Question

What is the reduction potential of a hydrogen electrode that is still at standard pressure, but has ph = 5.65 , relative to the she?

Answer 1

Answer:

$\boxed{\text{-0.275 V}}$

Explanation:

We must use the Nernst equation

$E = E^{\circ} - \dfrac{RT}{zF} \ln Q$

1. Write the equation for the cell reaction

If you want the reduction potential, the pH 5.65 solution is the cathode, and the cell reaction is

                                                                      E°/V

  Anode: H₂(1 bar) ⇌ 2H⁺(1 mol·L⁻¹) + 2e⁻;     0

Cathode: 2H⁺(pH 5.65) + 2e⁻ ⇌ H₂(1 bar);     0

 Overall: 2H⁺ (pH 5.65) ⇌ 2H⁺(1 mol·L⁻¹);      0

Step 2. Calculate E°

(a) Data

   E° = 0

   R = 8.314 J·K⁻¹mol⁻¹

   T = 25 °C

   n = 2

   F = 96 485 C/mol

pH = 5.65

Calculations:  

T = 25 + 273.15 = 298.15 K

$\text{H}^{+} = 10^{\text{-pH}} = 2.24 \times 10^{-5}\text{ mol/\L}\\\\Q = \dfrac{\text{[H}^{+}]_{\text{prod}}^{2}}{\text{[H}^{+}]_{\text{react}}^{2}} = \dfrac{(1.00)^{2}}{(2.24 \times 10^{-5})^{2}} =2.00 \times 10^{9}\\\\\\E = 0 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln{2.00 \times 10^{9}}\\\\= -0.01285 \times 21.41 = \textbf{-0.275 V}\\\text{The cell potential for the cell as written is }\boxed{\textbf{-0.275 V}}$