## 14.39 (a)

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

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

### 14.39 (a)

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
Posts: 18022
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 418 times

### Re: 14.39 (a)

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)

I'm working through this problem and the solutions manual uses the equation:
$E = E^{o} - (\frac{0.025693 V}{n})ln[H+]^{2}$

However, I'm confused because I'm using this equation (since the Nernst equations using log and ln are interchangeable I think):
$E = E^{o} - \frac{0.0592}{n}log[H+]^{2}$

and I'm getting a different answer to find the pH.
I'm confused about where this part, $(\frac{0.025693 V}{n})$ , comes from?

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

### Who is online

Users browsing this forum: No registered users and 1 guest