Chemistry - reduction potential help

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

SereneAurora

New Member
10+ Year Member
Joined
Jul 11, 2011
Messages
6
Reaction score
0
My question: the higher the reduction potential value, the more it's likely to be reduced and is more inclined to be the cathode reaction? Sometimes this isn't the case, why?

Thanks

Members don't see this ad.
 
My question: the higher the reduction potential value, the more it's likely to be reduced and is more inclined to be the cathode reaction? Sometimes this isn't the case, why?

Thanks

yes, the higher (more positive) the reduction potential, the greater the tendency for that atom to be reduced. Compounds that want to be reduced will gain electrons at the cathode. Electrons will flow from the atom being oxidized at the anode (something with a large, negative reduction potential, which means it favors oxidation) and the electrons will flow to the cathode, which is where what's in solution grabs those electrons and becomes reduced.
 
Members don't see this ad :)
Here's a way (at least that I use) to keep the signs (+/-) straight when working with reduction potentials.

Remember the formula for the Gibbs Free Energy of an electrochemical cell:

ΔG°= -nFE°

where n is the number of electrons transferred per mole of reactant and E° is the standard reduction potential. n, and F are positive.

When ΔG° is negative, the reaction is thermodynamically favourable (K>1), so the reactant will 'want' to be reduced (though a high energy barrier can prevent this, like not hooking up the wire in your beakers, for example, but this is a kinetic issue, not a thermodynamic one).

Anyway, for ΔG° to be negative, E° must be positive, and the more positive it is the more negative ΔG° is, and thus the more thermodynamically favourable the reduction half-reaction is.

One other thing that I used to mess up all the time, which isn't that relevant but worth saying - when you're adding two half-reactions together to get an overall oxidation/reduction reaction, the standard reduction potentials simply add together, regardless of whether or not you have to multiply one reaction by 2 or 3 so the electrons on both sides cancel out. Basically, multiplying a half-reaction by a number N doesn't change its reduction potential by a factor of N; it's still just E°.
 
yes, the higher (more positive) the reduction potential, the greater the tendency for that atom to be reduced. Compounds that want to be reduced will gain electrons at the cathode. Electrons will flow from the atom being oxidized at the anode (something with a large, negative reduction potential, which means it favors oxidation) and the electrons will flow to the cathode, which is where what's in solution grabs those electrons and becomes reduced.

+1

PS- there are MANY websites online that can explain this. It sometimes help to not be lazy and google it yourself.
 
The best conceptual treatment of this material on the web is at Steven Lower's Chem1 virtual textbook.

http://www.chem1.com/acad/webtext/elchem/index.html

He's really good at explaining the idea of the 'fall of the electron' in redox in a common sense way involving electrostatics and thermochemistry. For example, one thing I remember reading in Linus Pauling's really great GENERAL CHEMISTRY http://www.amazon.com/General-Chemi...2985/ref=sr_1_1?ie=UTF8&qid=1310666534&sr=8-1 is that if you graph reduction potential vs. electronegativity it's pretty much a straight line. Try to make your common sense understand that the difference in electronegativity between the elements creates 'electron pressure'. This is the underlying coherence of redox and electrochemistry.

http://www.chem1.com/acad/webtext/elchem/index.html
 
Top