Answer:
[tex]2BrO_{3} ^-(aq) +6H_2O+5Zn(s)+8OH^-\rightarrow 2Br^-(aq)+{5Zn(OH)_{4}}^{2-} (aq)[/tex]
Explanation:
Let's divide the process of balancing a redox reaction in basic conditions into steps:
- Split the equation into half-equations.
- Balance for elements other than H or O.
- Balance for O by adding H₂O.
- Balance for H by adding H⁺.
- Add the same # of OH⁻ as H⁺ because we are balancing in basic conditions.
- Combine the H⁺ and OH⁻ to make H₂O.
- Balance charges.
- Multiply to cancel out the electrons.
- Add the half-equations together and simplify.
Now, let's write our equation out to start.
[tex]BrO_{3} ^-(aq) + Zn(s) \rightarrow Br^-(aq)+ {Zn(OH)_{4}}^{2-} (aq)[/tex]
Now, we can separate our equation into it's half equations.
[tex]BrO_{3} ^-(aq) \rightarrow Br^-(aq)\\Zn(s)\rightarrow {Zn(OH)_{4}}^{2-} (aq)[/tex]
Now that we have two separate equations, we must make sure we are performing each step on both equations.
Step 2 does not apply to these equations because all non-O elements are already balanced.
Step 3: balancing for oxygen.
Let's first focus on our bromine half-equation. There are three oxygen ions that only appear on the left side, so we must add H₂O to the right side to make up for it. We'll add 3 H₂O molecules because each molecule contains one oxygen ion. That leaves us with:
[tex]BrO_{3} ^-(aq) \rightarrow Br^-(aq)+3H_2O[/tex]
For the zinc equation, there are 4 oxygen ions on the right side, so we must add 4 H₂O to the left side to balance. This gives us:
[tex]Zn(s)+4H_2O\rightarrow {Zn(OH)_{4}}^{2-} (aq)[/tex]
Now that we've added H₂O, we've unbalanced the hydrogens in the equation. Starting with the bromine half equation, We'll add 6 hydrogens to the left side because in each H₂O molecule, there are 2 hydrogens, and we have 3 H₂O molecules.
[tex]BrO_{3} ^-(aq) +6H^+ \rightarrow Br^-(aq)+3H_2O[/tex]
The zinc equation now has 8 hygrogens on the left side. However there are still 4 hydrogen ions on the right side. This means that we should add 4 hydrogen ions to the right side.
[tex]Zn(s)+4H_2O\rightarrow {Zn(OH)_{4}}^{2-} (aq)+4H^+[/tex]
Moving onto step 5: Because we are in basic conditions, we must add OH⁻ on both sides to neutralize the acidic H⁺. There are 6H⁺ ions, so add 6OH⁻ ions to the bromine half equation.
[tex]BrO_{3} ^-(aq) +6H^+ +6OH^-\rightarrow Br^-(aq)+3H_2O+6OH^-[/tex]
The zinc half equation has 4 H⁺ ions, so add 4OH⁻ions to counteract the H⁺.
[tex]Zn(s)+4H_2O+4OH^-\rightarrow {Zn(OH)_{4}}^{2-} (aq)+4H^++4OH^-[/tex]
Now that we added the OH⁻, the OH⁻ and H⁺ can be combined to make H₂O.
[tex]BrO_{3} ^-(aq) +6H_2O\rightarrow Br^-(aq)+3H_2O+6OH^-[/tex]
[tex]Zn(s)+4H_2O+4OH^-\rightarrow {Zn(OH)_{4}}^{2-} (aq)+4H_2O[/tex]
Now, we'll check the charges and add electrons accordingly.
The bromine half-equation has a charge of -1 on the left side and -6 on the right. We'll add 5 electrons to the left side to balance the charges so both sides have charges of -6.
[tex]BrO_{3} ^-(aq) +6H_2O+5e^-\rightarrow Br^-(aq)+3H_2O+6OH^-[/tex]
The left side of the zinc half-equation has a charge of 4- from the 4 OH⁻s, and the right side has a charge of 2-. We'll add 2 electrons to the right side to give both sides a charge of -4.
[tex]Zn(s)+4H_2O+4OH^-\rightarrow {Zn(OH)_{4}}^{2-} (aq)+4H_2O+2e^-[/tex]
We've added 5 electrons tp the first equation, and 2 to the other. the least common multiple of 5 and 2 is 10, so multiply the first equation by 2 and the second equation by 5.
[tex]2(BrO_{3} ^-(aq) +6H_2O+5e^-\rightarrow Br^-(aq)+3H_2O+6OH^-)=\\2BrO_{3} ^-(aq) +12H_2O+10e^-\rightarrow 2Br^-(aq)+6H_2O+12OH^-[/tex]
[tex]5(Zn(s)+4H_2O+4OH^-\rightarrow {Zn(OH)_{4}}^{2-} (aq)+4H_2O+2e^-)=\\5Zn(s)+20H_2O+20OH^-\rightarrow {5Zn(OH)_{4}}^{2-} (aq)+20H_2O+10e^-[/tex]
These equations have gotten lengthy and there is an excessive amount of H₂O, but once we add them back together, we can start cancelling things out.
[tex]2BrO_{3} ^-(aq) +12H_2O+10e^-+5Zn(s)+20H_2O+20OH^-\rightarrow 2Br^-(aq)+6H_2O+12OH^-+{5Zn(OH)_{4}}^{2-} (aq)+20H_2O+10e^-[/tex]
We can add together 12H₂O and 20H₂O on the left side, and 6H₂O and 20H₂O on the right side to make the next substraction simpler.
[tex]2BrO_{3} ^-(aq) +32H_2O+10e^-+5Zn(s)+20OH^-\rightarrow 2Br^-(aq)+26H_2O+12OH^-+{5Zn(OH)_{4}}^{2-} (aq)+10e^-[/tex]
There are OH⁻ions on both sides, so we can subtract 12OH⁻ leaving us with 8 OH⁻ on the left side . We can also subtract 26H₂O from both sides leaving us with 6H₂O on the left side. We'll also subtract the 10 electrons from both sides.
[tex]2BrO_{3} ^-(aq) +6H_2O+5Zn(s)+8OH^-\rightarrow 2Br^-(aq)+{5Zn(OH)_{4}}^{2-} (aq)[/tex]
This is the fully simplified, balanced redox reaction.
