Balancing acidic and basic equations

[reedy]

0__+5_-6____+2_+2_-2
Zn + NO3 --> Zn + N2O

The oxidation states should be the above - it's the balancing that is difficult.

We're going from -1 (+5 - 6= -1) to 0 (+2 -2=0)

$$NO_3^- \rightarrow N_2O +e^-$$
Now, the charge is correct, but the number of nitrogen atoms isn't. How do I fix that?

$$2NO_3^- \rightarrow N_2O +2e^-$$

How does that look?

Let's put it together

$$Zn + 2NO_3^- \rightarrow Zn^{2+} + 2e^- + N_2O +2e^-$$

The electrons end up on one side. Shouldn't there be equally as much on both sides so they can be cancelled?

 

[chemisttree]

What is the oxidation # for nitrogen in N2O? (its not +2)

 

[reedy]

Isn't it +2 in total? I read that it's +1, but since there are two of them and since the entire molecule has a neutral charge, I assumed N2 was +2 and O was -2.

But maybe I shouldn't add them together.

 

[reedy]

Let's not add them together.

0__+5_-6____+2_+1_-2
Zn + NO3 --> Zn + N2O

And if this is true, we have a sum of -1 on both sides. Meaning that the electrons are balanced. But is this even possible -> NO3- has a negative charge while N2O is neutral.

 

[chemisttree]

Let's not add them together.

0__+5_-6____+2_+1_-2
Zn + NO3 --> Zn + N2O

And if this is true, we have a sum of -1 on both sides. Meaning that the electrons are balanced. But is this even possible -> NO3- has a negative charge while N2O is neutral.

So each nitrogen is going from a +5 to a +1. How would you write the half reaction for that reduction?

 

[reedy]

Umm.

$$NO_3^- +4e^- \rightarrow N_2O$$

$$2NO_3^- +8e^- \rightarrow N_2O$$

and the other one goes from

$$Zn \rightarrow Zn^{2+} + 2e^-$$

$$4Zn \rightarrow 4Zn^{2+} + 8e^-$$

with balanced number of electrons

$$4Zn + 2NO_3^- +8e^- \rightarrow 4Zn^{2+} + N_2O + 8e^-$$Edit: Which would give

$$4Zn + 2NO_3^- \rightarrow 4Zn^{2+} + N_2O$$

and finally

$$4Zn + 2NO_3^- + 10H^+ \rightarrow 4Zn^{2+} + N_2O + 5H_2O$$

 

[chemisttree]

Umm.

$$NO_3^- +4e^- \rightarrow N_2O$$

$$2NO_3^- +8e^- \rightarrow N_2O$$

and the other one goes from

$$Zn \rightarrow Zn^{2+} + 2e^-$$

$$4Zn \rightarrow 4Zn^{2+} + 8e^-$$

with balanced number of electrons

$$4Zn + 2NO_3^- +8e^- \rightarrow 4Zn^{2+} + N_2O + 8e^-$$


Edit: Which would give

$$4Zn + 2NO_3^- \rightarrow 4Zn^{2+} + N_2O$$

and finally

$$4Zn + 2NO_3^- + 10H^+ \rightarrow 4Zn^{2+} + N_2O + 5H_2O$$

I wish you could see the smile on my face...

 

[reedy]

I wish I could show you my appreciation.

Moving on to the next problem-

0___+6-2_-1_______x_y___+2-2
Au + NO3-+Cl- ---> AuCl4- + NO

the oxidation numbers of Au and Cl4 in AuCl4 which kind of makes me wonder: what is the proper way of finding the oxidation numbers? Until now, I've used an online reference. These are great and ready to use, but it still seems kind of useless (atleast in this case since I can't find the compound I need). I've also noticed that Wikipedia has posted the oxidation states for all elements.

Oxidation states −1, 1, 2, 3, 4, 5, 6, 7
(amphoteric oxide)

How can I use these numbers to my advantage?

I need x and y to move on to the next level.

 

[chemisttree]

For chlorides it is generally the case that the oxidation state is -1. There may be a case in which chlorine has a different oxidation state as in sodium hypochlorite, NaClO.

I would assume that chlorine has a -1 oxidation state and then look at the metal's oxidation state. I would then refer to a table of oxidation states that have been observed for the metal and if that fits, go with it...

 

[reedy]

"http://library.thinkquest.org/C004970/atoms/oxidation.htm"
So assuming that chlorine is -1, I'd guess that the Au is +3 in AuCl4- to give the negative charge.

"All compounds have a net oxidation state of zero. The oxidation state of all of the atoms add up to zero."
Now if this is true, that would make the golds oxidation number +4, right?

English not being my first language, maybe I'm mixing the terms a bit. What should I believe? What is true? What is correct in this case?

 

[chemisttree]

"http://library.thinkquest.org/C004970/atoms/oxidation.htm"
So assuming that chlorine is -1, I'd guess that the Au is +3 in AuCl4- to give the negative charge.

"All compounds have a net oxidation state of zero. The oxidation state of all of the atoms add up to zero."
Now if this is true, that would make the golds oxidation number +4, right?

English not being my first language, maybe I'm mixing the terms a bit. What should I believe? What is true? What is correct in this case?

Both are correct. The second rule assumes a neutral complex whereas the first rule assumes an ion complex.

 

[reedy]

This being an ion, I say Au in the compound is +3.

$$_{0} \ \ \ \ \ \ \ _{+6} _{-2}\ \ \ \ \ _{-1}\ \ \ \ \ \ \ \ _{+3}\ _{-1}\ \ \ \ \ \ _{+2}\ _{-2}$$
$$Au + NO_3^- + Cl^- \rightarrow AuCl_4^- + NO$$

I'm guessing this one would need three half reactions.

$$Au \rightarrow AuCl_4^- + 3e^-$$

$$NO_3^- + 4e^- \rightarrow NO$$

$$4Cl^- \rightarrow AuCl_4^-$$
--------
Balancing the electrons:

$$4Au \rightarrow 4AuCl_4^- + 12e^-$$

$$3NO_3^- + 12e^- \rightarrow 3NO$$

$$16Cl^- \rightarrow 4AuCl_4^-$$
--------
Putting it all together:

$$4Au + 3NO_3^- + 16Cl^- \rightarrow 4AuCl_4^- + 3NO$$
--------
Finishing it off:

$$4Au + 3NO_3^- + 16Cl^- + 12H^+ \rightarrow 4AuCl_4^- + 3NO + 6H_2O$$

What's the verdict?

 

[chemisttree]

Looks good.