Balancing 14.5 part d

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Daniel Dobrin 2F
Posts: 12
Joined: Sat Jul 09, 2016 3:00 am

Balancing 14.5 part d

Postby Daniel Dobrin 2F » Sat Feb 11, 2017 9:14 pm

Hey everyone,

I'm having trouble with chapter 14 question 5 part d. We have to balance the equation using half-reactions. The reaction occurs in a basic solution. We are given:

P4(s) -> H2PO2-(aq) + PH3(g)

In the solutions manual, PH3 is gone when doing it. I really don't understand how to the whole part. If anyone can explain the problem, I will be very grateful!!

Thank you!! :)

Michelle_Li_1H
Posts: 32
Joined: Wed Sep 21, 2016 3:00 pm

Re: Balancing 14.5 part d

Postby Michelle_Li_1H » Mon Feb 13, 2017 4:52 pm

Hi!

I think you might have read the solution manual wrong, but to approach the question, we would split the redox equation into two half equations.

P4---> H2PO2- and P4---> PH3

Then to balance in a basic solution, we would first balance out the oxygen atoms then use OH- rather than H+ when balancing out the hydrogen atoms. Then we would balance out the charges on both sides. We would get the following:

P4 + 8H2O +8OH- ---> 4H2PO2- + 8H20 +4e-
P4 +12H2O +12e- ---> 4PH3 +12OH-

After adding both equations, we would get:

12OH- +4P4 + 12H2O ---> 12H2PO2- + 4PH3 + 4OH-

And we could simplify even further by dividing by 4 on both sides to get our final balanced equation.

3OH- +P4 +3H2O ---> 3H2PO2- + PH3

Hope this helps!


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