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^-
Zn(s) + BrO4^-(aq) = Zn(OH)4^2+ + Br^-
Answers
GPT 3.5
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
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
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