# Balancing acidic and basic equations

1. Jun 18, 2007

### reedy

The task is to add the missing particles ($$H_3O,OH,H_2O$$) and balance .
(1) is a basic solution while (2) and (3) are acidic.

1. $$MnO_4 + SO_3 \rightarrow MnO_4 + SO_4$$
2. $$Zn + NO_3 \rightarrow Zn + N_2O$$
3. $$Au + NO_3 + Cl \rightarrow AuCl_4 + NO$$
I think that they should all have $$H_2O$$ on the left side but only the first one has $$OH$$ on the right side, while the other two, that are acidic have $$H_3O$$ on the right side? And after that, it's just pure balancing.

The biggest concern is pretty much to place the ions and the water molecule (if any) on the correct spot.

Last edited: Jun 18, 2007
2. Jun 18, 2007

### Vagrant

It's OH- and H3O+

3. Jun 18, 2007

### reedy

Oh, I know the charge. I just don't know where to put them and how many of them to use.

All equations should have atleast one H2O as a reactant, right? The basic should have an OH in the product and the acidic should have H2O. Correct me if I'm wrong.

Where do I go from here?

4. Jun 18, 2007

### chemisttree

Did you copy the reactions properly? You should show charges and be careful with subscripts (MnO4 vs. MnO2).

5. Jun 18, 2007

### reedy

Oh, I didn't know that they were relevant. Why are the charges important? Are they always important when balancing?

1. $$MnO_4^- + SO_3^{2-} \rightarrow MnO_4^{2-} + SO_4^{2-}$$

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

3. $$Au + NO_3^- + Cl^- \rightarrow AuCl_4^- + NO$$

6. Jun 18, 2007

### chemisttree

Yes the charges are important. For example, as you originally wrote out the second equation,

$$Zn + NO_3^- \rightarrow Zn + N_2O$$,

it reads "Zinc metal is treated with nitrate anion and produces zinc metal and $$N_2O$$."
In this case, zinc metal is unchanged in the reaction

The corrected version,

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

reads "Zinc metal is oxidized by nitrate anion to produce Zn(+2) ion and $$N_2O$$."

The first step is to write out the half reactions. You should already know how to do this. For example, for sulfide being oxidized to sulfite:

$$S^{-2} \rightarrow SO_3^{-2} + 6e^-$$

Try it for each of the problems. You will need to determine the oxidation state of each of the metals (and sulfur and nitrogen) for the reactions. Assume that oxygen in an oxide is always in an oxidation state of -2. When the oxidation state changes, show the electrons either being added to or being produced by the reaction.

That should get you started if you need more help, show your work and ask your question(s).

Last edited: Jun 18, 2007
7. Jun 18, 2007

### reedy

I guess I need to do some extra reading.

But how will this help me in balancing the equation and correctly placing the different types of particles mentioned in the first post? Why is determining the oxidation state important to finish this task?

8. Jun 19, 2007

### chemisttree

Lets look at an example.

$$S^{-2} + NO_{3}^{-1} -> NO_{2} + S_{8}$$

For this example we see that, for reactants, the sulfur is in an oxidation state of -2 and the nitrogen is in an oxidation state of +5. For the products, the oxidation state of the sulfur is 0 and that of nitrogen is +4. In this reaction sulfur has gone from -2 to zero; a two electron oxidation. The half reaction is:

$$S^{-2} \rightarrow S_8 + 2e^-$$
Balancing the sulfur on left and right yields:
$$8S^{-2} \rightarrow S_8 + 16e^-$$

The nitrogen (from nitrate) has gone from an oxidation state of +5 to +4; a one electron reduction. It's half reaction is :

$$NO_3^- + e^- \rightarrow NO_2$$
We will worry about the oxygen later.

We need to add these two half reactions together to balance the electrons. This requires that we multiply the nitrate half reaction by 16. It is rewritten as:

$$16NO_3^- + 16e^- \rightarrow 16NO_2$$
Now we add the two half reactions together to get:
$$8S^{-2} + 16NO_3^- + 16e^- \rightarrow S_8 + 16NO_2 + 16e^-$$

Cancelling the $$16e^-$$ from both sides, we see that the resulting equation is not balanced with respect to oxygen. 16 oxygens are missing from the products side. These oxygens will be in a -2 oxidation state. Adding 32 acidic protons ($$H^+$$) to the left side and 16 water molecules to the right side completes the balancing.

$$8S^{-2} + 16NO_3^- + 32H^+ \rightarrow S_8 + 16NO_2 + 16H_2O$$

So if this were one of the questions you were asked, the answer would be $$H_3O^+$$. Do this analysis for your examples.

9. Jul 2, 2007

### reedy

Well chemistree, I've been doing a lot of reading. Hear me out.

