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Parker Smith
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Joined: Thu Jul 25, 2019 12:15 am


Postby Parker Smith » Sat Feb 22, 2020 4:21 pm

I am confused at how the solution book worked this problem, specifically the making of the reaction quotient (Q). Also, the book doesn't use Faraday's constant in the Nernst Equation. Why is this, or is this suppose to happen?

Elena Bell 1C
Posts: 64
Joined: Thu Jul 11, 2019 12:16 am

Re: 6N.3

Postby Elena Bell 1C » Sat Feb 22, 2020 5:58 pm

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.

Posts: 100
Joined: Sat Sep 14, 2019 12:16 am

Re: 6N.3

Postby 805291863 » Sat Feb 22, 2020 10:13 pm

Also, how is the n value determined here?

Posts: 180
Joined: Wed Sep 18, 2019 12:20 am

Re: 6N.3

Postby 805097738 » Sun Feb 23, 2020 4:26 pm

805291863 wrote:Also, how is the n value determined here?

n value is determined by the moles of electrons in each of the balanced half rxn equations

chari_maya 3B
Posts: 108
Joined: Sat Sep 07, 2019 12:18 am

Re: 6N.3

Postby chari_maya 3B » Sun Mar 01, 2020 8:33 pm

I am confused as to how the cathode/anode were determined. I thought they were interchangeable but if they were flipped, wouldn't this cause Q to be flipped as well?

Harry Zhang 1B
Posts: 101
Joined: Sat Sep 14, 2019 12:16 am

Re: 6N.3

Postby Harry Zhang 1B » Sun Mar 01, 2020 8:58 pm

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.

Megan Vu 1J
Posts: 101
Joined: Thu Jul 25, 2019 12:15 am

Re: 6N.3

Postby Megan Vu 1J » Mon Mar 02, 2020 12:13 pm

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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