Chemical Forums

Chemistry Forums for Students => High School Chemistry Forum => Topic started by: /// on November 21, 2006, 06:01:51 PM

Title: Balancing Using Half RXN method under basic conditions
Post by: /// on November 21, 2006, 06:01:51 PM
This particular question is odd:

Ag(s) + CN- + O2(g) > Ag(CN)2-

How would I break it down into half reactions. I can do the rest on my own.
I was told by my teacher to just balance it without turning into half reactions, but I dont know, he barely looked at the question.

I tried turning it into net ionic equation to cancel out spectator, but the polyatomic product is weird.
Help?
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Dan on November 21, 2006, 06:20:25 PM
The equation is incomplete, which is why it won't cancel (notice no O on the right hand side).

What do you think happens to the oxygen?
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: /// on November 21, 2006, 06:24:46 PM
The equation is incomplete, which is why it won't cancel (notice no O on the right hand side).

What do you think happens to the oxygen?

Would it react with the Ag?
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Borek on November 21, 2006, 06:26:29 PM
1. It is not that difficult to balance without treating it as redox, just remember to balance charge as well.

2. Basic conditions means that you use OH- and H2O to balance oxygen/hydrogen (as described on this balancing redox reactions (http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method) page).

3. If you really, really, really need half reactions, think how to reduce oxygen to OH-

Hmm, Dan was faster ;)
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: /// on November 21, 2006, 06:53:23 PM
1. It is not that difficult to balance without treating it as redox, just remember to balance charge as well.

How can I balance the charges when the two half reactions are mixed
I balanced it properly but the charge on the left doesnt = the charge on the right. I can't get a common number in charges and multiply because I dont know what to multiply.
So far I got,

Ag(s) + 2CN-(aq) + O2 + 2H2O  >> Ag(CN)2- + 4OH-

What I did up there was just follow the rules as if it was a half reaction altogther. But in the half reactions, before we can combine them, we get a LCM of electrons. So do I just add 3 electrons to the left side of that equation? But then is that right? To have an equation with electrons in the equation?
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Borek on November 21, 2006, 07:17:09 PM
How can I balance the charges when the two half reactions are mixed

Just like atoms. The only difference is that atoms don't come in two flavors (negative and positive).

Quote
I balanced it properly but the charge on the left doesnt = the charge on the right.

So it is not balanced yet.

Quote
I can't get a common number in charges and multiply because I dont know what to multiply.
So far I got,

Ag(s) + 2CN-(aq) + O2 + 2H2O  >> Ag(CN)2- + 4OH-

What I did up there was just follow the rules as if it was a half reaction altogther. But in the half reactions, before we can combine them, we get a LCM of electrons. So do I just add 3 electrons to the left side of that equation? But then is that right? To have an equation with electrons in the equation?

IF it is complete reaction electrons cancel out, so you can' use them to balance charge. This equation is relatively easy because it contains only negative charges. If the charges make it difficult for you, try to balance this:

Ag(s) + 2CNCh(aq) + O2 + 2H2O  >> AgCh(CN)2 + 4OHCh

assuming Ch is an element.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: /// on November 21, 2006, 07:27:05 PM

IF it is complete reaction electrons cancel out, so you can' use them to balance charge. This equation is relatively easy because it contains only negative charges. If the charges make it difficult for you, try to balance this:

Ag(s) + 2CNCh(aq) + O2 + 2H2O  >> AgCh(CN)2 + 4OHCh

assuming Ch is an element.

I'm confused :/

I got that equation after balancing the original equation they gave me, then adding the appropriate number of H2O, H+ and OH-;;; was my problem balancing it in the first place?
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Dan on November 21, 2006, 07:32:08 PM
The equation Borek gave you with the fictional element Ch isnot balanced, as there is more Ch on the right than the left.
In the same way, your equation is not balanced as there is more negative charge on the right than the left.

Try to balance Borek's equation, and it should start to make sense.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: /// on November 21, 2006, 07:44:00 PM
I dont know:/ I'm still not sorta understanding how that relates to it.
Is that equation meant to be incomplete, or did we have to complete it on our own. It's weird we would get a question like that, because it's nothing like what we learned through the text books.

Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Borek on November 21, 2006, 07:53:11 PM
In redox quite often you will have to balance hydrogen/oxygen adding water/OH-/H+. That's the way they are usually treated. Besides, that's the way these reactions take place ;)
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: /// on November 21, 2006, 07:59:15 PM
In redox quite often you will have to balance hydrogen/oxygen adding water/OH-/H+. That's the way they are usually treated. Besides, that's the way these reactions take place ;)

I did that; thats how I balanced it in the beginning, but I still got different charges. :(
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Dan on November 21, 2006, 08:00:35 PM
The equation you were originally given was meant to be incomplete, and you are supposed to complete and balance it. What you are learning in your books will apply to this question, but this question is a bit more demanding. You are not far off from getting this I assure you.

