6K. D

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BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

6K. D

Postby BeylemZ-1B » Wed Feb 19, 2020 10:40 am

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 ?

Joseph Saba
Posts: 154
Joined: Thu Jul 11, 2019 12:16 am

Re: 6K. D

Postby Joseph Saba » Wed Feb 19, 2020 4:30 pm

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

JChen_2I
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am

Re: 6K. D

Postby JChen_2I » Mon Feb 24, 2020 9:37 pm

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?

Joseph Saba
Posts: 154
Joined: Thu Jul 11, 2019 12:16 am

Re: 6K. D

Postby Joseph Saba » Tue Feb 25, 2020 1:37 pm

Oh yes, my bad.


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