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?
pairs of redox couples
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Re: pairs of redox couples
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.
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Re: pairs of redox couples
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
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.
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