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REDOX Half Reaction and Net Ionic

  • Thread starter djh101
  • Start date
  • #1
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Just need some clarification as to whether or not I'm doing this right. 1 drop HCl and several drops KMnO4 added to 2mL NaHSO3.

HSO3- → HSO4- + 2e-
MnO42- + 4e- → MnO22-
2HSO3-(aq) + MnO42-(aq) → 2HSO4-(aq) + MnO22-(aq)

A few questions:
-Since NaHSO3 is not an acid, the hydrogen atom would not be removed in the half reaction, correct?
-Are H+ and H2O part of the half reaction or are they only included as a step to help in creating/balancing the equation(s)?
-Should I have done something with the HCl? There were 4 H+ ions on each side of the reaction, which would cancel out and result in no reaction with the HCl.
-If there is no reaction with the HCl, why did I have to add it to the solution?
 

Answers and Replies

  • #2
Borek
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HSO3- → HSO4- + 2e-
Not balanced.

MnO42- + 4e- → MnO22-
Where did you get MnO42- and MnO22- from?

Your questions have to wait till the reaction equation is balanced. As for now - half reactions must be correctly balanced in charge and atoms, so far they are not.
 
  • #3
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In the reaction it is stated that MnO4 is oxidized to MnO2.
Would MnO42- + 4H+ + 4e- → MnO22- + 2H2O be correct (and hydrogens and waters don't get cancelled until doing the net ionic)?
 
  • #4
Borek
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In the reaction it is stated that MnO4 is oxidized to MnO2.
Would MnO42- + 4H+ + 4e- → MnO22- + 2H2O be correct (and hydrogens and waters don't get cancelled until doing the net ionic)?
Check charges (-2 is incorrect in both cases), and it is not oxidation, but reduction.
 
  • #5
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Wow, so very many errors I found after starting to really understand what I'm doing. I fixed it, thanks for your help.

HSO3- + H2O → SO42- + 3H+ + 2e-
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
 
  • #6
Borek
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Half reactions are OK, now balance overall. We will see what is left of your original questions.
 
  • #7
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5HSO3-(aq) + 2MnO4-(aq) + H+(aq) → 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)
5NaHSO3(aq) + 2KMnO4(aq) + HCl(aq) → 5NaSO4(aq) + + 3H2O(l)

Not sure about the complete molecular equation, although it isn't actually required. Chlorine seems to be the only anion and there's not enough of that to bond with the potassium and manganese...
 
Last edited:
  • #8
Borek
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Still some typos, but you are almost there.

Molecular reaction is a difficult one, but doable:

5NaHSO3 + 2KMnO4 + HCl -> 2Na2SO4 + K2SO4 + 2MnSO4 + 3H2O + NaCl

I admit it looks artificial. Note that the main part (5HSO3- and 2MnO4-) is identical.
 
  • #9
160
4
Interesting. I probably would have never figured that out, and I'm glad I didn't have to (but this is only Chem 101 stuff). Thanks for your help. And I fixed the typo, I had a 2 instead of a -.
 

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