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Author Topic:   Help on redox reactions
yenmor
Member (Idle past 3655 days)
Posts: 145
Joined: 07-01-2013


Message 1 of 8 (707601)
09-28-2013 11:09 PM


Hello, I've been writing a program to help students and chemists with calculations and stuff. My app already has a formula calculation, concentraiont, balancer, etc. The two big things I'm working on right now is calculating the redox reactions (balance, determine which is the reducing agent and which is the oxidating agent) and 3-D modeling of molecules. It's so cool, you can choose a carbon atom and attach hydrogen to it to give it a tetrahedral shape. Because the app renders the model in 3-D, you will be able to see it from any angle. I'm sure once people get used to this they will be able to create very creative and complex molecular models.
Anyway, what I'm having trouble on is the damn redox reactions. The algo I have for it right now solves very complex redox reactions. Like the following.
What my algo has trouble with is when to add an H+ ion and on what side and when to add H2O and on what side.
I know how to do it by hand. Human intuition is always a plus. But I'm trying to figure out a programmatic way to solve it.
Just exactly when do I add an H+ and how do I know which side to add? And when and where for H2O?
Help in this will be greatly appreciated.

Replies to this message:
 Message 2 by NoNukes, posted 09-29-2013 12:12 AM yenmor has replied

  
NoNukes
Inactive Member


Message 2 of 8 (707602)
09-29-2013 12:12 AM
Reply to: Message 1 by yenmor
09-28-2013 11:09 PM


Write half reactions.
Balance elements in each half reactions except H and O's last
.you add h20 when you need to balance oxygen on the other side.
you add h+ when you need to balance H's on the other side. This can happen because of adding those h2Os.
Then balance charges by adding electrons to each half reaction.
etc...
In basic soln' once finished add OH to each side to form water with each H+.
Edited by NoNukes, : No reason given.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
I believe that a scientist looking at nonscientific problems is just as dumb as the next guy.
Richard P. Feynman
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

This message is a reply to:
 Message 1 by yenmor, posted 09-28-2013 11:09 PM yenmor has replied

Replies to this message:
 Message 3 by yenmor, posted 10-02-2013 12:12 AM NoNukes has replied

  
yenmor
Member (Idle past 3655 days)
Posts: 145
Joined: 07-01-2013


Message 3 of 8 (707904)
10-02-2013 12:12 AM
Reply to: Message 2 by NoNukes
09-29-2013 12:12 AM


Ok, please demonstrate for us how you would go about balancing the following with half reactions like you said.
Cr7N66H96C42O24+MnO4(-1)+H(+1)⇌Cr2O7(-2)+Mn(+2)+CO2+NO3(-1)+H2O
Edited by yenmor, : No reason given.

This message is a reply to:
 Message 2 by NoNukes, posted 09-29-2013 12:12 AM NoNukes has replied

Replies to this message:
 Message 4 by NoNukes, posted 10-02-2013 1:13 AM yenmor has replied

  
NoNukes
Inactive Member


Message 4 of 8 (707908)
10-02-2013 1:13 AM
Reply to: Message 3 by yenmor
10-02-2013 12:12 AM


Ok, please demonstrate for us how you would go about balancing the following with half reactions like you said.
Who is us?
What is giving you trouble? Can you not identify the two half reactions? You ignore the h+ and the water and then write down the two half reactions using what is left. one half reaction involves Mn, the other has Cr, C, and N.
Sure the problem tedious, but aren't you using a computer?
Balance every element except H,O
2Cr7N66H96C42O24 = 7Cr2O7-2 + 84CO2 + 132NO3-
MnO4 = Mn+2
Now balance the O using H20.
Then balance the H using H+
2Cr7N66H96C42O24 + 565 H2O = 7Cr2O7-2 + 84CO2 + 132NO3- + 1322 H+
MnO4- + 8H+ = Mn+2 + 4H2O
The rest requires a bunch of tedious arithmetic and nothing more. Do you really need me to do that stuff?
Then add e- as needed to each half reaction so each half reaction has balanced charges. If electrons are on the same side in the two half reactions, then it wasn't an ox-redox. Quit.
2Cr7N66H96C42O24 + 565 H2O = 7Cr2O7-2 + 84CO2 + 132NO3- + 1322 H+ + 1176e-
MnO4- + 8H+ +5e- = Mn+2 + 4H2O
Multiplty each equation by whatever required so that each half reaction involves the same number of electrons. (5 and 1176 respectively) then add two half reactions, remove species that are on two sides of equation. This should clear out all electrons. Clear fractions as needed. Reduce if possible. Done
I believe the answer in your display is correct, but I did not do a complete check.
Edited by NoNukes, : No reason given.
Edited by NoNukes, : No reason given.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
I believe that a scientist looking at nonscientific problems is just as dumb as the next guy.
Richard P. Feynman
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

