Pt(s) | H2(g, 2.0 bar) | H+ (pH = 3.0) || Cl -(aq, 0.85 mol·L-1)) | Hg2Cl2(s) | Hg(l)
I need to calculate the voltage of the cell. I have a pretty good idea of how to use the Nernst equation for this, but am unsure if I am calculating the concentration of H+ correctly. I calculated the concentration to be 10^-3 given that pH=[H3O+]
6 answers
If it says pH = 3, then (H3O+) = 1 x 10^-3 M.
Yes that's what I used to to try to find the voltage. But my answer is still coming out to be -.1833 which is incorrect
Are you squaring it?
H2 ==> 2H^+ + 2e
E = Eo-(0.0592/2)log(1/(H^+)^2
As I wrote this I remember I may have contradicted myself about how the Q is written. When working with this equation, it is log (red/ox). If working with the full equation, it is K.
H2 ==> 2H^+ + 2e
E = Eo-(0.0592/2)log(1/(H^+)^2
As I wrote this I remember I may have contradicted myself about how the Q is written. When working with this equation, it is log (red/ox). If working with the full equation, it is K.
where does the .0592/2 come from? I've been calculating my RT/nF to be .025693/2 ....this is all so confusing...
ln RT/F = 2.303*log(RT/F)is 0.0592 at 25 degrees C(298.15 K) So RT/nF becomes 0.0592/n at 25 C. To get 0.02569... you aren't converting ln to log. Take your number and multiply by 2.303. I THINK that gives 0.05916 and most texts round to 0.0592/n
[2.303*8.314*298.15/96,485]=0.0591669.
[2.303*8.314*298.15/96,485]=0.0591669.
Pt(s) | H2(g, 1.0 bar) | H+ (pH =?) || Cl -(aq, 1,0mol·L-1)) | AgCl(s) | Ag(s).
Como calcular o Ph?
Como calcular o Ph?