Question

What would be the potential in volts of a standard hydrogen (S.H.E.) electrode under the following conditions?: [H+] = 0.57M P(H2)=4.9atm T=298

Answer 1

The reduction reaction is given as

2 H⁺ (aq) + 2 e^- --------> H₂ (g); E⁰ = 0.00 V

When [H⁺] = 0.57 M and P(H₂) = 4.9 atm at T = 298 K, the reduction potential of the standard hydrogen electrode is given by

E = E⁰ – (2.303*R*T/n*F)*log P(H₂)/[H⁺]²

= -(2.303*R*T/n*F)*log P(H₂)/[H⁺]² where n = 2 = number of electrons involved

Plug in values and obtain

E = -2.303*(8.314 J/mol.K)*(298 K)/(2*96485 C/mol)*log (4.9)/(0.57)²

= -2.303*(8.314 J/mol.K)*(298 K)/(2*96485 J/V.mol)*log (4.9)/(0.57)²

= -(0.0296 V)*log (15.08156)

= -(0.0296 V)*(1.17845)

= -0.03488 V ≈ -0.035 V (ans).