Answer:
[tex]\boxed{\text{-0.275 V}}[/tex]
Explanation:
We must use the Nernst equation
[tex]E = E^{\circ} - \dfrac{RT}{zF} \ln Q[/tex]
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
[tex]\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}}[/tex]
