How do you balance redox equation under acidic and basic solution.give seven examples with explanation on each ?

Answer 1

Balancing redox equations under acidic and basic conditions involves adding H+ ions and balancing the charges to ensure that the number of electrons lost equals the number gained. Here are seven examples, each explained for both acidic and basic solutions:

1. Acidic solution:
Unbalanced equation: MnO4- + H2O2 → Mn2+ + O2
Steps:
a) Balance the atoms: MnO4- + H2O2 → Mn2+ + O2
b) Balance oxygen by adding H2O: MnO4- + H2O2 → Mn2+ + 2H2O + O2
c) Balance hydrogen by adding H+: MnO4- + H2O2 + 8H+ → Mn2+ + 2H2O + O2
d) Balance charge by adding electrons: MnO4- + H2O2 + 8H+ + 5e- → Mn2+ + 2H2O + O2

Basic solution:
Unbalanced equation: MnO4- + H2O2 → Mn2+ + O2
Steps:
a) Balance the atoms: MnO4- + H2O2 → Mn2+ + O2
b) Add OH- ions to neutralize H+ ions: MnO4- + H2O2 + 4OH- → Mn2+ + 2H2O + O2
c) Balance oxygen by adding H2O: MnO4- + H2O2 + 4OH- → Mn2+ + 2H2O + O2
d) Balance charge by adding electrons: MnO4- + H2O2 + 4OH- + 5e- → Mn2+ + 2H2O + O2

2. Acidic solution:
Unbalanced equation: Cr2O7^2- + SO2 → Cr^3+ + SO4^2-
Balanced equation: Cr2O7^2- + 3SO2 + 8H+ → 2Cr^3+ + 3SO4^2- + 4H2O

Basic solution:
Unbalanced equation: Cr2O7^2- + SO2 → Cr^3+ + SO4^2-
Balanced equation: Cr2O7^2- + 6SO2 + 14OH- → 2Cr^3+ + 6SO4^2- + 7H2O

3. Acidic solution:
Unbalanced equation: H2O2 + MnO2 → O2 + Mn^2+
Balanced equation: H2O2 + MnO2 + 2H+ → O2 + Mn^2+ + 2H2O

Basic solution:
Unbalanced equation: H2O2 + MnO2 → O2 + Mn^2+
Balanced equation: H2O2 + MnO2 + 4OH- → O2 + Mn^2+ + 2H2O

4. Acidic solution:
Unbalanced equation: Fe2+ + MnO4- → Fe3+ + Mn^2+
Balanced equation: 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn^2+ + 4H2O

Basic solution:
Unbalanced equation: Fe2+ + MnO4- → Fe3+ + Mn^2+
Balanced equation: 5Fe2+ + MnO4- + 4OH- → 5Fe3+ + Mn^2+ + 2H2O

5. Acidic solution:
Unbalanced equation: ClO3- + I- → IO3- + Cl-
Balanced equation: 2ClO3- + 10I- + 16H+ → 10IO3- + 8Cl- + 8H2O

Basic solution:
Unbalanced equation: ClO3- + I- → IO3- + Cl-
Balanced equation: 6ClO3- + 11I- + 6OH- → 5IO3- + 6Cl- + 3H2O

6. Acidic solution:
Unbalanced equation: Cr2O7^2- + Fe^2+ → Cr^3+ + Fe^3+
Balanced equation: Cr2O7^2- + 6Fe^2+ + 14H+ → 2Cr^3+ + 6Fe^3+ + 7H2O

Basic solution:
Unbalanced equation: Cr2O7^2- + Fe^2+ → Cr^3+ + Fe^3+
Balanced equation: Cr2O7^2- + 14Fe^2+ + 8OH- → 2Cr^3+ + 7H2O + 14Fe^3+

7. Acidic solution:
Unbalanced equation: MnO4- + SO2 → Mn^2+ + SO4^2-
Balanced equation: 5MnO4- + 3SO2 + 8H+ → 5Mn^2+ + 3SO4^2- + 4H2O

Basic solution:
Unbalanced equation: MnO4- + SO2 → Mn^2+ + SO4^2-
Balanced equation: 5MnO4- + 3SO2 + 6OH- → 5Mn^2+ + 3SO4^2- + 3H2O

By following the steps of balancing redox equations under acidic and basic solutions in the above examples, it is possible to obtain balanced equations for various redox reactions.