NOTE: This is NOT the same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY!
These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction:
CuS(s) + NO3-(aq) ---> Cu2+(aq) + SO42-(aq) + NO(g)
Step 1:
Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product.
Here is the unbalanced half-reaction involving CuS:
CuS(s) ---> Cu2+(aq) + SO42-(aq)
And the unbalanced half-reaction for NO3- is:
NO3-(aq) --> NO(g)
Step 2:
Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction.
In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here.
Step 3:
Balance oxygen by adding H2O to one side of each half-reaction.
CuS + 4 H2O ---> Cu2+ + SO42-
NO3- --> NO + 2 H2O
Step 4:
Balance hydrogen atoms. This is done differently for acidic versus basic solutions.
For acidic solutions: Add H3O+ to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H2O to the other side.
For basic solutions: add H2O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH- to the other side.
Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have:
CuS + 12 H2O ---> Cu2+ + SO42- + 8 H3O+
NO3- + 4 H3O+ --> NO + 6 H2O
Step 5:
Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.
Oxidation: CuS + 12 H2O ---> Cu2+ + SO42- + 8 H3O+ + 8 e-
Reduction: NO3- + 4 H3O+ + 3 e- --> NO + 6 H2O
Step 6:
Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H3O+, H2O, or OH- appears on both sides of the final equation, cancel out the duplication also.
Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed).
3 CuS + 36 H2O ---> 3 Cu2+ + 3 SO42- + 24 H3O+ + 24 e-
8 NO3- + 32 H3O+ + 24 e- ---> 8 NO + 48 H2O
Adding these two together gives the following equation:
3 CuS + 36 H2O + 8 NO3- + 8 H3O+ ---> 3 Cu2+ + 3 SO42- + 8 NO + 48 H2O
Step 7:
Finally balancing both sides for excess of H2O
(On each side -36) This gives you the following overall balanced equation at last:
3 CuS(s) + 8 NO3-(aq) + 8 H3O+(aq) ---> 3 Cu2+(aq) + 3 SO42-(aq) + 8 NO(g) + 12 H2O(l)
3Cd(s) + 2Al3+(aq) → 3Cd2+(aq) + 2Al(s)
Termochemical reactions include the enthalpy of reactants and products.
Burning of propane:CH3H8 + 5 O2 = 3 CO2 + H2OBurning of butane:2 CH4H10 + 13 O2 = 8 CO2 + 10 H2O
The mass of reactants is equal to the mass of products.
Activity series are useful for single displacement reactions.
To understand the coupling and decoupling of compounds in transit based on chemical reaction.
Chemical reactions are abbreviated by their chemical equations.
If you can solve the beam reactions by the equations of equilibrium, then it is statically deterrminate. If not, that is, more unknown reactions than the equations of equilibrium, then it is indeterminate, and you need to know something about its deformation to solve the reactions.
Both atoms and charge
chemical equations
Termochemical reactions include the enthalpy of reactants and products.
Chemical equations describe the products and reactants in a chemical reaction.
Oxygen is often the "problem child" of balancing, so do oxygen last. In combustion reactions, it's usually easiest to do carbon first, then hydrogen, then oxygen.
because yes!
Chemical equations are representative for chemical reactions.
Burning of propane:CH3H8 + 5 O2 = 3 CO2 + H2OBurning of butane:2 CH4H10 + 13 O2 = 8 CO2 + 10 H2O
The mass of reactants is equal to the mass of products.
Balancing redox reactions is different from balancing other reactions because both the number of atoms and the amount of charge must be balanced.