14.5d

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Ya Gao
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Joined: Fri Sep 29, 2017 7:04 am
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14.5d

Postby Ya Gao » Sun Feb 18, 2018 7:27 pm

Hey guys, I'm having trouble getting the right answer for question 14.5d. A little clarification would be helpful. Thanks!
P4=H2PO2- + PH3 under basic condition.

Gevork 2E
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Re: 14.5d

Postby Gevork 2E » Sun Feb 18, 2018 8:07 pm

Hi,

can you maybe be a bit more specific about your question? Are you asking for how this reaction occurs or how the process is to get the products? I think the main idea here is that P4 is acting as both the oxidizing AND the reducing agent. Was that the part that was confusing?

Glendy Gonzalez 1A
Posts: 52
Joined: Fri Sep 29, 2017 7:07 am

Re: 14.5d

Postby Glendy Gonzalez 1A » Sun Feb 18, 2018 9:14 pm

In this problem, P4 is both the reducing and oxidizing agent. Therefore, you can start by separating and balancing both half-reactions. Remember, you start with P4 in both reactions. After balancing them you can combine them in order to get the final equation.


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