Balance by oxidation method the SO2 + MnO4+H2O --MnSO4+H2SO4+SO4(-2)
Will this get you started.
S on the left is +4; on the rght is +6
Mn on the left is +7 and on the right is +2. I think you failed to add the charge to MnO4. It should be MnO4^-.
Yes I get it thanks Dr.
To balance the given chemical equation using the oxidation number method, follow these steps:
Step 1: Identify the elements and their oxidation states
Write down the oxidation states of each element above their respective symbols in the equation.
SO2: S = +4 and O = -2
MnO4-: Mn = +7 and O = -2
H2O: H = +1 and O = -2
MnSO4: Mn = +2 and S = +6
H2SO4: H = +1, S = +6, and O = -2
SO4(2-): S = +6 and O = -2
Step 2: Identify the elements that undergo oxidation and reduction
In this equation, the Mn atom is reduced from +7 to +2, while the S atom is oxidized from +4 to +6. Therefore, Mn is the reducing agent, and S is the oxidizing agent.
Step 3: Write the half-reactions for oxidation and reduction
Write two separate equations for the oxidation and reduction half-reactions based on the changes in the oxidation states:
Oxidation (S):
SO2 -> SO4(2-) (S goes from +4 to +6)
Reduction (Mn):
MnO4- -> MnSO4 (Mn goes from +7 to +2)
Step 4: Balance the atoms other than hydrogen and oxygen in each half-reaction
Balance the atoms other than hydrogen and oxygen in each half-reaction. Here's how:
Oxidation (S):
There are already an equal number of sulfur (S) atoms on both sides.
Reduction (Mn):
To balance the number of Mn atoms, add a coefficient of 5 in front of the MnSO4 in the product side:
MnO4- + 8H+ + 5e- -> MnSO4 + 4H2O
Step 5: Balance oxygen atoms by adding water (H2O)
Balance the oxygen atoms by adding water (H2O) to the side that lacks oxygen.
Oxidation (S):
Add 3 H2O to the product side to balance the oxygen atoms:
SO2 + 3H2O -> SO4(2-) + 8H+ + 6e-
Reduction (Mn):
There are already an equal number of oxygen (O) atoms on both sides.
Step 6: Balance hydrogen atoms by adding hydrogen ions (H+)
Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that lacks hydrogen.
Oxidation (S):
There are already an equal number of hydrogen (H) atoms on both sides.
Reduction (Mn):
Add 8 H+ to the reactant side to balance the hydrogen atoms:
MnO4- + 8H+ + 5e- -> MnSO4 + 4H2O
Step 7: Balance the charges by adding electrons (e-)
Balance the charges by adding electrons (e-) to the side that has a higher positive charge.
Oxidation (S):
Add 6e- to the reactant side to balance the charges:
SO2 + 3H2O -> SO4(2-) + 8H+ + 6e-
Reduction (Mn):
There are already an equal number of electrons on both sides.
Step 8: Multiply the half-reactions, if necessary
Check if the number of electrons in both half-reactions cancels each other out. If not, multiply one or both of the half-reactions by suitable coefficients to balance the electrons.
In this case, the electrons (e-) on both sides are already balanced.
Step 9: Add the half-reactions together
Add the oxidation and reduction half-reactions together.
3SO2 + 6H2O + 2MnO4- + 16H+ -> 3SO4(2-) + 2MnSO4 + 8H2O
Step 10: Simplify the equation, if necessary
If possible, simplify the equation by canceling out common terms.
In this case, you can divide all coefficients by 2 to simplify the equation:
2SO2 + 4H2O + MnO4- + 8H+ -> 2SO4(2-) + MnSO4 + 4H2O
Therefore, the balanced equation is:
2SO2 + 4H2O + MnO4- + 8H+ -> 2SO4(2-) + MnSO4 + 4H2O