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Topic: balancing redox reactions?  (Read 8567 times)

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Offline AWK

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Re: balancing redox reactions?
« Reply #15 on: January 25, 2016, 12:44:32 AM »
At last. Now use correct coefficient for both reactions to cancel electrons, add both reactions and move H+ to the left side
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Offline magicalatom

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Re: balancing redox reactions?
« Reply #16 on: January 25, 2016, 12:49:21 AM »
Multiply top by 4 and second one by 5 to cancel 20e-?  Then do I multiply again by something to cancel out more?

At last. Now use correct coefficient for both reactions to cancel electrons, add both reactions and move H+ to the left side

Offline AWK

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Re: balancing redox reactions?
« Reply #17 on: January 25, 2016, 12:51:48 AM »
After multiplying add both reactions by sides and calculate H+ on the left side and water on the right one
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Offline magicalatom

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Re: balancing redox reactions?
« Reply #18 on: January 25, 2016, 12:59:43 AM »
I got 18.   I first multiplied the second by 2 to get 8H+ and cancel those out.  Then I did the new second equation by 2 to get 4H2O and cancel those out.

Then I had to divide by 2 to get 8e- and multiply the first one by 8 and the second one by 5 to cancel out 40e-

I ended up with products side coeffiecients as 18.  Do you think the book made an error?  No text is perfect.  I have caught some errors in it.

After multiplying add both reactions by sides and calculate H+ on the left side and water on the right one

Offline AWK

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Re: balancing redox reactions?
« Reply #19 on: January 25, 2016, 01:01:20 AM »
I got 4, 5 and 11. Check your calculations
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Offline magicalatom

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Re: balancing redox reactions?
« Reply #20 on: January 25, 2016, 01:08:22 AM »
Are you canceling the water too or just e- and H+?

I got 4, 5 and 11. Check your calculations

Offline AWK

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Re: balancing redox reactions?
« Reply #21 on: January 25, 2016, 01:14:20 AM »
Electrons should be canceled. All H+ on the left side, all H2O on the right side
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Offline magicalatom

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Re: balancing redox reactions?
« Reply #22 on: January 25, 2016, 01:19:24 AM »
we have 5e- and 4e- so you use common factor 20 correct then cancel?

Electrons should be canceled. All H+ on the left side, all H2O on the right side

Offline AWK

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Re: balancing redox reactions?
« Reply #23 on: January 25, 2016, 01:22:59 AM »
Write down the final equation with electrons canceled and check charge balance and material balance
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Offline magicalatom

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Re: balancing redox reactions?
« Reply #24 on: January 25, 2016, 01:26:11 AM »
I have

1.  5e-  +  8H+  +  MnO4- --->  Mn+2   +   4H2O
2.  H2O  +  C3H8O ---->  C3H6O2  +  4H+  +  4e-         is this correct so far?

Write down the final equation with electrons canceled and check charge balance and material balance

Offline AWK

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Re: balancing redox reactions?
« Reply #25 on: January 25, 2016, 01:28:36 AM »
both equations are correct
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Offline magicalatom

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Re: balancing redox reactions?
« Reply #26 on: January 25, 2016, 01:29:35 AM »
So I multiplied the top by 4 and the bottom one by 5 to cancel out 20e-

Is that correct first step?

both equations are correct

Offline AWK

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Re: balancing redox reactions?
« Reply #27 on: January 25, 2016, 01:33:50 AM »
Yes, and at this moment you can find 16-5 = 11H2O on the right side
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Offline magicalatom

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Re: balancing redox reactions?
« Reply #28 on: January 25, 2016, 01:37:04 AM »
Nailed it BAAM  :)

12H+  +  5C3H8O  +  4MnO4-  ---->  11 H2O  +  4Mn+2  +  5 C3H6O2     


So I multiplied the top by 4 and the bottom one by 5 to cancel out 20e-

Is that correct first step?

both equations are correct

Offline magicalatom

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Re: balancing redox reactions?
« Reply #29 on: January 25, 2016, 01:38:39 AM »
Well worth the 4 hours spent.  Every moment taught me something.  This is what I try to tell other students what studying is.  Some will just never get it.

And thank you for not just showing me, but guiding me.  Brain grows new pathways.  It is like weightlifting.  One must do problems mentally to grow brain power :)



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