Question

Write the net ionic equation for the reaction under acidic conditions (H30+) that releases O2 from sulfate and forms hydrogen sulfide gas (H2S). Use water as the reactant (!) in the half-reaction that describes the formation of oxygen.

I’ve tried so many things and cannot get this question right. Please help!

Answer 1

\[H_3O _{(aq)}^{+} + SO _{4(aq)}^{-} \rightarrow H_2S _{(g)} + O _{2(g)} \]

Is this the molecular equation?

Answer 2

That wasn't it either. As feedback from a wrong answer, it said, "Oxygen and hydrogen sulfide are the principle products of this reaction. Remember, for aqueous redox reactions, we can freely add water and H+ to balance the half-reactions. For the two half-reactions, the reduction is of SO42- to H2S. The oxidation is that of oxygen in water, from H2O to O2."

Answer 3

I understand that hydronium ions are released in the acidic reactions, but is it necessary to have in this net ionic equation?

Answer 4

I made a mistake. The H3O^+ is the solution.

First let's get the molecular equation which I think would be

\[SO _{4(aq)}^{2-} \rightarrow H_2S _{(g)} + O _{2(g)}\]

Answer 5

Yeah that looks good to me, except the oxygen would have to be balanced with a 2 and hydrogen somewhere in the reactants (not sure where)

Answer 6

half reaction # 1:
\[H_2O \rightarrow O_2\]
half reaction #2:
\[SO _{4}^{2-} \rightarrow H_2S\]

Now we have to balance both reactions. Did you try this?

Answer 7

Yeah. I came up with \[2S O_{4}^{2-}+2H _{2}O \rightarrow5O _{2}+2H _{2}S\]

Answer 8

Let's first balance half reaction # 1

Answer 9

H2O => O2

What do we need to do to balance the right side?

Answer 10

I'd balance it with 2H2O --> O2 + 4H+

Answer 11

Good. How would you balance the charges?

Answer 12

\[O _{2}^{2-}\] ?

Answer 13

You can't add 2- to the O2. O2 is oxygen in its neutral state, so it has no charge. You need to add 2 electrons

Answer 14

Typo. I meant 4 electrons

Answer 15

Isn't that the 4 H+?

Answer 16

The 4H+ has a total of 4 positive charges. The left side has H2O which has a zero charge. For the right side to have a zero charge the 4+ (from the H+) needs to be balanced by 4 electrons. (Electrons have a - charge.)

Answer 17

Yes, I understand that. So we literally just add 4 e- to the right side?

Answer 18

Correct. Can you write it out?

Answer 19

2H2O --> O2 + 4H+ + 4 e-

Answer 20

Good. Now let's balance half reaction #2
\[SO _{4}^{2-} \rightarrow H_2S\]
What needs to be added to the right side?

Answer 21

\[2H ^{+}+SO _{4}^{2-} + 2e ^{-}--> H _{2}S+2O _{2}\]

Answer 22

The O2 is not added to the right because that came from H2O in the first half reaction. To balance the right side H2O is added.

SO4^2- => H2S + 4H2O

Now what needs to be added to the left side?

Answer 23

10 H+ and 10 e-, I think

Answer 24

10 H+ but not 10 e-

Write out the second half reaction without the e- for now.

Answer 25

10 H+ + SO4^2- --> H2S + 4H2O

Answer 26

So what is the total charge on the left and the charge on the right?

Answer 27

Left will be +8, right will be I believe 0

Answer 28

So we add 8 e- to the right?

Answer 29

That's right.

Answer 30

We want to be able to cancel out electrons when we re-write the full reaction. So what can we do so that both reactions should have the same number of e- ?

Answer 31

We could multiply the coefficients of the first half-reaction by 2 to get both reactions to have 8 e-

Answer 32

Correct. So do that and re-write the entire equation (combining both half reactions).

Answer 33

\[10H ^{+}+SO _{4}^{2-} + 4H _{2}O --> H _{2}S+2O _{2}+4H _{2}O+8H ^{+}\]

Answer 34

So, I assume the 4 water molecules on each side would cancel, as would the 8 H+ from the right, leaving 2 H+ on the left. Yes?

Answer 35

Yes. So what's your final answer?

Answer 36

\[2H ^{+}(aq)+SO _{4}^{2-}(aq) --> H _{2}S(g)+2O _{2}(g)\]

Answer 37

Correct. At this point it's a good idea to double check that everything is actually balanced.

Answer 38

Looks balanced to me. Unfortunately, according to this system that's still incorrect...

Answer 39

Can you email your professor, and let me know what the correct answer is?

Answer 40

Haha sure thing. This online thing can be so fickle, I really hate it sometimes. I'll ask a friend tomorrow (or the professor) and get back to you. Thanks for working with me on it!

Answer 41

I'm wondering if the answer should be

2H3O+(aq) + SO4^2−(aq)−−>H2S(g)+2O2(g) + 2H2O

Answer 42

Yes, that's it!! What did you change about the approach to get that?

Answer 43

Originally we came up with

2H+(aq) + SO4(aq)^2- => H2S(g) + 2O2(g)

Technically H+ does not exist. What is H2O + H+ ?