# Balancing acidic and basic equations

• reedy
In summary, the task is to add the missing particles (H_3O,OH,H_2O) and balance the charges on the left and right sides of the basic solution.
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

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

Last edited:
reedy said:
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...

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.

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...

Last edited by a moderator:
reedy said:
"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.

Last edited by a moderator:
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?

Looks good.

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