Using the oxidation number method to balanced the given reaction equation, for a redox reaction in a basic solution:

Question

Using the oxidation number method to balanced the given reaction equation, for a redox reaction in a basic solution: 
Zn(s) + BrO4^-(aq) = Zn(OH)4^2+ + Br^-

GPT 3.5 · Answer

To balance this redox reaction in a basic solution using the oxidation number method, follow these steps:

1. Write down the unbalanced equation:
Zn(s) + BrO4^-(aq) = Zn(OH)4^2+ + Br^-

2. Assign oxidation numbers to all elements in the equation. The oxidation number of oxygen is usually -2, except in peroxides where it is -1. The oxidation number of hydrogen is usually +1. Use the oxidation number rules to find the oxidation numbers for other elements.

Zn(s) + BrO4^-(aq) = Zn(OH)4^2+ + Br^-
0 + ? = ? + ?

3. Identify the elements that undergo changes in oxidation number. In this case, zinc (Zn) and bromine (Br) are oxidized or reduced.

4. Write down the oxidation half-reaction for each element. The half-reactions show the loss or gain of electrons for each element.

Oxidation half-reaction: Zn(s) → Zn(OH)4^2+

5. Balance the electrons in the half-reaction by adding the appropriate coefficients.

Oxidation half-reaction: Zn(s) → Zn(OH)4^2+ + 8e^-

6. Write down the reduction half-reaction(s) for the other element(s) undergoing a change in oxidation number.

Reduction half-reaction: BrO4^- → Br^-

7. Balance the electrons in the reduction half-reaction by adding the appropriate coefficients.

Reduction half-reaction: BrO4^- + 10e^- → Br^-

8. Multiply each half-reaction by a coefficient to make the number of electrons gained equal to the number of electrons lost.

Multiply the oxidation half-reaction by 10 and the reduction half-reaction by 8 to equalize the electrons:

10Zn(s) → 10Zn(OH)4^2+ + 80e^-
8BrO4^- + 80e^- → 8Br^-

9. Combine the half-reactions to cancel out the electrons.

10Zn(s) + 8BrO4^- + 80e^- → 10Zn(OH)4^2+ + 8Br^-

10. Balance the remaining elements in the equation.

10Zn(s) + 8BrO4^- + 80e^- → 10Zn(OH)4^2+ + 8Br^-

11. Add water (H2O) to balance the oxygen atoms.

10Zn(s) + 8BrO4^- + 80e^- → 10Zn(OH)4^2+ + 8Br^- + 40H2O

12. Add hydroxide ions (OH^-) to balance the hydrogen atoms.

10Zn(s) + 8BrO4^- + 80e^- + 40OH^- → 10Zn(OH)4^2+ + 8Br^- + 40H2O

13. Double-check that the charges are balanced on both sides of the equation.

The balanced equation for the redox reaction in the basic solution is:
10Zn(s) + 8BrO4^- + 80e^- + 40OH^- → 10Zn(OH)4^2+ + 8Br^- + 40H2O