Question

1. Balance the following redox equations by the ion-electron method: (25 points) a. H2O2 + Fe+2 → Fe+3 + H2O (in acidic solution) b. Cu + HNO3 → Cu+2 + NO + H2O (in acidic solution) c. CN– + MnO4– → CNO– + MnO2 (in basic solution) d. Br2 → BrO3– + Br– (in basic solution) e. S2O3–2 + I2 → I– + S4O62– (in acidic solution)

Answer 1

To balance redox reactions, you can use the ion-electron method. By adding species such as H₂O, H⁺, and OH⁻, and adjusting the coefficients and charges, you can balance the equations. The specific steps for each reaction are provided in the detailed answer.

Step-by-step explanation:

For question 1, let's balance the redox equations using the ion-electron method:

1. a. H2O2 + Fe+2 -> Fe+3 + H2O (in acidic solution)
2. Add H2O and H+ on the product side. Then, balance the O and H atoms and finally balance the charges by adding electrons.
3. b. Cu + HNO3 -> Cu+2 + NO + H2O (in acidic solution)
4. Add H2O and H+ on the product side. Then, add water to balance the oxygen atoms and finally balance the charges by adding electrons.
5. c. CN- + MnO4- -> CNO- + MnO2 (in basic solution)
6. Add OH- on the product side. Then, balance the O and H atoms and finally balance the charges by adding electrons.
7. d. Br2 -> BrO3- + Br- (in basic solution)
8. Add OH- on both sides. Then, balance the O and H atoms and finally balance the charges by adding electrons.
9. e. S2O3-2 + I2 -> I- + S4O62- (in acidic solution)
10. Balance the I atoms by adding I- on both sides. Then, balance the H atoms and finally balance the charges by adding electrons.