This message is a reply to:
 Message 3 by yenmor, posted 10-02-2013 12:12 AM yenmor has replied

Replies to this message:
 Message 5 by yenmor, posted 10-02-2013 5:34 AM NoNukes has replied

  
yenmor
Member (Idle past 3655 days)
Posts: 145
Joined: 07-01-2013


Message 5 of 8 (707912)
10-02-2013 5:34 AM
Reply to: Message 4 by NoNukes
10-02-2013 1:13 AM


Well, the algo I got right now works for that equation just fine. The answer in the display matches what is published in the literature, so it's correct.
I guess working on several projects at the same time is taking its toll. I'm developing a 3-D modeling software, crunching some data for my company, checking someone's design, and this redox thing at the same time. I also have to prepare for a presentation next week to a bunch of college kids. These critters can smell blood!
Edit.
As I was saying, I used a different approach with the programming to get that result. I'll have some time tomorrow to work on this, so I'll try to write out another subroutine using the half reduction thingy you speak of.
One question. Do you always add the H+ on the left side and the H2O on the right side?
Edited by yenmor, : No reason given.
Edited by yenmor, : No reason given.

This message is a reply to:
 Message 4 by NoNukes, posted 10-02-2013 1:13 AM NoNukes has replied

Replies to this message:
 Message 6 by NoNukes, posted 10-02-2013 8:22 AM yenmor has not replied

  
NoNukes
Inactive Member


Message 6 of 8 (707917)
10-02-2013 8:22 AM
Reply to: Message 5 by yenmor
10-02-2013 5:34 AM


One question. Do you always add the H+ on the left side and the H2O on the right side?
No. That is not even true for the example I worked. You add the H2O on the side opposite from the excess oxygen. Then you add H+ to balance the excess Hs.
And don't forget the extra rule at the end for basic solutions. The steps I used above assumed an acidic solution.
Good luck with your presentation.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
I believe that a scientist looking at nonscientific problems is just as dumb as the next guy.
Richard P. Feynman
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

This message is a reply to:
 Message 5 by yenmor, posted 10-02-2013 5:34 AM yenmor has not replied

Replies to this message:
 Message 7 by NoNukes, posted 10-02-2013 1:58 PM NoNukes has not replied

  
NoNukes
Inactive Member


Message 7 of 8 (707945)
10-02-2013 1:58 PM
Reply to: Message 6 by NoNukes
10-02-2013 8:22 AM


Rule exceptions...
I thought I would mention at least one rule exception.
When you have something like this
O- ==> O2
You don't balance it by adding H2O. You just adjust the coefficients.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
I believe that a scientist looking at nonscientific problems is just as dumb as the next guy.
Richard P. Feynman
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

This message is a reply to:
 Message 6 by NoNukes, posted 10-02-2013 8:22 AM NoNukes has not replied

Replies to this message:
 Message 8 by onifre, posted 10-02-2013 11:59 PM NoNukes has not replied

  
onifre
Member (Idle past 2950 days)
Posts: 4854
From: Dark Side of the Moon
Joined: 02-20-2008


(2)
Message 8 of 8 (707966)
10-02-2013 11:59 PM
Reply to: Message 7 by NoNukes
10-02-2013 1:58 PM


Re: Rule exceptions...
When you have something like this
O- ==> O2
What about when you have something like this:
8====D~~~~
Do you also have to adjust the coefficients?
- Oni
Edited by onifre, : No reason given.

This message is a reply to:
 Message 7 by NoNukes, posted 10-02-2013 1:58 PM NoNukes has not replied

  
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