Difference between balancing Basic and Acidic

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Rita Chen 1B
Posts: 112
Joined: Sat Jul 20, 2019 12:15 am

Difference between balancing Basic and Acidic

Postby Rita Chen 1B » Sat Feb 29, 2020 3:39 pm

Is the only difference between balancing basic and acidic reactions whether you add H+ or OH?

Also, do you guys have any tips for balancing?

Rebecca Remple 1C
Posts: 137
Joined: Wed Sep 18, 2019 12:16 am
Been upvoted: 1 time

Re: Difference between balancing Basic and Acidic

Postby Rebecca Remple 1C » Sat Feb 29, 2020 3:54 pm

Rita Chen 1B wrote:Is the only difference between balancing basic and acidic reactions whether you add H+ or OH?

Also, do you guys have any tips for balancing?

Hi Rita,

From what I understand, that is the most important difference between the two!
I spoke with Matthew Tran, a UA, regarding balancing tips. I thought his most useful advice was to treat basic solutions the same way you do acidic solutions. I find it fairly simple to balance in acidic solutions using H2O and H+, as H+ is a single atom. For example, given the half reaction ClO- -> Cl-, you would add water so that ClO- -> Cl- + H2O (balances oxygens). Then you would add H+ so that 2H+ + ClO- -> Cl- + H2O (balances hydrogens). Now, given that the charge of the left is +1 and the right is -1, you would add 2 electrons to the left to balance the charges. Your final answer is 2e- + 2H+ + ClO- -> Cl- + H2O.
However, if this was a basic solution H+ would not be present. You can take the same equation (without the electrons) 2H+ + ClO- -> Cl- + H2O. Then, you can add OH- to both sides. OH- combines with H+ to form water, H2O, so you would want to add 2 OH- on both sides. After doing so, you should get 2H2O + ClO- -> Cl- + H2O + 2OH-. The charge is -1 on the left and -3 on the right, so you would balance it as 2e- + 2H2O + ClO- -> Cl- + H2O + 2OH-. I hope I explained this alright! I pulled my example from this site, and they have some other problems you can practice with: https://www.chemteam.info/Redox/Balance-Redox-Base.html. I hope this helps! :)

-Rebecca

RRahimtoola1I
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

Re: Difference between balancing Basic and Acidic

Postby RRahimtoola1I » Sat Feb 29, 2020 7:47 pm

This is how you balance for both:

1. determine the oxidation and reduction half reactions.
2. Balance both half reactions:
a) Balance elements other than H or O
b) Balance O elements by adding H2O molecules to match the number of O molecules you have.
c) Balance H elements by adding H+ (aq) FOR BASES: add OH- molecules to cancel all H+ molecules which will become H2O and most likely cancel with H2Os on the other side of the reaction
d) Balance the overall charge on both sides by adding electrons to the more positive side.
e) Combine both of the balanced half rxns (make sure the # of electrons in the half rxns match before you do this). Cancel any electrons or molecules that appear on both sides.

Rita Chen 1B
Posts: 112
Joined: Sat Jul 20, 2019 12:15 am

Re: Difference between balancing Basic and Acidic

Postby Rita Chen 1B » Sat Feb 29, 2020 8:59 pm

Rebecca Remple 1C wrote:
Rita Chen 1B wrote:Is the only difference between balancing basic and acidic reactions whether you add H+ or OH?

Also, do you guys have any tips for balancing?

Hi Rita,

From what I understand, that is the most important difference between the two!
I spoke with Matthew Tran, a UA, regarding balancing tips. I thought his most useful advice was to treat basic solutions the same way you do acidic solutions. I find it fairly simple to balance in acidic solutions using H2O and H+, as H+ is a single atom. For example, given the half reaction ClO- -> Cl-, you would add water so that ClO- -> Cl- + H2O (balances oxygens). Then you would add H+ so that 2H+ + ClO- -> Cl- + H2O (balances hydrogens). Now, given that the charge of the left is +1 and the right is -1, you would add 2 electrons to the left to balance the charges. Your final answer is 2e- + 2H+ + ClO- -> Cl- + H2O.
However, if this was a basic solution H+ would not be present. You can take the same equation (without the electrons) 2H+ + ClO- -> Cl- + H2O. Then, you can add OH- to both sides. OH- combines with H+ to form water, H2O, so you would want to add 2 OH- on both sides. After doing so, you should get 2H2O + ClO- -> Cl- + H2O + 2OH-. The charge is -1 on the left and -3 on the right, so you would balance it as 2e- + 2H2O + ClO- -> Cl- + H2O + 2OH-. I hope I explained this alright! I pulled my example from this site, and they have some other problems you can practice with: https://www.chemteam.info/Redox/Balance-Redox-Base.html. I hope this helps! :)

-Rebecca


OMG this is so great thank you so much!


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