Question

What would the potential of a standard hydrogen (S.H.E) electrode be if it was under the following condition ?
[H+]=0.56 M , P H2=4.3 atm, T=298. How do I solve ?

Answer 1

The cell reaction that occurs in a hydrogen electrode is: 2H+ (aq) + 2e- -> H2(g)
Under standard conditions ([H+] = 1M, Pressure H2 = 1 atm), the electrode has a potential of 0.00 V when coupled with the S.H.E.

The potential for a nonstandard cell is: E = E° - (RT/nF) ln Q

where E° is the standard cell potential, R is the gas constant (8.314 J/mol-K), n is the number of electrons involved in the reaction, F is Faraday's constant (96485 C/mol). Q is the reaction quotient. For the cell reaction above, Q = (Pressure H2]) / [H+]^2 , n = 2 two electrons transferred

E = E° - (2.303 RT/nF) log (Pressure H2]) / [H+]^2

E = 0.00 - [(2.303 x 8.314 x 298) / 2 x 96485] log [(4.3) / (0.56)²]

E = - 0.03362 volts

Thank you.