are 2 H ions^{+}. This means that 2 OH ions have to be added^{} on both sides of the equation.
 H_{2}O + 2Ag + Zn^{2+} + 2OH^{}  Ag_{2}O + Zn + 2H^{+} + 2OH^{}
 The H^{+} combines with OH^{} to form a water molecule (H_{2}O), so the equation will look like this: H_{2}O + 2Ag + Zn^{2+} + 2OH^{}  Ag_{2}O + Zn + 2H_{2}OR
 You can cancel a water molecule on the right side and you will get the following final balanced equation: 2Ag + Zn^{2+} + 2OH^{}  Ag_{2}O + Zn + H_{2}OR
Step 9. Check that each side of the equation has a charge equal to zero
Once you've finished balancing, doublecheck the equation to make sure the loads are balanced on both sides. The charges (oxidation state of all elements) on each side of the equation must equal zero.
 On the left side of the equation: Ag has an ODE of 0. The Zn ion^{2} it has an ODE of +2. In the OH ion^{} the ODE is 1, but since there are 2, the total charge is 2. The +2 of Zn and the 2 of the OH ion^{} they cancel and go to zero.
 On the right side of the equation: in the Ag_{2}Or, Ag has an ODE of +1, while that of O is 2. Multiplying by the number of atoms of Ag = +1 x 2 = +2, the 2 of O cancels out. Zn has an ODE of 0. The water molecule also has an ODE of 0.
 Because all charges are equal to zero, the equation is correctly balanced.