14.39 (a)


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Maria1E
Posts: 61
Joined: Sat Jul 22, 2017 3:01 am

14.39 (a)

Postby Maria1E » Tue Feb 20, 2018 5:27 pm

Determine the unknown quantity in each of the following cells.
a) Pt(s) I H2 (g, 1 bar) I H+ (pH = ?) II Cl- (aq, 1 M) I Hg2Cl2 (s) I Hg (l)
E = .33 V

Can someone please write out and explain the calculations? I'm confused about how the pH value was obtained at the end.

Chem_Mod
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Re: 14.39 (a)

Postby Chem_Mod » Wed Feb 21, 2018 7:12 am

Hi, Maria,
So, firstly, you need to write anode and cathode separately and calculate the standard cell potential from E = E(cathode) - E(anode). Remember to use only reduction potentials for this formula and that cathode is always on the right side of the diagram.
Then, plug the resulting value into the Nernst equation, where Q = [H+]^2*[Cl-]^2/[H2]. You can figure out the Q by combining cathode and anode together and balancing the resulting redox reaction (remember, there are no solids and liquids in Q). As you know the concentration of both Cl- = 1M and H2 = 1, your only unknown is [H+]. (E is given in the problem as well). After rearranging the equation and taking -log[H+] = pH, you will get your answer.

Rhay Flores 4K
Posts: 11
Joined: Fri Sep 29, 2017 7:04 am

Re: 14.39 (a)

Postby Rhay Flores 4K » Sat Mar 09, 2019 6:23 pm

I'm working through this problem and the solutions manual uses the equation:


However, I'm confused because I'm using this equation (since the Nernst equations using log and ln are interchangeable I think):


and I'm getting a different answer to find the pH.
I'm confused about where this part, , comes from?


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