Balancing basic redox reactions

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NatBrown1I
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Joined: Fri Sep 28, 2018 12:17 am

Balancing basic redox reactions

Postby NatBrown1I » Wed Mar 13, 2019 1:47 pm

How do you balance basic redox reactions if OH and H2O both have O in them? (as opposed to acidic where you can just use H+ and automatically balance the equations)?

Mona Lee 4L
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

Re: Balancing basic redox reactions

Postby Mona Lee 4L » Wed Mar 13, 2019 2:14 pm

For redox reactions in basic solutions, you should first balance it using H+, assuming an acidic solution. Then, you should add OH- to both sides, creating H2O on the side with H+ and OH- on the other. This will result in a redox reaction in basic solutions. There's a table on this in the textbook if you want more details.

Kyither Min 2K
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Joined: Wed Oct 03, 2018 12:15 am

Re: Balancing basic redox reactions

Postby Kyither Min 2K » Wed Mar 13, 2019 2:36 pm

I like to balance it first like an acidic reaction with H+ and then to remove the H+, add OH- on both sides to neutralize the H+ to H2O. That way, you'll have H2O and OH-.

Ethan Breaux 2F
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Joined: Sat Sep 29, 2018 12:16 am

Re: Balancing basic redox reactions

Postby Ethan Breaux 2F » Thu Mar 14, 2019 1:25 am

Khan academy has a great video on it but yeah like they said you have to do it as if it were in acidic then you add the HO-

Marina Gollas 1A
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Joined: Fri Sep 28, 2018 12:20 am

Re: Balancing basic redox reactions

Postby Marina Gollas 1A » Thu Mar 14, 2019 4:19 am

NatBrown1I wrote:How do you balance basic redox reactions if OH and H2O both have O in them? (as opposed to acidic where you can just use H+ and automatically balance the equations)?


Basically, what you do is you add the H2O first in the redox half reaction to balance out the Os on the two sides. For example, in the image provided, the half reaction on the right side of the board, the "Cr3+(aq)---> CrO4^2-(aq), has 4 O's on the right side, but no O's on the left side. Therefore, you need to add 4 H2O molecules to the left side of the equation to get the 4 O's you needed to match the right. However, now you have an imbalance of H's, since now the right side doesn't have any H's. Henceforth, you add H20 to the right side now. Lastly, add OH- to the left side to provide the left side with the last O's and H's it needed to match both the product and reactant side with the same number of molecules of each element in both sides.
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Lynsea_Southwick_2K
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Joined: Fri Sep 28, 2018 12:25 am

Re: Balancing basic redox reactions

Postby Lynsea_Southwick_2K » Thu Mar 14, 2019 10:06 am

Acidic solutions: balance O by using H2O, then H using H+
Basic solutions: balance O using H2O; then H using H2O (FOR EACH H) to the side of each half rxn that needs H and adding OH to the other side

CristinaMorales1F
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Joined: Fri Feb 16, 2018 3:01 am

Re: Balancing basic redox reactions

Postby CristinaMorales1F » Sun Mar 17, 2019 12:03 am

How do you know which reaction is being oxidized and which is being reduced in a question where you are only given a the final reaction?

SydBenedict2H
Posts: 60
Joined: Wed Nov 08, 2017 3:00 am

Re: Balancing basic redox reactions

Postby SydBenedict2H » Sun Mar 17, 2019 12:10 am

CristinaMorales1F wrote:How do you know which reaction is being oxidized and which is being reduced in a question where you are only given a the final reaction?

Whichever reactant has a value that is becoming less positive, as more electrons 'reduces' the number, (+7 to +5 etc.) is being reduced. Whichever reactant is getting more positive (+5 to +7 etc.) is being oxidized.


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