How do you balance this redox rxn?
Al + HNO3 ----> Al(NO3)3 + NH4NO3 +H2O
Two things are wrong with that equation:
(a) Aluminum does not react with nitric acid.
(b) If we assume the reaction does occur, aluminum metal oxidizes (goes from an oxidation state of 0 to +3) but no element is reduced as required (experiences a lowering of its oxidation state).
It would be pointless to balance a chemical equation that is totally wrong.
To balance a redox reaction, you need to follow a set of steps. I will explain the process step by step to help you balance the given redox reaction.
Step 1: Assign oxidation numbers
Start by assigning oxidation numbers to each element in the reaction. The oxidation number is a positive or negative number that represents the number of electrons an atom has gained, lost, or shared during the formation of a compound or molecule.
In this reaction:
Al is undergoing oxidation from its elemental state, so its oxidation number changes from 0 to +3.
H is also undergoing oxidation from its elemental state, so its oxidation number changes from 0 to +1.
N in HNO3 has an oxidation number of +5.
O in HNO3 has an oxidation number of -2.
Step 2: Identify the half-reactions
Next, separate the redox reaction into two half-reactions: one for the oxidation half-reaction and one for the reduction half-reaction.
Oxidation half-reaction: Al -> Al³⁺
Reduction half-reaction: HNO3 -> NH4NO3
Step 3: Balance the atoms other than hydrogen and oxygen
Balance the atoms in each half-reaction other than hydrogen and oxygen by adding appropriate coefficients.
For the oxidation half-reaction:
Al -> Al³⁺ (already balanced)
For the reduction half-reaction:
HNO3 -> NH4NO3
Since the number of nitrogen atoms is the same on both sides, we only need to balance the hydrogen and oxygen atoms.
HNO3 -> NH4NO3
Add 8 H⁺ to the left side:
HNO3 + 8H⁺ -> NH4NO3
Step 4: Balance the oxygen atoms
Balance the oxygen atoms in each half-reaction by adding water (H2O) molecules to the side that needs oxygen.
For the oxidation half-reaction:
Al -> Al³⁺ (no oxygen atoms)
For the reduction half-reaction:
HNO3 + 8H⁺ -> NH4NO3
Add 3 H2O to the left side, as there are 3 oxygen atoms on the right side:
HNO3 + 8H⁺ + 3H2O -> NH4NO3
Step 5: Balance the hydrogen atoms
Balance the hydrogen atoms by adding H⁺ ions to the side that needs hydrogen.
For the oxidation half-reaction:
Al -> Al³⁺ (no hydrogen atoms)
For the reduction half-reaction:
HNO3 + 8H⁺ + 3H2O -> NH4NO3
Add 7 H⁺ to the right side:
HNO3 + 8H⁺ + 3H2O -> NH4NO3 + 7H⁺
Step 6: Balance the charge
Balance the overall charge by adding electrons (e⁻) to the side that needs charge balancing.
For the oxidation half-reaction:
Al -> Al³⁺ (no charge imbalance)
For the reduction half-reaction:
HNO3 + 8H⁺ + 3H2O -> NH4NO3 + 7H⁺
Add 5e⁻ to the left side to balance the charge, as the nitrogen in NH4NO3 has a charge of -3 while nitrogen in HNO3 has a charge of +5:
HNO3 + 8H⁺ + 3H2O + 5e⁻ -> NH4NO3 + 7H⁺
Step 7: Equalize the number of electrons
Make the number of electrons equal in both half-reactions by multiplying one or both of the half-reactions by appropriate coefficients.
For the oxidation half-reaction:
Multiply by 5:
5Al -> 5Al³⁺ + 15e⁻
For the reduction half-reaction:
Multiplication is not required as both sides of the half-reaction already have the same number of electrons.
Step 8: Add the half-reactions and simplify
Add the two balanced half-reactions together and simplify by canceling out common terms.
5Al + 3HNO3 + 8H⁺ + 3H2O -> 5Al(NO3)3 + NH4NO3 + 8H⁺ + 15e⁻
Simplify by canceling out 8H⁺ from both sides:
5Al + 3HNO3 + 3H2O -> 5Al(NO3)3 + NH4NO3 + 15e⁻
Step 9: Finalize the balanced equation
To write the final balanced equation, combine all terms and cancel out any common terms:
5Al + 3HNO3 + 3H2O -> 5Al(NO3)3 + NH4NO3
The balanced redox equation is:
5Al + 3HNO3 + 3H2O -> 5Al(NO3)3 + NH4NO3
By following these steps, you can balance the given redox reaction.