6K. 5a)

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Labiba Sardar 2A
Posts: 103
Joined: Sat Jul 20, 2019 12:15 am

6K. 5a)

Postby Labiba Sardar 2A » Sat Feb 22, 2020 9:24 pm

I need help trying the balance the half reaction Br-(aq) --> BrO3-(aq) in a basic solution.

I've balanced the reaction so then the O's are balanced on both sides:
Br-(aq) + 3H2O (l) --> BrO3-

But I don't understand what to do after this part, especially where you're supposed to put more H2O molecules on the products side and OH- molecules on the reactants side. Can someone please explain?

Emma Popescu 1L
Posts: 105
Joined: Wed Sep 11, 2019 12:16 am

Re: 6K. 5a)

Postby Emma Popescu 1L » Sat Feb 22, 2020 9:33 pm

You have to first balance the Os and then balance the Hs giving 6OH- + 3H2O +Br- ---> BrO3- + 6H2O. Then you need to balance the charge by adding electrons in order to ensure the charge is equal on both sides of the equations. This gives the half equation 6OH- + 3H2O +Br- ---> BrO3- + 6H2O +6e-

Dina Marchenko 2J
Posts: 54
Joined: Thu Jul 25, 2019 12:16 am

Re: 6K. 5a)

Postby Dina Marchenko 2J » Sat Feb 22, 2020 9:42 pm

Section 6K has a useful table that explains how to balance reactions in acidic and basic solutions.

When it comes to balancing a reaction in a basic solution, you balance oxygen first with water on either side and then to balance hydrogens, you add water to the side that needs more hydrogens and add hydroxide (the same amount) to the other side. Since hydroxide has one hydrogen and water has 2, when you add one of each to opposite sides, there is a net increase of 1 hydrogen on whichever side has the water.

Anyway, not that you've balanced the oxygen you balance the hydrogens using hydroxide and water:

6OH- + Br- + 3H2O --> BrO3- + 6 H2O
(I started by adding the 6 water to the right to balance the hydrogens, then added the 6 hydroxide to balance the added hydrogens on the right)
Then if you cancel the waters you get:

6OH- + Br- --> BrO3- + 3 H2O + 6e-

(electrons used to balance charges since bromine goes from a charge of -1 to a charge of +5)

Labiba Sardar 2A
Posts: 103
Joined: Sat Jul 20, 2019 12:15 am

Re: 6K. 5a)

Postby Labiba Sardar 2A » Sun Feb 23, 2020 11:54 am

Dina Marchenko 2J wrote:Section 6K has a useful table that explains how to balance reactions in acidic and basic solutions.

When it comes to balancing a reaction in a basic solution, you balance oxygen first with water on either side and then to balance hydrogens, you add water to the side that needs more hydrogens and add hydroxide (the same amount) to the other side. Since hydroxide has one hydrogen and water has 2, when you add one of each to opposite sides, there is a net increase of 1 hydrogen on whichever side has the water.

Anyway, not that you've balanced the oxygen you balance the hydrogens using hydroxide and water:

6OH- + Br- + 3H2O --> BrO3- + 6 H2O
(I started by adding the 6 water to the right to balance the hydrogens, then added the 6 hydroxide to balance the added hydrogens on the right)
Then if you cancel the waters you get:

6OH- + Br- --> BrO3- + 3 H2O + 6e-

(electrons used to balance charges since bromine goes from a charge of -1 to a charge of +5)


Did you add 6 H2O on the right because you had a total of 6 H's from the 3H2O on the left?

Dina Marchenko 2J
Posts: 54
Joined: Thu Jul 25, 2019 12:16 am

Re: 6K. 5a)

Postby Dina Marchenko 2J » Thu Feb 27, 2020 6:49 pm

Labiba Sardar 2A wrote:
Dina Marchenko 2J wrote:Section 6K has a useful table that explains how to balance reactions in acidic and basic solutions.

When it comes to balancing a reaction in a basic solution, you balance oxygen first with water on either side and then to balance hydrogens, you add water to the side that needs more hydrogens and add hydroxide (the same amount) to the other side. Since hydroxide has one hydrogen and water has 2, when you add one of each to opposite sides, there is a net increase of 1 hydrogen on whichever side has the water.

Anyway, not that you've balanced the oxygen you balance the hydrogens using hydroxide and water:

6OH- + Br- + 3H2O --> BrO3- + 6 H2O
(I started by adding the 6 water to the right to balance the hydrogens, then added the 6 hydroxide to balance the added hydrogens on the right)
Then if you cancel the waters you get:

6OH- + Br- --> BrO3- + 3 H2O + 6e-

(electrons used to balance charges since bromine goes from a charge of -1 to a charge of +5)


Did you add 6 H2O on the right because you had a total of 6 H's from the 3H2O on the left?



Yes, that's why I added 6 waters. I started by balancing the oxygens found in BrO3- by adding 3 waters to the elft side of the equation. Now there are equal amounts of oxygen on each side but 3 more hydrogens on the left than right. I wanted to end up with a net increase of 3 hydrogens on the right so i added 6 waters on the right and 6 hydroxides on the left (which canceled out half the hydrogens in water).

Anthony Hatashita 4H
Posts: 103
Joined: Wed Sep 18, 2019 12:21 am

Re: 6K. 5a)

Postby Anthony Hatashita 4H » Thu Feb 27, 2020 8:55 pm

Because it's a basic solution, in order to add more H to one side you need to add H2O to that side and OH- to the other, resulting in one more H per H2O molecule on that side while leaving the O balanced.


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