6K. 5

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Gisela F Ramirez 2H
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Joined: Fri Sep 28, 2018 12:27 am

6K. 5

Postby Gisela F Ramirez 2H » Tue Feb 26, 2019 11:38 pm

d) P4 (s) ---> H2PO2(^-1) (aq) + PH3 (aq)

oxidizing: 8OH(^-1) +P4 ---> 4H2PO2(^-1) +4e(^-1)
reduction: P4 ---> 4PH3

I have been trying to balance these reactions for so long and I don't get how to balance the reduction half equation? Can someone help?

Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

Re: 6K. 5

Postby Albert_Luu3K » Wed Feb 27, 2019 5:02 am

For the reduction, you should end up with : 12e- +12H2O + P4 ---> 4PH3 + 12OH-
You got the start correct but you just got to keep going and trying out different things to balance. First I saw that there were 12 H on the right side, so I needed to balance that. All I can add is H2O to add more H but that would add more OH. So I figured out you had to add equal amounts of both.

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