Balance using oxidation Number Methode

A. MnO4- + C2H4 + H20 --> MnO2 + OH- + C2H6
B. ClO- --> Cl- + ClO3-

To balance the given chemical equations using the oxidation number method, follow these steps:

A. MnO4- + C2H4 + H2O → MnO2 + OH- + C2H6

1. Assign oxidation numbers to each element in the equation. Look for the rules and guidelines for assigning oxidation numbers if you are unsure.
MnO4-: Mn = +7, O = -2
C2H4: C = -2, H = +1
H2O: H = +1, O = -2
MnO2: Mn = +4, O = -2
OH-: O = -2, H = +1
C2H6: C = -3, H = +1

2. Determine the elements undergoing oxidation and reduction.
In this case, manganese (Mn) is being reduced from +7 to +4, while carbon (C) is being oxidized from -2 to -3.

3. Balance the atoms that are not hydrogen or oxygen. In this equation, only the carbon and manganese atoms are involved.
Mn: 1 Mn on the left side and 1 Mn on the right side, so they are balanced.
C: 2 C on the left side and 2 C on the right side, so they are balanced.

4. Balance oxygen atoms by adding water (H2O) to the side that needs more oxygen.
On the left side, there are 4 oxygen atoms from MnO4- and only 2 oxygen atoms from H2O, so you need to add 2 H2O on the right side.

MnO4- + C2H4 + H2O → MnO2 + OH- + C2H6 + 2H2O

5. Balance hydrogen atoms by adding H+ ions to the side that needs more hydrogen.
On the left side, there are 4 hydrogen atoms from H2O and 6 hydrogen atoms from C2H6, so you need to add 10 H+ ions on the left side.

MnO4- + C2H4 + H2O + 10H+ → MnO2 + OH- + C2H6 + 2H2O

6. Now, balance the charges by adding electrons to the side that is more positive.
On the left side, there are 10 positive charges from H+ ions and 1 negative charge from MnO4-, so you need to add 9 electrons to the left side.

MnO4- + C2H4 + H2O + 10H+ + 9e- → MnO2 + OH- + C2H6 + 2H2O

7. Finally, perform a check to ensure all the atoms and charges are balanced.
Mn: 1 on both sides
O: 4 on both sides
H: 10 on both sides
C: 2 on both sides
Charge: -1 on both sides

The balanced equation is: MnO4- + C2H4 + H2O + 10H+ + 9e- → MnO2 + OH- + C2H6 + 2H2O

B. ClO- → Cl- + ClO3-

1. Assign oxidation numbers to each element:
ClO-: Cl = +1, O = -2
Cl-: Cl = -1
ClO3-: Cl = +5, O = -2

2. Determine the elements undergoing oxidation and reduction.
In this case, chlorine (Cl) is being reduced from +1 to -1, while oxygen (O) is being oxidized from -2 to -2 and +5.

3. Balance the atoms that are not hydrogen or oxygen. In this equation, only the chlorine atoms are involved.
Cl: 1 Cl on the left side and 1 Cl on the right side, so they are balanced.

4. Balance oxygen atoms by adding water (H2O) to the side that needs more oxygen.
On the left side, there are 1 oxygen atom from ClO-, while on the right side, there are 1 oxygen atom from ClO3-. They are already balanced.

ClO- → Cl- + ClO3-

5. Finally, perform a check to ensure that all atoms and charges are balanced.
Cl: 1 on both sides
O: 1 on both sides
Charge: -1 on both sides

The balanced equation is: ClO- → Cl- + ClO3-