Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: CKabes on December 16, 2011, 10:34:17 AM
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A mixture of Potassium Permanganate (KMnO4) and Potassium Iodide (KI) are mixed with solid Sodium Hydrogen Sulfate (NaHSO4) and a few ml's of water are added. The Solution is then heated and a purple vapor (Iodine) travels thrugh the curved glass pipe affixed at the top of the testube, into a solution of Alcohol. Then Alcohol becomes brown, and tests positive for Iodine.
What's going on here?
KMnO4 + KI + NaHSO4 --> ???
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So, you know one of your products is iodine...
You are in an acidic solution (sodium hydrogen sulfate [a.k.a. sodium bisulfate] is a fairly strong acid). Which oxidation state is Mn(VII) reduced to in acidic solution?
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I believe permanganates in acid are reduced to Mn2+
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Good, you also observe iodine vapors - that means you already should know both half reactions.
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Ok, so
KMnO4 + 2KI + NaHSO4 + H2O --> I2 + H2O (because of the permangante in acid) + Mn2+ + NaHSO4 (this doesn't play a role in the reaction except to change the ph maybe?) + ??
What happens to the potassium?
And if the acid gives up it's hydrogen, what are we left with?
-I also think maybe this is two reactions or more?-
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Do you know what a net ionic reaction is?
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NaHSO4 (this doesn't play a role in the reaction except to change the ph maybe?) + ??
It does play a role - you have identified that Mno4- is reduced to Mn2+, but what happened to the oxygen?
Hint: What's the half-reaction for the reduction of permanganate in acid (http://en.wikipedia.org/wiki/Permanganate)?
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8 H+ + MnO4- + 5 eā ā Mn2+ + 4 H2O
We're getting there! :) but now what to do with the potassium? Could it form K2SO4? and then we're only left with a sodium compound, if that were the case.
And what is a net ionic reaction?
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Ok, I'm sure this is wrong, but so far, this is what I would hypothosise:
2KMnO4 + 14KI + 16NaHSO4 --> 2Mn2+ + 8H2O + 8K2SO4 + 8Na2SO4 + 7 I2
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Ok, I'm sure this is wrong, but so far, this is what I would hypothosise:
2KMnO4 + 14KI + 16NaHSO4 --> 2Mn2+ + 8H2O + 8K2SO4 + 8Na2SO4 + 7 I2
You are right it is wrong - it is not balanced. Not only the mass is conserved (which means atoms must be balanced), also the charge is conserved (so the charge on the left must be identical to the charge on the right). Apparently this is not the case - you have +2 on the right, but 0 on the left. In the net ionic reaction Mn2+ would be a perfectly valid product, but as you are listing uncharged compounds as everything else, you need to find a counterion to list Mn2+ as an electrically neutral salt.
Net ionic reaction is the one that doesn't list all spectator species (ones that don't change during reaction). In the net ionic reaction you don't care about exact compounds reacting, but about ions reacting. For example
NaOH + HCl -> NaCl + H2O
is an overall reaction, while
H+ + OH- -> H2O
is a net ionic reaction. Note that if you mix any strong acid with any strong base counterions (like Na+ and Cl- in this case) don't change - before and after reaction they are freely floating in the solution. Net ionic reaction shows what is really happening, ignoring everything else.
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hmmm,
The first counter Ion that comes to mind would be Oxygen,
Manganese(II) could from Manganese Oxide right? MnO? But there would need to be and Oxygen ion present.
What about Manganese(II) Sulfate? There is already a sulfate ion present, would that work?
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Oxygen as a counterion in water solution? No way.
Manganese(II) sulfate is a reasonable product.
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Could somebody please provide me with the final equation?
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Could somebody please provide me with the final equation?
No. But you were given so many hints it is the highest time yo try to solve it. You know iodine is being produced, you know you should expect MnSO4 between products - that's already half of the work.