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!! :)
Balancing 14.5 part d
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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!
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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