Question

What would the potential of a standard hydrogen electrode (S.H.E.) be if it was under the following conditions?

[H+] = 0.88M Ph2= 1.7atm T=298k

Answer 1

`\boxed{\text{-8.5 mV}}`

Step-by-step explanation:

The SHE is not under standard conditions, so we must use the Nernst Equation, which relates the cell potential to the standard potential and to the activities of the electroactive species.

`E = E^(\circ) - (RT)/(zF)lnQ`

Step 1. Write the equation for the cell reaction

Anode: H₂(1 atm) ⇌ 2H⁺(1 mol·L⁻¹) + 2e⁻

Cathode: 2H⁺(0.88 mol·L⁻¹) + 2e⁻ ⇌ H₂(1.7 atm)

Overall: H₂(1 atm) + 2H⁺(0.88 mol·L⁻¹) ⇌ 2H⁺(1 mol·L⁻¹) + H₂(1.7 atm)

Step 2. Calculate Q

`Q = (1 * 1.7 )/( 1 * 0.88) = 1.93`

Step 3. Apply the Nernst equation

Data:

E° = 0 V

R = 8.314 J·K⁻¹mol⁻¹

T = 298 K

n = 2

F = 96 485 C/mol

Calculation:

`E = 0 - (8.314 * 298)/(2 * 96 485)\ln1.93`

= -0.0128 × 0.658

= -0.0085 V

=
`\boxed{\text{-8.5 mV}}`