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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: 0rion on March 25, 2009, 11:33:50 PM

Title: Tricky Redox reaction
Post by: 0rion on March 25, 2009, 11:33:50 PM
hey guys, can anyone help me out with this?

Heres the reaction:

Au + CN- + O2 :rarrow: Au(CN4)- + OH-

My problem is i dont know what to do with the half reactions...

The one is simple....

3e- + 2H+ + O2  :rarrow: 2OH-

But what to do with the other one? ive tried balancing
 Au  :rarrow: Au3+
and Au + CN-  :rarrow: Au(CN4)- and neither worked for me

What do i do?
Thanks in advance :)
Title: Re: Tricky Redox reaction
Post by: AWK on March 26, 2009, 02:55:52 AM
Not so tricky, firstly correct formula Au(CN4)-
correctly should be Au(CN)2-  or in your case Au(CN)4-
Then in your case
Au = Au3+ +e-
O2 + H2O +  e- = OH-
both half reaction should be balanced
Title: Re: Tricky Redox reaction
Post by: 0rion on March 26, 2009, 08:39:07 AM
Huh? the charge on the left is 0 and on the right is 2+? how does that work

lol yeah soz about the typos with CN- :S
Title: Re: Tricky Redox reaction
Post by: 0rion on March 26, 2009, 08:41:51 AM
Sorry to spread this over two posts but i only just realised what you put...

O2 + H2O +  e- = OH-
4 O and 2 H            1 O and 1 H

zar?
Title: Re: Tricky Redox reaction
Post by: Borek on March 26, 2009, 08:46:32 AM
What AWK posted are skeletal reactions, you have to balance them. All reactants are in place, just stoichiometry is not correct yet.
Title: Re: Tricky Redox reaction
Post by: 0rion on March 26, 2009, 08:58:16 AM
OK obviously the way you guys do redox and the way i have been taught are different :( im lost by what you're saying
Title: Re: Tricky Redox reaction
Post by: ARGOS++ on March 26, 2009, 09:17:36 AM

Dear Orion;

Do you know how to balance red- and ox- reaction equations?:
        http://www.science.uwaterloo.ca/~cchieh/cact/c123/balance.html 

Good Luck!
                    ARGOS++
Title: Re: Tricky Redox reaction
Post by: Borek on March 26, 2009, 10:50:52 AM
Quote from: 0rion on March 26, 2009, 08:58:16 AM
OK obviously the way you guys do redox and the way i have been taught are different :( im lost by what you're saying

This is skeletal reaction:

H2 + O2 -> H2O

This is balanced reaction:

2H2 + O2 -> 2H2O

This is skeletal half-reaction:

O2 + H2O +  e- = OH-

Now you have to balance it - balance both atoms and charge, that's not different from normal balancing.
Title: Re: Tricky Redox reaction
Post by: lucas89 on March 26, 2009, 10:52:30 PM
Balance your half reaction by placing the correct coefficients in the reaction if nessecary, and then balance the oxygens by adding water and continuing from there in either acidic or basic conditions. Ring any bells?