Midterm 2014 Question 8

Moderators: Chem_Mod, Chem_Admin

ckim_1I
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

Midterm 2014 Question 8

Postby ckim_1I » Tue Feb 14, 2017 11:42 pm

In the 2nd portion of the question, it mentions that ln k = nFE(standard) / RT and the answer is -12.46.
I used n=2 and we use Faraday's constant and E (standard) cell all over R (Boltzman's constant) and T (as 295 K), the units don't cancel out.
Am I doing this problem incorrectly?
Pleas help!!!

Liam Giffin 2B
Posts: 34
Joined: Fri Jul 22, 2016 3:00 am
Been upvoted: 1 time

Re: Midterm 2014 Question 8

Postby Liam Giffin 2B » Wed Feb 15, 2017 12:07 am

So I'm not sure exactly where your issue is occurring, but I am not running into the same issue as you. When I use the formula the units cancel and I am left with a unitless K with the same value as listed in the book. In this case we use n as a pure number, so it has no units. The units for the Faraday constant are C/mol, the units for the cell potential are volts which is the same as 1 J/C. The units for R are J/K*mol and the units for T are Kelvin. If you use these units the should cancel and you should be left with the correct value.

Arie Hakimi 1L
Posts: 25
Joined: Wed Sep 21, 2016 2:57 pm
Been upvoted: 2 times

Re: Midterm 2014 Question 8

Postby Arie Hakimi 1L » Wed Feb 15, 2017 12:22 am

Why is Ka=sqrt(K)??

Sangita_Sub_3H
Posts: 23
Joined: Sat Jul 09, 2016 3:00 am

Re: Midterm 2014 Question 8

Postby Sangita_Sub_3H » Wed Feb 15, 2017 1:42 am

Ka = sqrt (K) because Ka written out is Ka = ([H][F])/[HF] and K = ([H]^2[F]^2)/[HF]^2 since the equation is 2HF(aq) -->2H(plus) + 2F(minus).

Jenna_Hakel_2A
Posts: 34
Joined: Fri Jul 15, 2016 3:00 am

Re: Midterm 2014 Question 8

Postby Jenna_Hakel_2A » Wed Feb 15, 2017 1:22 pm

I'm still confused on why Ka=(K)^(1/2). I understand that when we solve for the balanced equation to solve for E, it ends up as 2HF-->2H+ + 2F- with a standard cell potential of -0.16V. However, if we were to divide this equation by 2 to get HF-->H+ + F-, wouldn't we then have the correct equation and our E would still be equal to -0.16? I may be mistaken, but isn't that a property of the standard cell potential? Why would this not work?


Return to “Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams”

Who is online

Users browsing this forum: No registered users and 11 guests