What would the potential of a standard hydrogen (S.H.E.) electrode be if it was under the following conditions?
[H+] = 0.38
P(H2) = 3.8
T = 298 K
I assume that's 3.8 atm.
2H^+ + 2e ==> H2
E = Eo - (0.0592/2)log[(pH2/(H^+)^2]
Eo is zero, right? :-)
The SHE under standard condition is always 0.
I had a similar question why doesn't Eo count is it because it is under standard conditions?
Wait but there is no Eo
Yeah then how would you calculate it
Well, under those conditions, the potential of the S.H.E. electrode would probably be asking itself: "Hey, why am I here? I mean, I'm just a humble electrode, hanging out in some solution, and now someone wants to know my potential? Talk about pressure! I'm feeling a little charged up, if you ask me." *inserts clown horn sound*
To determine the potential of a Standard Hydrogen Electrode (S.H.E.) under given conditions, we need to use the Nernst equation. The Nernst equation relates the electrode potential to the concentrations of species involved in the reaction and the temperature. The equation is given as:
E = E° - (RT / nF) * ln(Q)
Where:
- E is the potential of the electrode under the given conditions
- E° is the standard electrode potential (which is 0 V for the S.H.E.)
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- n is the number of electrons transferred during the reaction (for the S.H.E., n = 2)
- F is the Faraday constant (96485 C/mol)
- Q is the reaction quotient, which is calculated as [H+] / P(H2)
Now, let's substitute the given values into the equation:
E = 0 V - (8.314 J/(mol·K) * 298 K / (2 mol * 96485 C/mol) * ln(0.38 / 3.8)
First, we need to calculate the natural logarithm of the reaction quotient:
ln(0.38 / 3.8) ≈ -2.77259
Substituting this value back into the main equation:
E = 0 V - (8.314 J/(mol·K) * 298 K / (2 mol * 96485 C/mol) * (-2.77259)
Calculating this expression:
E ≈ -0.0592 V
Therefore, the potential of the S.H.E. under the given conditions would be approximately -0.0592 V.