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

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Discussion Overview

The discussion revolves around the ionic equation for the reaction between zinc and sulfuric acid, specifically addressing the inclusion of water in the products. Participants explore the chemical processes involved, the role of sulfate ions, and the nature of the reactants and products in this reaction.

Discussion Character

  • Homework-related
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant presents an initial attempt at writing the ionic equation, questioning the appearance of water in the final products.
  • Another participant suggests that the oxygen in water might come from the sulfate ion, raising concerns about the accuracy of this interpretation.
  • A different participant argues that the sulfate ion should appear on both sides of the equation, indicating that it does not participate in the reaction.
  • One participant proposes that sulfur dioxide might be a product, suggesting a misunderstanding of the reaction's stoichiometry.
  • Another participant questions whether the original question might have involved zinc oxide instead of zinc, implying a potential error in the textbook.
  • There is a discussion about the implications of sulfate ions potentially converting to sulfite ions and donating oxygen, which is contested by others.
  • One participant provides half-reactions to clarify the process, emphasizing that sulfate does not participate in the reaction and discussing the conditions under which sulfate could be reduced.

Areas of Agreement / Disagreement

Participants express differing views on the role of sulfate ions and the products of the reaction, with no consensus reached on the correct interpretation of the ionic equation or the involvement of water.

Contextual Notes

Participants highlight potential misunderstandings regarding the reactants and products, as well as the conditions required for certain reactions to occur. There is also uncertainty about the definitions and roles of various ions in the reaction.

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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?
 
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From the O4? Why is this a problem?
 
DaveC426913 said:
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.
 
Last edited:
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.
 
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!
 
What about (SO4)2- turning to (SO3)2- which will donate the oxygen atom.

The textbook worded zinc and not zincoxide in the reactants.
 
Last edited:
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)
 

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