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magicalatom

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balancing redox reactions?
« on: January 24, 2016, 09:04:05 PM »
Hi I need help with this problem below:

The question asks in acidic conditions, the sum of the coefficients in the products of the balanced reaction is:

MnO4-+C3H7OH---->Mn+2+C2H5COOH

I think this is a redox reaction balancing problem.

So far

I got 5e-+8H++MnO4-   ------> Mn+2  +  4H2O

I can't seem to figure out how to do this with C3H7OH and then multiply each equation to help cancel out terms.

The answer is D. 20 but I don't know how to get this answer.

« Last Edit: January 24, 2016, 11:00:32 PM by magicalatom »

AWK

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Re: balancing redox reactions?
« Reply #1 on: January 24, 2016, 11:38:00 PM »
C3H8O + H2O = C3H6O2 +.
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magicalatom

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Re: balancing redox reactions?
« Reply #2 on: January 24, 2016, 11:40:48 PM »
What is this and how do you arrive at 20 as the answer?  Did you even read the problem?

C3H8O + H2O = C3H6O2 +.

AWK

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Re: balancing redox reactions?
« Reply #3 on: January 24, 2016, 11:47:46 PM »
Just finish this reaction and add properly two half reactions!
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magicalatom

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Re: balancing redox reactions?
« Reply #4 on: January 24, 2016, 11:53:47 PM »
That is where I am having trouble.  I got 6H+  +  H2O  + 2C3H7OH --->  3C2H5COOH +  6e-

then I tried multiplying first equation by 6 and this second one by 5 so I could sum and cancel 30e-, but still can't get 20 on products side as coefficients?

Just finish this reaction and add proprly two half reactions!

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Re: balancing redox reactions?
« Reply #5 on: January 25, 2016, 12:06:11 AM »
H+ should be on the right side of equation!
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magicalatom

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Re: balancing redox reactions?
« Reply #6 on: January 25, 2016, 12:12:48 AM »
Right side has 16 H's so the left side needed 6 more.  Why do you say 6H's should be on the right side?

I looked up the rules on this.  First you deal with metals then with H and O at the end.

I just seem to not see if propanol is able to split up ionically.  I know it has a OH and H can come off.

H+ should be on the right side of equation!

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Re: balancing redox reactions?
« Reply #7 on: January 25, 2016, 12:15:35 AM »
Check charge and material balance of your equation.
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magicalatom

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Re: balancing redox reactions?
« Reply #8 on: January 25, 2016, 12:19:39 AM »
I am still lost.  I have been on this problem for 3 hours.  I have to move on sorry.

Check charge and material balance of your equation.

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Re: balancing redox reactions?
« Reply #9 on: January 25, 2016, 12:22:34 AM »
My hint reaction have correct coefficients on the left side. Nothing more is needed on this side.
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magicalatom

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Re: balancing redox reactions?
« Reply #10 on: January 25, 2016, 12:27:09 AM »
In your hint what happen to C3H7OH?

My hint reaction have correct coefficients on the left side. Nothing more is needed on this side.

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Re: balancing redox reactions?
« Reply #11 on: January 25, 2016, 12:29:13 AM »
C3H7OH = C3H8O just for easier addition and subtraction
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magicalatom

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Re: balancing redox reactions?
« Reply #12 on: January 25, 2016, 12:34:12 AM »
Just learned another trick.  Ok so I have

H2O +  C3H8O  ---->  C3H6O2 + 2H+    +   1e-          is this correct?

C3H7OH = C3H8O just for easier addition and subtraction

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Re: balancing redox reactions?
« Reply #13 on: January 25, 2016, 12:36:48 AM »
Still wrong - both charges and hydrogens
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magicalatom

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Re: balancing redox reactions?
« Reply #14 on: January 25, 2016, 12:39:22 AM »
Oops I see

H2O +  C3H8O  ---->  C3H6O2 + 4H+    +   4e-      how about now?

Still wrong - both charges and hydrogens