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November 01, 2025, 01:12:56 pm

Author Topic: need help with redox problems  (Read 1619 times)  Share 

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methodsboy

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need help with redox problems
« on: January 07, 2009, 04:22:55 pm »
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hi. i need help with two redox questions:
 
1. a solution containing sulfite ions (SO32–) reacts with an acidified solution of permanganate ions (MnO4–) to produce a colourless solution contain¬ing sulfate ions and manganese(II) ions.
S(O3)^(2–)(aq) + H2O(l) ---> S(O4)^(2–)(aq) + 2H+(aq) + 2e–
MnO4–(aq) + 8H+(aq) + 5e– ----> Mn^(2+)(aq) + 4H2O(l)

^I have done the ionic equations but i don't know how to do the overall equation

2. Potassium permanganate reacts with hydrogen peroxide:
2MnO4–(aq) + 5H2O2(aq) + 6H+(aq) ---> 2Mn^(2+)(aq) + 8H2O(l) + 5O2(g)
25.0 mL of 0.02 M KMnO4 solution is reduced by 20.0 mL of H2O2 solution. What is the concentration of the hydrogen peroxide solution?
I have no idea  :(
Thanks in advance
« Last Edit: January 07, 2009, 04:27:55 pm by methodsboy »

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Re: need help with redox problems
« Reply #1 on: January 07, 2009, 04:37:26 pm »
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1.





You have to make the electrons on both sides cancel out. So multiply equation 1 by 5 and multiply equation 2 by 2:





And add them together:





I think that's how it goes...

2.

(ionic equation, K+ is spectator, [thanks edmund372])





Or , to 1 significant figure.
« Last Edit: January 07, 2009, 04:55:06 pm by /0 »

methodsboy

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Re: need help with redox problems
« Reply #2 on: January 07, 2009, 04:38:54 pm »
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^ i got everything up to the second last line. Please explain what you did for the very last line.

shinny

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Re: need help with redox problems
« Reply #3 on: January 07, 2009, 04:43:30 pm »
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He just cancelled things which were common on both sides. It's just like maths. If you have +10x on one side, and +16x on the other, you simplify it down to +6x. Same goes here; you're just cancelling identical substances.
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methodsboy

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Re: need help with redox problems
« Reply #4 on: January 07, 2009, 04:45:25 pm »
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i dont understand this part for question 2 :
KMnO4 isnt even in the equation  or just potassium itself???
« Last Edit: January 07, 2009, 04:48:06 pm by methodsboy »

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Re: need help with redox problems
« Reply #5 on: January 07, 2009, 04:46:40 pm »
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Sorry maybe I'm misunderstanding something here...

In the given equation there is no Potassium, so why does it say Potassium permanganate reacts?

methodsboy

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Re: need help with redox problems
« Reply #6 on: January 07, 2009, 04:48:28 pm »
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no idea ???

methodsboy

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Re: need help with redox problems
« Reply #7 on: January 07, 2009, 04:49:10 pm »
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and why did u use 5/2?

Edmund

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Re: need help with redox problems
« Reply #8 on: January 07, 2009, 04:50:51 pm »
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K+ is a spectator ion and does not react, thus it can be ignored...

and 5/2 is the mole ratio...
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Re: need help with redox problems
« Reply #9 on: January 07, 2009, 04:53:54 pm »
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Oh... oops, it's an ionic equation isn't it?

@ methodsboy: I used 5/2 because that's the mole ratio of to

i.e.

Edmund

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Re: need help with redox problems
« Reply #10 on: January 07, 2009, 04:55:56 pm »
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We know the moles of permanganate, so from this, we can find the moles of peroxide using the mole ratio, which is 5/2

then Working as above.


EDIT: hmmm how do you type fraction?? would have been much easier to explain  ;D
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methodsboy

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Re: need help with redox problems
« Reply #11 on: January 07, 2009, 04:57:21 pm »
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Thanks guys, i fully get it now :)