How do you balance redox equation under acidic and basic solution.give seven examples with explanation on each ?
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.
To balance a redox equation, follow these steps:
1. Write the unbalanced equation: Start by writing the unbalanced equation for the redox reaction.
2. Assign oxidation numbers: Determine the oxidation numbers of each element in the reactants and products.
3. Separate each half-reaction: Split the redox equation into two half-reactions - one for reduction and one for oxidation.
4. Balance the atoms: Balance the atoms in each half-reaction by adjusting coefficients.
5. Balance the charges: Add electrons (e-) to one side of each half-reaction to balance the charges.
6. Equalize the number of electrons: Multiply each half-reaction by the necessary coefficients so that the number of electrons transferred is equal in both half-reactions.
7. Combine the half-reactions: Add the balanced half-reactions together and simplify if necessary.
Here are seven examples, with step-by-step explanations for balancing redox equations under both acidic and basic conditions:
Example 1: Redox reaction under acidic conditions
Equation: Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
Explanation:
Step 1: Write the unbalanced equation: Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
Step 2: Assign oxidation numbers: Fe2+ is oxidized to Fe3+, while MnO4- is reduced to Mn2+.
Step 3: Separate each half-reaction:
Oxidation: Fe2+(aq) → Fe3+(aq)
Reduction: MnO4-(aq) → Mn2+(aq)
Step 4: Balance the atoms:
Oxidation: Fe2+(aq) → Fe3+(aq) + e-
Reduction: MnO4-(aq) + 8H+(aq) → Mn2+(aq) + 4H2O(l)
Step 5: Balance the charges:
Oxidation: 2Fe2+(aq) → 2Fe3+(aq) + 2e-
Reduction: MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
Step 6: Equalize the number of electrons:
Oxidation: 5Fe2+(aq) → 5Fe3+(aq) + 5e-
Reduction: MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
Step 7: Combine the half-reactions:
5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
Example 2: Redox reaction under basic conditions
Equation: Cl2(g) + OH-(aq) → Cl-(aq) + ClO-(aq)
Explanation:
Step 1: Write the unbalanced equation: Cl2(g) + OH-(aq) → Cl-(aq) + ClO-(aq)
Step 2: Assign oxidation numbers: Cl2 is reduced to Cl-, while OH- is oxidized to ClO-.
Step 3: Separate each half-reaction:
Oxidation: OH-(aq) → ClO-(aq)
Reduction: Cl2(g) → Cl-(aq)
Step 4: Balance the atoms:
Oxidation: 2OH-(aq) → ClO-(aq) + 2e-
Reduction: Cl2(g) + 2e- → 2Cl-(aq)
Step 5: Balance the charges:
Oxidation: 2OH-(aq) + 2e- → ClO-(aq)
Reduction: Cl2(g) + 2e- → 2Cl-(aq)
Step 6: Equalize the number of electrons:
Oxidation: 3OH-(aq) + 3e- → ClO-(aq)
Reduction: Cl2(g) + 2e- → 2Cl-(aq)
Step 7: Combine the half-reactions:
3Cl2(g) + 6OH-(aq) → 5Cl-(aq) + ClO-(aq) + 3H2O(l)
Similarly, you can follow these steps to balance redox equations for other examples under acidic and basic conditions.