I took the following equation

$$MnO_4^- + SO_3^{-2} \rightarrow MnO_4^{-2} + SO_4^{-2}$$

for practice.

I start off with the first half reaction:

$$MnO_4^- + e^- \rightarrow MnO_4^{-2}$$

Second:

$$SO_3^{-2} \rightarrow SO_4^{-2}$$

No change in oxidation state, right?

This gives

$$MnO_4^- + SO_3^{2-} + e^- \rightarrow MnO_4^{2-} + SO_4^{2-}$$

Here I'm stuck. You said the electrons should be balanced first, but how do I do that when there aren't any among the products? Where do I go from here?

10. Jul 2, 2007

### chemisttree

Wrong. Look at the sulfurs again...

11. Jul 3, 2007

### reedy

I'm not seeing it. The charge is unchanged - how can there be an exchange of electrons?
----------------------------
I've tried a different approach - have a look:

$$_{+7} \ \ _{-8}\ \ \ \ _{+4}\ _{-6}\ \ \ \ \ _{+6}\ \ _{-8}\ \ \ \ _{+6}\ _{-8}$$
$$MnO_4 + SO_3 \rightarrow MnO_4 + SO_4$$

This should be correct.

Writing out new half reactions according to the above:

$$MnO_4^- + e^- \rightarrow MnO_4^{-2}$$

This is still true - manganese is, as as a reactant, in an oxidation state of +7, while as a product in +6.

Number two:

$$SO_3^{-2} \rightarrow SO_4^{-2} + 2e^-$$

Yes? Now what? Put them together?

That gives:

$$MnO_4^- + SO_3^{-2} + e^- \rightarrow MnO_4^{-2} + SO_4^{-2} + 2e^-$$

But the electrons aren't balanced.

$$2MnO_4^- + 2SO_3^{-2} + 2e^- \rightarrow MnO_4^{-2} + SO_4^{-2} + 2e^-$$

Electrons are balanced - cancelling electrons.

$$2MnO_4^- + 2SO_3^{-2} \rightarrow MnO_4^{-2} + SO_4^{-2}$$

Or am I moving ahead of myself?

12. Jul 3, 2007

### chemisttree

Multiply the manganese oxide half reaction by two before you add it to the sufite/sulfate half reaction. Try it from there...

13. Jul 3, 2007

### reedy

Aha, alright.

$$MnO_4^- + e^- \rightarrow MnO_4^{-2}$$
$$2MnO_4^- + 2e^- \rightarrow 2MnO_4^{-2}$$

gives

$$2MnO_4^- + 2SO_3^{-2} + 2e^- \rightarrow 2MnO_4^{-2} + SO_4^{-2} + 2e^-$$

$$2MnO_4^- + 2SO_3^{-2} \rightarrow 2MnO_4^{-2} + SO_4^{-2}$$

Still nothing. What do I do about the sulfur and the oxygen?

14. Jul 3, 2007

### chaoseverlasting

The sulfur is balanced, you balance the oxygen by adding H30+ or OH- depending on the medium to one side of the equation. (Your equation should be $$2MnO_4^- + 2SO_3^{-2} \rightarrow 2MnO_4^{-2} + 2SO_4^{-2}$$

15. Jul 3, 2007

### reedy

But if $$2SO_4^{-2}$$ is correct, where did I go wrong? Wasn't multiplying the manganese reaction, to get the correct number of electrons, enough?

16. Jul 3, 2007

### chemisttree

This is what the sulfite/sulfate half reaction should look like. When you add the manganese half reaction to it why did you put a '2' in front of the sulfite?

Try it again... you are sooo close!

17. Jul 4, 2007

### chaoseverlasting

My bad. I just copied the tex directly. Almost there.

18. Jul 4, 2007

### reedy

Aight, from the top.

$$MnO_4^- + e^- \rightarrow MnO_4^{-2}$$

$$SO_3^{-2} \rightarrow SO_4^{-2} + 2e^-$$

Balacing on the manganese side:
$$2MnO_4^- + 2e^- \rightarrow 2MnO_4^{-2}$$

I put both balanced half reactions together.
$$2MnO_4^- + SO_3^{-2} + 2e^- \rightarrow 2MnO_4^{-2} + SO_4^{-2} + 2e^-$$

Cancelling electrons.
$$2MnO_4^- + SO_3^{-2} \rightarrow 2MnO_4^{-2} + SO_4^{-2}$$

Beautiful. But the oxygens need some fine-tuning.

19. Jul 4, 2007

### chemisttree

To which side of the equation do you need to add oxygen?

20. Jul 4, 2007

### reedy

Reactants have 8 + 3 = 11 while
products have 8 + 4 = 12

so the left side needs oxygen.

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