Balancing acidic and basic equations

[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.

 

[Vagrant]

It's OH- and H3O+

 

[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?

 

[chemisttree]

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

 

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

 

[chemisttree]

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

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).

 

[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?

 

[chemisttree]

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?

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.

 

[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?

 

[chemisttree]

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

No change in oxidation state, right?

Wrong. Look at the sulfurs again...

 

[reedy]

Look at the sulfurs again...

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?

 

[chemisttree]

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

 

[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?

 

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

 

[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?

 

[chemisttree]

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

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!

 

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

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

 

[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.

 

[chemisttree]

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

 

[reedy]

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

so the left side needs oxygen.

 

[chemisttree]

It seems that you have a choice of reagents to accomplish this. Which do you choose?

 

[reedy]

I would add some $$H_2O$$ among the reactants. That would balance the oxygen, but require some hydrogen among the products.

I read, in the assignment, that this reaction was basic. That means we need $$OH^-$$ as a product. But we would still be one hydrogen short.

Or maybe adding the water was wrong in the first place. Tell me - how do I know what to add? And what if I didn't know that it was a basic solution - how would I have found out?

 

[chemisttree]

I would add some $$H_2O$$ among the reactants. That would balance the oxygen, but require some hydrogen among the products.

Show me what you mean here.

I read, in the assignment, that this reaction was basic. That means we need $$OH^-$$ as a product. But we would still be one hydrogen short.

How would you show this hydrogen in your example?


Hint: Think of water as 2H+ and O-2 and think of hydroxide as H+ and O-2.

 

[reedy]

$$2MnO_4^- + SO_3^{-2} + xH_2O \rightarrow 2MnO_4^{-2} + SO_4^{-2} + xY$$

The above gives a balanced number of oxygen atoms. But the added hydrogen has to be compensated.

So by just adding water and hydroxide molecules, this would be impossible to balance.

 

[chemisttree]

$$2MnO_4^- + SO_3^{-2} + xH_2O \rightarrow 2MnO_4^{-2} + SO_4^{-2} + xY$$

The above gives a balanced number of oxygen atoms. But the added hydrogen has to be compensated.

So by just adding water and hydroxide molecules, this would be impossible to balance.

Try this...

$$2MnO_4^- + SO_3^{-2} + H_2O \rightarrow 2MnO_4^{-2} + SO_4^{-2} + 2H^+$$

or this...

$$2MnO_4^- + SO_3^{-2} + 2OH^- \rightarrow 2MnO_4^{-2} + SO_4^{-2} + H_2O$$

 

[reedy]

$$2MnO_4^- + SO_3^{-2} + H_2O \rightarrow 2MnO_4^{-2} + SO_4^{-2} + 2H^+$$

This would work, but wouldn't the hydrogen ions make the solution acidic?

$$2MnO_4^- + SO_3^{-2} + 2OH^- \rightarrow 2MnO_4^{-2} + SO_4^{-2} + H_2O$$

That would actually work great, but I assumed that the solution goes from neutral to basic, and not the other way around.

Edit: One other thing - why aren't we taking the charges into consideration? When balancing before, charges were important, but now we can add and remove however we please?

 

[chemisttree]

This would work, but wouldn't the hydrogen ions make the solution acidic?

Yes.

...why aren't we taking the charges into consideration? When balancing before, charges were important, but now we can add and remove however we please?

If you prefer:

$$2KMnO_4 + H_2SO_3 + H_2O \rightarrow K_2MnO_4 + H_2MnO_4 + H_2SO_4$$

and

$$2KMnO_4 + K_2SO_3 + 2KOH \rightarrow 2K_2MnO_4 + K_2SO_4 + H_2O$$

 

[reedy]

Oh now you're just showing off.

But hey, chemisttree, thank you for all your help. I couldn't have done it without you. I'm going to give the other two a try- don't go anywhere.

 

[reedy]

Well, it didn't take long until I came back. Let's see.

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

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

But - the nitrous oxide / nitrate seems a bit tricky.

$$NO_3^- \rightarrow N_2O$$

I'm stuck.

 

[chemisttree]

Well, it didn't take long until I came back. Let's see.

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

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

But - the nitrous oxide / nitrate seems a bit tricky.

$$NO_3^- \rightarrow N_2O$$

I'm stuck.

Follow the method... start by indicating the oxidation state of the atom in question and write the balanced half reaction. (balanced with respect to the atom in question)

 

[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?