So in Borek's equation, there is not enough Ch on the left, so it is not balanced. How can we sort that out?

At this point, don't worry yourself too much about how this relates to the charge problem in your equation - it will become clear soon. Focus on balancing Borek's equation, it is the key.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: /// on November 21, 2006, 08:26:07 PM
Ag(s) + CNCh(aq) + O2 + H2O  >> AgCh(CN)2 + OHCh


I can't balance it :( Too hard!!!
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Dan on November 21, 2006, 08:32:09 PM
Don't give up!

Start from Borek's unbalanced equation:

Ag(s) + 2CNCh(aq) + O2 + 2H2O  >> AgCh(CN)2 + 4OHCh

You need more Ch on the left side. Your goal is 4 Ch atoms on both sides. Once you have that, you will need to balance the Ag as well. Have another go.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: /// on November 21, 2006, 08:36:40 PM
Everytime I balance the CH, the other atoms lose their balance


Ag(s) + 5CNCh(aq) + O2 + 2H2O  >> AgCh(CN)2 + 4OHCh


Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Dan on November 21, 2006, 08:53:33 PM
Everytime I balance the CH, the other atoms lose their balance

Sorry, I made a mistake, your goal is 8 Ch on both sides. It is the Ag and CNCh that need looking at in Borek's equation.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: /// on November 21, 2006, 09:04:48 PM
4Ag(s) + 8CNCh(aq) + O2 + 2H2O  >> 4AgCh(CN)2 + 4OHCh
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Borek on November 22, 2006, 03:39:50 AM
4Ag(s) + 8CNCh(aq) + O2 + 2H2O  >> 4AgCh(CN)2 + 4OHCh

Correct! Now - put negative charge in the place of Ch:

4Ag + 8CN- + O2 + 2H2O -> 4Ag(CN)2- + 4OH-

Check whether the charge is balanced.

Got it?
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: AWK on November 22, 2006, 04:42:46 AM
My addition - balancing under basic conditions means that a base is on the left side of  an equation. When you add water (or nothing) to the left side then we have neutral conditions.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Borek on November 22, 2006, 05:13:17 AM
My addition - balancing under basic conditions means that a base is on the left side of  an equation.

So how do you balance half reaction ClO- -> Cl- in basic conditions? Looks like you need OH- on the right, not on the left.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: AWK on November 22, 2006, 05:45:19 AM


So how do you balance ClO- -> Cl- in basic conditions? Looks like you need OH- on the right, not on the left.

Sorry, but I do not understand your qustion. Redox reactions need reductors and oxidizers.The only reactions I know when CLO- fulfills both reagents is eg
3ClO- = ClO3- + 2Cl-
which proceeds in neutral conditions.
Oxidations by hypochlorite  proceed usually in acidic conditions, ie HClO ia the oxidizing agent.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Borek on November 22, 2006, 06:40:42 AM
I suppose you have started answering my post and you have not noticed I have modfied it in the meantime (even if you answered about 12 minutes after I have modified the post). I was referring to the half reaction. Sorry for not being clear.

Oxidations by hypochlorite  proceed usually in acidic conditions, ie HClO ia the oxidizing agent.

I don't think so. Bleach, which is nothing else but hypochlorite solution, is basic (with pH in the range of 12/13) yet is is a strong oxidizer - and to use it you don't have to add any acid. With pKa 7.5 at so high pH almost all hypochlorite is in dissociated form.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: AWK on November 22, 2006, 08:24:43 AM


I don't think so. Bleach, which is nothing else but hypochlorite solution, is basic (with pH in the range of 12/13) yet is is a strong oxidizer - and to use it you don't have to add any acid. With pKa 7.5 at so high pH almost all hypochlorite is in dissociated form.

Well, though bleach is basic in solution, for classification conditions we use stoichiometry, not actual pH of the reactants, hence oxidation by bleach itself proceeds in neutral conditions. When we add acid, or base to the reaction, conditions are acidic or basic, respectively.
Sometimes reaction is hold up by an  acid or a base formed during reaction, but even then we do not take into account the base or the acid added, since they are not at the left side of chemical equation.
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: Borek on November 22, 2006, 08:35:29 AM
Oxidations by hypochlorite  proceed usually in acidic conditions

oxidation by bleach itself proceeds in neutral conditions

You've lost me here  :-\
Title: Re: Balancing Using Half RXN method under basic conditions
Post by: AWK on November 22, 2006, 08:49:23 AM
Oxidations by hypochlorite  proceed usually in acidic conditions

oxidation by bleach itself proceeds in neutral conditions

You've lost me here  :-\
There is no contradiction in these two statements, oxidation by hypochlorite can be done in neutral or acidic conditions depending on the reaction. I never used a word: exclusively