Why Does the Ionic Equation for Zinc and Sulfuric Acid Reaction Include Water?

[pivoxa15]

Homework Statement

Consider the reaction between Sulfiric acid and zinc:
H2SO4 + Zn ->
What would be the ionic equation for this reaction?

Homework Equations

The Attempt at a Solution

In solution it would be
H2+ + (SO4)2- + Zn -> Zn2+ + 2e- + (SO4)2- + H2+ -> Zn2+ + H2 (in ionic form)

but instead the answers gave Zn2+ + H2O
where did the O come from?

 

[DaveC426913]

From the O4? Why is this a problem?

 

[pivoxa15]

From the O4? Why is this a problem?

If that is the case than shouldn't they show (SO4)2- on the left and (SO3)2- on the right?

ions can only be canceled if they do not participate in the reaction.

 

[DaveC426913]

But your answer to the equation seems to have LOST the SO4.

Zn + (SO4) + H2 -> Zn + H2

Wouldn't one of the products be SO2? leaving the rest of the O to combine with the H2?

I guess I didn't study this. Nevermind me.

 

[chemisttree]

Are you sure the question wasn't:

H2SO4 + ZnO --> Zn+2 + (SO4)-2 + H2O?

If it isn't, you have found an error in your textbook!

As you wrote it, Zn + H2SO4 --> H2(g) + ZnSO4
is the correct answer. You might write the zinc sulfate as Zn+2 and (SO4)-2, but zinc will not react with sulfuric acid to produce water!

Remember that bases react with acids to produce water (bases like ZnO) not zero valent metals!

 

[pivoxa15]

What about (SO4)2- turning to (SO3)2- which will donate the oxygen atom.

The textbook worded zinc and not zincoxide in the reactants.

 

[chemisttree]

If that were true, zinc sulfate solutions would either spontaneously produce sulfurous acid and zinc oxide (H2SO3 and Zn+2) or excess zinc would react with sulfate solutions to produce sulfurous acid.

Neither of these two happens.


You can write all of the half reactions and see for yourself:

Zn --> Zn+2 + 2e-
2H+ + 2e- --> H2(g)

(SO4)-2 doesn't participate in this reaction so,

Zn + 2H+ + (SO4)-2 --> Zn+2 + H2(g) + (SO4)-2

If you want to reduce the products further (SO4 to SO3 for example) you would have to apply a reducing potential to the products. Applying a reducing potential to the products would reduce Zn+2 to Zn metal first since it is the most easily reduced species in solution.

ZnSO4 7H20 + 2e- --> Zn + (SO4)-2 + 7H20 (-0.7993V)