Balancing Redox equations in Basic conditions

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005115864
Posts: 64
Joined: Fri Sep 28, 2018 12:15 am

Balancing Redox equations in Basic conditions

Postby 005115864 » Sun Feb 24, 2019 3:29 pm

Hi, I am having trouble knowing where to put the OH- and H2O when making my half reactions. I tried completing 14.5 in the 6th edition, but I just can't seem to figure it out. Can someone please explain?

An example is P4(s) > H2PO2^- +PH3

Anita Wong 1H
Posts: 68
Joined: Fri Sep 28, 2018 12:27 am
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Re: Balancing Redox equations in Basic conditions

Postby Anita Wong 1H » Sun Feb 24, 2019 4:56 pm

they both go on the left side

almaochoa2D
Posts: 67
Joined: Fri Sep 28, 2018 12:23 am

Re: Balancing Redox equations in Basic conditions

Postby almaochoa2D » Sun Feb 24, 2019 6:56 pm

You have to make everything is balanced. P4 is the reactant for both equations. After you've split the equations balance the O's then balance the H's and lastly balance the charges. In a basic solution, you add H2O and OH to balance the reactions. Then you add elections to whichever side has the least negative charge.

Carlos De La Torre 2L
Posts: 60
Joined: Tue Oct 09, 2018 12:16 am

Re: Balancing Redox equations in Basic conditions

Postby Carlos De La Torre 2L » Sun Feb 24, 2019 8:14 pm

You have to add water to balance out the oxygens once you balance all other elements, so in this case it would be on the left. Then you add OH^- to balance the hydrogens and oxygens. Then lastly you balance the charges. Since you can't raise the charge of one side, you can only lower it you add electrons to the most positive side s.t. it matches the most negative side.


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