[TheMoose]

I finished my last exam for the semester about an hour ago, and one problem really bothered me.

SCE || H+ (1.0 x 10 ^ -5 M) | pH electrode

Cell was found to have a potential of 0.215 V.  H+ was taken out and replaced with an unknown solution.  Potential was found to be -0.158 V.  What is the pH of the unknown solution?

[Borek]

Please read forum rules.

And what was your approach?

[TheMoose]

I was totally stumped.  We hadn't done a problem like that in class, or at least the classes I had gone to, which is all but one, so I just winged it with the only equation I could think of that might apply, the Nernst equation.

I used E(H half cell) = E(standard) - 0.0592/1 * log[H+].  Hydrogen standard is 0 by definition, so E(H half cell) comes out to be 0.296 V.  E(cell) = E(SCE) - E(H+) -> E(SCE) = 0.511 V.  I then used E(cell) = E(SCE) - E(unknown) -> E(unknown) = 0.669.  Applying Nernst equation again E(unknown) = 0 - 0.0592/1 * log[H+] -> E(unknown) = 0.0592 * pH -> pH = 11.3

Actually, I have no clue what I did.  I remember that my answer was around 2.3, but now that I think about it, since the voltage decreased, there must have been a decrease in H+ concentration at the cathode, which would result in a higher pH, meaning that the final pH would have to be greater than 5.

EDIT: Sorry about the rules breaking.  I had an account on these forums quite a while ago and had forgotten about that little requirement.

[Borek]

Strange. For obvious reasons you should use Nernst equation, with the most convenient form to use

E = E₀ + 0.059 pH

(see http://www.ph-meter.info/pH-Nernst-equation for example)

But when solving I get negative pH value (not impossible, but -2.5 is just too far for being realistic).

Are you sure about the numbers?

[TheMoose]

I'm not sure about the numbers, those were just what I thought they were.  I had never seen a problem like that before the test, so I had to just guess where to begin.  I'm guessing that link will tell me how to solve this.  Thanks.

[mdlhvn]

I was totally stumped.  We hadn't done a problem like that in class, or at least the classes I had gone to, which is all but one, so I just winged it with the only equation I could think of that might apply, the Nernst equation.

I used E(H half cell) = E(standard) - 0.0592/1 * log[H+].  Hydrogen standard is 0 by definition, so E(H half cell) comes out to be 0.296 V.  E(cell) = E(SCE) - E(H+) -> E(SCE) = 0.511 V.  I then used E(cell) = E(SCE) - E(unknown) -> E(unknown) = 0.669.  Applying Nernst equation again E(unknown) = 0 - 0.0592/1 * log[H+] -> E(unknown) = 0.0592 * pH -> pH = 11.3

Actually, I have no clue what I did.  I remember that my answer was around 2.3, but now that I think about it, since the voltage decreased, there must have been a decrease in H+ concentration at the cathode, which would result in a higher pH, meaning that the final pH would have to be greater than 5.

EDIT: Sorry about the rules breaking.  I had an account on these forums quite a while ago and had forgotten about that little requirement.

Your result is right, but you have some confusals.

1. E(H half cell) = E(standard) - 0.0592/1 * log[H+] ---->wrong. It must be:
 E(H half cell) = E(standard) +0.0592/1 * log[H+]

2. E(cell) = E(SCE) - E(H+) -------> wrong. It must be:
E(cell) = E(H) - E(SCE).

3. And finally E(cell) = E(unknown) - E(SCE).

If you apply these equations, you will get the pH 11.3. It is the good result