Reaction Quotient and Keq for galvanic cells

[umdnjmed]

What compounds are used in the reaction quotient and Keq in electrochemistry? Wouldn't you have 2 reactants (the oxidizing agent and the reducing agent), and two products, the products of oxidation and reduction? For some reason, I don't think the concentrations of all four are used, but i'm not sure.

In the concentration cell example in EK chem, the concentration of Fe2+ ions in the oxidation half cell was used as the product while that of Fe2+ in the reduction half cell was used as the reactant. I don't see the logic behind it.

Secondly, since oxidation always occurs at the anode and electrons move from the less concentrated half cell to the more concentrated half cell (in concentration cells), does this mean the less concentrated solution will always be at the anode?

Thanks in advance.

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[PlumbLine]

You are correct that (in most cases) there are two reactants and two products. However one of the reactants is solid and one of the products is solid, therefore they aren't included in the reaction quotient. Think of the standard galvanic cell, you will see solid "probes". When Fe2+ gets reduced to Fe(s) think of it becoming part of the solid probe, or vice versa for reduction reactions.

One way of understanding concentration cells is through the Nernst equation. You'll notice that the voltage of concentration cells is always given as positive. This tells you something about the conventions used for concentration cells. I'll try to explain:

Look at the simplified Nernst equation E = E0 - 0.0592logQ. We know that for concentration cells E0 is always zero (look up basic definitions of concentration cells to learn why). And as mentioned before, E (Voltage) is always positive. If E is always positive then logQ must be a negative number, right? So for a log to be negative, it must be taking the log of a number that is positive and less than 1. Less than 1 tells us that the concentration of the numerator is less than the concentration of the denominator. The numerator represents the products (anode) so it will always have the lower concentration.

I doubt this cleared everything up for you, but hopefully it gave you a new way of looking at it. Also if you ever need to solve the voltage of a concentration cell, don't worry about making sure you have the right numerator and denominator. If you have them swapped on accident, you will just get a negative voltage, so just take the absolute value of that answer. That's just another Log rule --> -LogQ = Log(1/Q)

Hopefully I didn't confuse you further.

 

[umdnjmed]

Thank you, I has forgotten that the concentrations of pure solids and liquids are never included in the equation for equilibrium constants.

Your tip about what to do if I get confused with the numerator and denominator was very helpful. However, I don't quite understand the use of the anode as the product. If anything, oxidation needs to occur before reduction, therefore in my mind it's the other way around.

 

[umdnjmed]

Actually never mind. I realized the answer to my last post was in the explanation you gave for the compounds used in Keq. The product at the cathode is a solid so only that at the anode is considered. Thanks again.