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I think they got Q from the concentrations given in the problem. Also, I was confused about why the book didn't use Faraday's constant either, so I checked and apparently RT/F is equal to 0.025693V at 298K, so that's what they used instead. The solutions manual can be a little confusing sometimes.
You should use appendix 2 to find the half-reactions happening in cathode and anode. Remember, cathode is the one with a higher E. Then balance the half-reactions and write them out to determine Q. Faraday's constant was not used because it was included in the formula given in the book after multiplying the constants together.
In the book, Faraday's constant is not used because the solutions manual utilizes a short cut. If you multiplied the values of the constants together and divided with temperature, R, and Faraday's constant, you will be able to get 0.025693 which is divided by the moles. This is only used when the temperature is at 25 degrees or 298 since this is standard value. n is the number of moles that you can get from the half reactions of cathode and anode.
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