14.5d

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Douglas Nguyen 2J
Posts: 71
Joined: Fri Sep 28, 2018 12:15 am

14.5d

Postby Douglas Nguyen 2J » Sat Feb 23, 2019 11:25 pm

What are the steps you take to balance this reduction reaction?

P4(s) -> H2PO2(-) +PH3(g)

I have:
12e- + P4(s) -> 4PH3(g)

and then:

12e- + P4(s) + 6 H2O -> 4PH3(g) + 6 OH-

but the solutions manual indicates that the reduction reaction requires 12 water and 12 hydroxide ions.

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

Re: 14.5d

Postby Matthew Tran 1H » Sun Feb 24, 2019 12:32 am

Your final equation is not balanced; you have 12 H on the left side (6*2), but 18 on the right side (4*3+6). In addition, your charges aren't balanced either; the left side has a -12 charge (12e-) and the right side has a -6 charge (6OH-)

almaochoa2D
Posts: 67
Joined: Fri Sep 28, 2018 12:23 am

Re: 14.5d

Postby almaochoa2D » Sun Feb 24, 2019 9:11 am

I had trouble with this one too. With 6OH and 6H2O the equation is still not balanced that's why you need 12OH and 12 H2O.


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