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6K. 5d

Posted: Sat Feb 29, 2020 3:37 am
by Lindsey Chheng 1E
Balance P4 → H2PO2- + PH3 in basic solution.

I know that the half reactions are:
3 x (P4 + 8OH- → 4H2PO2- + 4e-)
12H2O + P4 + 12e- → 4PH3 + 12OH-

and the answer is 3OH- + P4 → 3H2PO2- + PH3

My question is how was the second half reaction found? I know that H2O and OH- have to be added for basic solutions, but I don't know what was added first or if anything was cancelled out or combined to end up with the final half reaction. Can someone go through it step by step please?

Re: 6K. 5d

Posted: Sat Feb 29, 2020 9:10 am
by 005391550
P4 ---> PH3
balance P:
P4 ---> 4PH3
balance H:
12H2O + P4 ---> 4PH3 + 12OH-
balance charge:
12 H2O + P4 + 12e- ---> 4PH3 + 12OH-

the H2O and OH- are added at the same time to balance H. because the left side needed 12 H, you can add 12 H2O and 12 OH- on the opposite side to balance the H.