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6K. D

Posted: Wed Feb 19, 2020 10:40 am
by BeylemZ-1B
d) P4 (s) ---> H2PO2(^-1) (aq) + PH3 (aq)

I know how to balance the reduction reaction, but can someone explain how to balance the oxidation reaction ?

Re: 6K. D

Posted: Wed Feb 19, 2020 4:30 pm
by Joseph Saba
The way i start this problem is by balancing the solution in acidic solution and then, at the end, adding OH- to both sides so the H+'s cancel out. The OH- + H+ turns into neutral H2O molecules (which may or may not cancel out. Therefore,
Oxidation: 8OH- +P4 --> H2PO2-+4e-
Reduction: 12e-+ 12H2O+ P4--> 4PH3 + 12OH-
Let me know if i made any mistakes

Re: 6K. D

Posted: Mon Feb 24, 2020 9:37 pm
by JChen_2I
Joseph Saba wrote:The way i start this problem is by balancing the solution in acidic solution and then, at the end, adding OH- to both sides so the H+'s cancel out. The OH- + H+ turns into neutral H2O molecules (which may or may not cancel out. Therefore,
Oxidation: 8OH- +P4 --> H2PO2-+4e-
Reduction: 12e-+ 12H2O+ P4--> 4PH3 + 12OH-
Let me know if i made any mistakes

For the oxidation, shouldn't it be 4H2PO2 because you have to balance the phosphorus?

Re: 6K. D

Posted: Tue Feb 25, 2020 1:37 pm
by Joseph Saba
Oh yes, my bad.