How do I balance this redox reaction?
MnO4- + Fe2+ + H+ + Mn2+ + Fe3+ + H2O
I don't understand. Don't I need a product?
You just need to put an arrow in the right place. I see so many equations on this board with no arrow. The arrow separates reactants from products.
MnO4- + Fe2+ + H+ ==> Mn2+ + Fe3+ + H2O
No
Yes, you are correct. In order to balance a redox reaction, you need both reactants and products. I apologize for the confusion.
In this case, the given reaction is not complete. It represents the oxidation-reduction half-reactions of the overall reaction. To balance the complete redox reaction, we need to combine these half-reactions and balance the number of atoms and charges on both sides of the equation.
The oxidation half-reaction is:
Fe2+ → Fe3+
The reduction half-reaction is:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
To balance the number of electrons transferred, we need to multiply the oxidation half-reaction by 5 and the reduction half-reaction by 1. This will allow us to have the same number of electrons on both sides.
5Fe2+ → 5Fe3+
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Now, we can combine both reactions and add them together:
5Fe2+ + MnO4- + 8H+ + 5e- → 5Fe3+ + Mn2+ + 4H2O
Now you have the balanced redox reaction.
To balance a redox reaction, you must observe that both the number of atoms and the charge are conserved. The given equation is the skeletal equation, which is not balanced.
In this case, you have a reactant on the left side and need to determine the product on the right side. The product will consist of the reduced form of the oxidizing agent (Mn2+) and the oxidized form of the reducing agent (Fe3+). Additionally, H2O molecules and H+ ions may be involved.
To balance the equation, follow these steps:
Step 1: Separate the half-reactions
Split the equation into two half-reactions, one showing the reduction and the other showing the oxidation:
Half-reaction 1: MnO4- → Mn2+
Half-reaction 2: Fe2+ → Fe3+
Step 2: Balance the atoms (except oxygen and hydrogen)
For both half-reactions, make sure the number of atoms is the same on both sides. Start by balancing the atoms other than oxygen and hydrogen. In this case, the elements involved are Mn and Fe.
Half-reaction 1: MnO4- → Mn2+
Here, there is one Mn atom on both sides, so it is already balanced.
Half-reaction 2: Fe2+ → Fe3+
There is one Fe atom on both sides, so it is balanced.
Step 3: Balance the electrons
Determine the number of electrons gained or lost in each half-reaction.
Half-reaction 1: MnO4- → Mn2+
Since the Mn atom undergoes a reduction, it gains 5 electrons. Add the electrons to the side needing balance.
MnO4- + 5e- → Mn2+
Half-reaction 2: Fe2+ → Fe3+
The Fe atom undergoes oxidation and loses 1 electron. Add the electrons to the side needing balance.
Fe2+ → Fe3+ + 1e-
Step 4: Balance the oxygen atoms
Count the oxygen atoms on both sides of each half-reaction. Add water (H2O) molecules to the side with fewer oxygen atoms. Each oxygen atom in the permanganate ion (MnO4-) requires one water molecule.
Half-reaction 1: MnO4- + 5e- → Mn2+
Since there are four oxygen atoms in MnO4-, add 4 H2O molecules to the product side:
MnO4- + 5e- → Mn2+ + 4H2O
Half-reaction 2: Fe2+ → Fe3+ + 1e-
No oxygen atoms are present, so no additional water molecules are needed.
Step 5: Balance the hydrogen atoms
Count the hydrogen atoms on both sides and add H+ ions to balance them.
Half-reaction 1: MnO4- + 5e- → Mn2+ + 4H2O
Since there are no hydrogen atoms on either side, you can skip this step.
Half-reaction 2: Fe2+ → Fe3+ + 1e-
Add 4H+ ions to the reactant side:
Fe2+ + 4H+ → Fe3+ + 1e-
Step 6: Balance the charges
Check that the charges on both sides of each half-reaction balance out. Each half-reaction should have the same overall charge.
Half-reaction 1: MnO4- + 5e- → Mn2+ + 4H2O
The charge is balanced: -1 + 5(-1) = +2
Half-reaction 2: Fe2+ + 4H+ → Fe3+ + 1e-
The charge is balanced: +2 + 4(+1) = +3 + 1(-1)
Step 7: Combine the half-reactions
To combine the half-reactions, multiply them by the necessary coefficients to ensure that the number of electrons is equal in both half-reactions.
Multiply Half-reaction 1 by 1 and Half-reaction 2 by 5 to achieve electron balance:
MnO4- + 5Fe2+ + 8H+ → 5Fe3+ + Mn2+ + 4H2O
Now, the equation is balanced in terms of atoms and charge for both the reactants and products.