How do you balance oxidation-Reduction equation.Demonstrate with 3 examples.
To balance oxidation-reduction equations, you need to follow these basic steps:
1. Identify the species being oxidized and reduced. This will help you determine the oxidation state of each element.
2. Write the half-reactions for the oxidation and reduction processes separately. Balance all the elements except H and O in each half-reaction.
3. Balance the number of oxygen atoms by adding water molecules (H2O) to the side that needs more oxygen.
4. Balance the number of hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen.
5. Balance the charge by adding electrons (e-) to the side that has a higher positive charge.
6. Equalize the number of electrons in both half-reactions by multiplying one or both half-reactions by a factor that will convert the number of electrons in one half-reaction to the same number in the other half-reaction.
7. Add the two half-reactions together and simplify the equation as needed.
Here are three examples:
Example 1:
Fe2+ + MnO4- → Fe3+ + Mn2+
Step 1: Fe is being oxidized from +2 to +3, and Mn is being reduced from +7 to +2.
Step 2: Half-reaction for oxidation: Fe2+ → Fe3+
Half-reaction for reduction: MnO4- → Mn2+
Step 3: MnO4- + 4H2O → Mn2+ + 8H+ + 5O2-
Step 4: Fe2+ → Fe3+ + e-
Step 5: Balancing the charges, we need to multiply the oxidation half-reaction by 5, and the reduction half-reaction by 1.
Step 6: 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
Step 7: Simplifying the equation gives us 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
Example 2:
KNO2 + KMnO4 + H2SO4 → K2SO4 + MnSO4 + HNO3 + H2O
Step 1: Mn is being reduced from +7 to +2, and N is being oxidized from +3 to +5.
Step 2: Half-reaction for oxidation: N → NO3-
Half-reaction for reduction: MnO4- → Mn2+
Step 3: MnO4- + 4H2O → Mn2+ + 8H+ + 5O2-
Step 4: N → NO3- + 2e-
Step 5: Balancing the charges, we need to multiply the oxidation half-reaction by 5, and the reduction half-reaction by 2.
Step 6: 10N + 2KMnO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5HNO3 + 8H2O
Step 7: Simplifying the equation gives us 5KNO2 + 2KMnO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5HNO3 + 3H2O
Example 3:
Zn + HNO3 → Zn(NO3)2 + NO + H2O
Step 1: N is being oxidized from +3 to +4, and Zn is being reduced from 0 to +2.
Step 2: Half-reaction for oxidation: N → NO
Half-reaction for reduction: Zn → Zn2+
Step 3: Not applicable, since there are no oxygen atoms in the reactants.
Step 4: N → NO + e-
Step 5: Balancing the charges, we need to multiply the oxidation half-reaction by 2, and the reduction half-reaction by 1.
Step 6: Zn + 2HNO3 → Zn(NO3)2 + 2NO + H2O
Step 7: Simplifying the equation gives us Zn + 2HNO3 → Zn(NO3)2 + 2NO + H2O
To balance an oxidation-reduction equation, you need to ensure that the number of atoms and charges are equal on both sides of the equation. Here are three examples of balancing oxidation-reduction equations:
Example 1: Balancing the reaction between iron(II) sulfate (FeSO4) and potassium permanganate (KMnO4) in an acidic solution to form iron(III) sulfate (Fe2(SO4)3) and manganese(II) sulfate (MnSO4):
Step 1: Identify the oxidation states of each element:
Fe2+ + MnO4- → Fe3+ + Mn2+
Step 2: Balance the atoms other than hydrogen and oxygen:
Fe2+ + MnO4- → Fe3+ + Mn2+
Step 3: Balance oxygen using water (H2O):
Fe2+ + MnO4- → Fe3+ + Mn2+ + H2O
Step 4: Balance hydrogen using H+ ions (in acidic solution):
Fe2+ + MnO4- + 4H+ → Fe3+ + Mn2+ + H2O
Step 5: Balance the charge using electrons:
Fe2+ + MnO4- + 4H+ → Fe3+ + Mn2+ + H2O + 5e-
Step 6: Ensure the number of electrons transferred is equal on both sides by multiplying the half-reactions:
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + H2O + 5e-
Example 2: Balancing the reaction between potassium iodide (KI) and chlorine (Cl2) in an aqueous solution to form potassium chloride (KCl) and iodine (I2):
Step 1: Identify the oxidation states of each element:
2I- + Cl2 → 2Cl- + I2
Step 2: Balance the atoms other than hydrogen and oxygen:
2I- + Cl2 → 2Cl- + I2
Step 3: Balance oxygen using water (H2O):
2I- + Cl2 → 2Cl- + I2
Step 4: Balance hydrogen using H+ ions (in acidic solution):
2I- + Cl2 → 2Cl- + I2
Step 5: Balance the charge using electrons:
I- + Cl2 → Cl- + I2 + 2e-
Step 6: Ensure the number of electrons transferred is equal on both sides by multiplying the half-reactions:
2I- + Cl2 → 2Cl- + I2 + 4e-
Example 3: Balancing the reaction between sodium hydroxide (NaOH) and chlorine gas (Cl2) to form sodium hypochlorite (NaOCl) and sodium chloride (NaCl):
Step 1: Identify the oxidation states of each element:
NaOH + Cl2 → NaOCl + NaCl
Step 2: Balance the atoms other than hydrogen and oxygen:
NaOH + Cl2 → NaOCl + NaCl
Step 3: Balance oxygen using water (H2O):
NaOH + Cl2 → NaOCl + NaCl + H2O
Step 4: Balance hydrogen using H+ ions (in acidic solution):
NaOH + Cl2 + 2H+ → NaOCl + NaCl + H2O
Step 5: Balance the charge using electrons:
NaOH + Cl2 + 2H+ → NaOCl + NaCl + H2O + 2e-
Step 6: Ensure the number of electrons transferred is equal on both sides by multiplying the half-reactions:
2NaOH + Cl2 + 4H+ → 2NaOCl + 2NaCl + 2H2O + 2e-
These are step-by-step examples of balancing oxidation-reduction equations.