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pairs of redox couples

Posted: Tue Feb 09, 2016 8:30 pm
by Kelly Duong 2D
When given pairs of redox couples, how do you know which one should be the reduction/oxidation half reaction?
For example, in WINTER 2011 Q6, given F2/F- with a standard reduction potential of +2.87V & RB+/RB with a standard reaction potential of -2.93V, they made fluorine the cathode/reduction and rubidium the anode/oxidation. Is it because when you do "cathode minus anode", it'll be positive? *2.87-(-2.93)=5.80V* So thats why fluorine is the cathode and rubidium is the anode?

Re: pairs of redox couples

Posted: Tue Feb 09, 2016 9:29 pm
by 004662171
Yes, this is correct. A positive cell potential is necessary for the reaction to be spontaneous. If you do not have a positive cell potential, then the half reactions are not spontaneous, meaning the cell will not produce a current.

Re: pairs of redox couples

Posted: Wed Feb 10, 2016 11:31 am
by amyjunus1A
The important thing to remember is the cathode is the better oxidizing agent (gains electrons), thus the reduction number (which is given) must be more positive. Hopefully that helps a little.

Re: pairs of redox couples

Posted: Thu Feb 11, 2016 12:07 pm
by 704628249
Basically, your reduction (cathode) equation will be the one that has the more positive cell potential. Therefore, your anode will be the one that is less positive. Since Ecell= Ecathode - Eanode, your Ecell would be positive, making the reaction spontaneous and favorable.