Problem: Balance each of the following redox reactions occurring in basic solution. a. H2O2(aq) + ClO2(aq) → ClO2-(aq) +O2(g)b. Al(s)+MnO4-(aq) → MnO2(s) + Al(OH)4-(aq)c. Cl2(g) → Cl−(aq)+ ClO−(aq)

When balancing redox reactions under acidic conditions, we will follow the following steps.

Step 1: Separate the whole reaction into two half-reactions.

Step 2: Balance the non-hydrogen and non-oxygen elements first.

Step 3: Balance oxygen by adding H₂O to the side that needs oxygen. (1 O: 1 H₂O)

Step 4: Balance hydrogen by adding H⁺ to the side that needs hydrogen. (1 H: 1 H⁺)

Step 5: Balance the charges: add electrons to the more positive side (or less negative side).

Step 6: Balance electrons on the two half-reactions.

Step 7: Get the overall reaction by adding the two reactions.

When balancing under basic conditions, additional steps are required:

Step 8: Balance remaining H⁺ by adding an equal amount of OH^– ions to both sides.

Step 9: H⁺(aq) will combine with OH^–(aq) to form H₂O(l).

Step 10: Cancel out common species.

a. H₂O₂(aq) + ClO₂(aq) → ClO₂^-(aq) +O₂(g)

(1-2): Separate and balance non-H and O atoms

H₂O₂(aq) →  O₂(g)

ClO₂(aq) → ClO₂^-(aq) 

(3): Add H₂O

H₂O₂(aq) →  O₂(g)

ClO₂(aq) → ClO₂^-(aq) 

(4): Add H⁺

H₂O₂(aq) →  O₂(g) + 2 H⁺ (aq) 

ClO₂(aq) → ClO₂^-(aq) 

(5): Balance charges

H₂O₂(aq) →  O₂(g) + 2 H⁺ (aq) + 2 e^-

ClO₂(aq) + 1 e^- → ClO₂^-(aq)