HOMEWORK 14.5 D

Elle_Mendelson_2K
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Joined: Fri Sep 28, 2018 12:28 am

HOMEWORK 14.5 D

Will someone explain how to balance d in chapter 14 problem 5 in the 6th edition. I don't understand why the OH is on the right side instead of the left when you would add H2O on the left and therefore OH- on the right

Elle_Mendelson_2K
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

Re: HOMEWORK 14.5 D

Hi again, can you explain both parts? I can't seem to get P4->Ph3 balanced!

705192887
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Joined: Fri Sep 28, 2018 12:18 am

Re: HOMEWORK 14.5 D

Elle_Mendelson_2K wrote:Will someone explain how to balance d in chapter 14 problem 5 in the 6th edition. I don't understand why the OH is on the right side instead of the left when you would add H2O on the left and therefore OH- on the right

First, you want to balance the oxygen by adding 8 H2O molecules on the left. Now, balance the hydrogen atoms by adding 8H2O and 8OH- on the right and left side respectively. Now the 8H2Os cancel and leaves the equation as:
P4(s) + 8OH- (aq) --> 4H2PO2- (aq)
Finally just add 4e- to the right to balance the charges to get your balanced oxidation half reaction.

For the reduction half-reaction, you're going to add 12H2O on the left and 12OH- on the fight, and 12e- on the left to balance the charge. This gives you:
P4(s)+12H2O+12e- --> 4PH3(g)+12OH-

Cancel like terms and multiply the reactions by constants to find final reaction which is:
P4(s)+3H2O+3OH- ---> 3H2PO2- + PH3(g)

Elle_Mendelson_2K
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

Re: HOMEWORK 14.5 D

Thank you so much! wouldn't it be 4P4? Or can you explain why there is only one P4? thank you!

Elle_Mendelson_2K
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

Re: HOMEWORK 14.5 D

Also, does the order of the elements matter? As long as they are on the correct side, is there a difference?