[musicman92]

Hey everybody, I'm new to the forum. I'm a junior in high school currently enrolled in a home school type program (but I'm at a school building).

I am nearing the end of my first chemistry course and I will say I enjoyed it very much. I've got a couple questions, and if your answers help me understand the concepts and with any luck I will finish my chemistry course tonight  (except for the final, of course)..so anyway, here are the questions:

1. Rather simple, but I can't find out why anywhere online...why is antifreeze so sweet?

2. I'm having trouble balancing some redox reactions. In my textbook, they want me to balance them using oxidation numbers (I completely understand oxidation numbers, btw). However, in my homework, the problems don't work with the oxidation number method. Also, they want me to balance the charges on either side of the reaction. I asked my "teacher" about it, and she told me just to balance them using trial and error instead of the oxidation numbers. But that is proving to be easier said than done.

Here is the first example:

__ P₄(s) + __ H₂O(l)    __ PH₃(g) + __ H₂PO₂(aq)

The problem is in an acidic solution, so I can add H⁺ ions to either side if I need them. I worked through the problem a few times...but I always ended up missing something.

Thanks in advance for the assistance! (I caught the "delete me" thing, btw  )

[Train]

__ P₄(s) + __ H₂O(l)    __ PH₃(g) + __ H₂PO₂(aq)

The problem is in an acidic solution, so I can add H⁺ ions to either side if I need them. I worked through the problem a few times...but I always ended up missing something.

Interesting problem.  I've so far come up with 4 different balanced equations - none of them multiples of each other.  One of them I got by using oxidation numbers and the other three just by playing around.  Did you not get any balanced results or just not the one in the book (if the book gives answers)?  Can you show some of your work?

Rather simple, but I can't find out why anywhere online...why is antifreeze so sweet?

They had to add extra sugar so you could still taste it when it's really cold . . . because the reactions on your tongue follow Arrhenius' law.

(You are joking, right?)

[musicman92]

Ok, I've been working on it and I think I got it:

21 H⁺ + 3 P₄ + 6 H₂O    9 PH₃ + 3 H₂PO₂

I'm not sure about the H⁺ ions on the right though...

[Borek]

Charge is not balanced. It is not enough to balance atoms (assuming they are balanced - I have not checked).

[musicman92]

Charge is not balanced. It is not enough to balance atoms (assuming they are balanced - I have not checked).

ok, could you help me understand how to balance the charges? or at least refer me to somewhere where I could learn? because in the original problem, none of the compounds have a charge. They tell me I HAVE TO add H⁺ ions to one side or the other. my textbook does not explain anything about this.

[Borek]

Try here:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-redox

http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method

They tell me I HAVE TO add H⁺ ions to one side or the other

No, you were closer in your first post:

I can add H⁺ ions to either side if I need them.

You don't need H⁺ here, you can balance this equation as is.

[musicman92]

Ok, what I meant to say was that the instructions of the problem tell me that I NEED to add some H⁺ to one side or the other. Oh well...I'll figure it out. thanks for the references...

[Train]

If you balance by half reactions you do add H+, but they end up canceling out.  If you balance by oxidation numbers then you don't add any H+.

Nevermind what I said about coming up with multiple answers.  I was making a mistake and I forgot that the charge has to balance.

[Borek]

Ok, what I meant to say was that the instructions of the problem tell me that I NEED to add some H⁺ to one side or the other.

As Train wrote - they will cancel out. But you don't have to add them. You may as well assume that P₄ reacts directly with water and that H⁺ and OH^- are between the products. In the final equation they will cancel out in exactly the same way.

[musicman92]

Final answer:

P₄ + 6 H₂O  PH₃ + 3 H₂PO₂ + 3 H⁺

I checked the answer with my teacher, and she said it was right.

[Train]

Final answer:

P₄ + 6 H₂O  PH₃ + 3 H₂PO₂ + 3 H⁺

I also got that answer the other night.  The problem is, I got 3 others just like it:

P₄ + 4H₂O + 2H+  2PH₃ + 2H₂PO₂

P₄ + 2H₂O + 7H+  3PH₃ + H₂PO₂

11P₄ + 24H₂O + 72H+  32PH₃ + 12H₂PO₂

They can't all be right, can they?  No, but they can all be wrong.

Here is the oxidation numbers method applied to this specific problem:

I observe that the phosphorus in P₄ has oxidation state of 0, the phosphorus in PH₃ has oxidation state of 3-, and the phosphorus in H₂PO₂ has an oxidation state of 2+. 

So for each molecule of PH₃ formed 3 electrons are gained, and for each molecule of H₂PO₂ 2 electrons are lost.  I wonder, how can I make the number of electrons gained equal the number of electrons lost?  This is essentially a least common multiple problem.  What is the least common multiple of 2 and 3?  Therefore what should I multiply PH₃ and H₂PO₂ by so that e- gained = e- lost?

From there I wonder what fraction I should multiply P₄ by so that the number of phosphorus atoms on the left is equal to those on the right.  Hydrogen and oxygen are easily balanced by multiplying H₂O by a whole number.  Finally I multiply the entire equation by the denominator of the fraction in front of P4, so that I'm left with only whole numbers.

Why is the resulting answer a better one than the one we came up with?

[Borek]

P₄ + 6 H₂O  PH₃ + 3 H₂PO₂ + 3 H⁺

You have posted not long ago equation that had 21H⁺ on the left - do you remember what I told you then?

I checked the answer with my teacher, and she said it was right.

No comments. But sad.

--------------

It is a little bit tricky and doesn't look logical at first, but the same equation can be balanced by half reaction method.

First - P₄ reacts with water to produce PH₃. That means we need hydrogen from somewehere. We can use H⁺ on the left, or water on the left and OH^- on the products side, net effect is identical:

P₄ + H₂O  PH₃ + OH^-

Balance atoms - charge will be not balanced, but that's how the half reaction method is supposed to work. Add electrons to balance charge.

Next, P₄ has to react with water producing H₂PO₂. Again, we have to balance hydrogen and oxygen somehow, let's do it adding water on the left and whatever will be necesary on the right - seems like we have to add one H and one O per each P atom, that leaves us with excess hydrogen on the right:

P₄ + H₂O  H₂PO₂ + H⁺

Again, balance atoms then balance charge with electrons.

Finally, do what is needed to balance the final equation - combine both half reactions multiplying them by coefficients that will guarantee that electrons cancels out. Take a look at the resulting equation - you have H⁺ and OH^- on the right. What is the next logical step?