Question

At 298 K, the following reaction takes place for a cell at the hydrogen electrode: \[ H^+(aq) + e^- \longrightarrow \frac{1}{2} H_2 (1 \text{ bar}) \] The solution pH is 10.0. What is the hydrogen electrode potential in volts? (\(\frac{2.303RT}{F} = 0.06\) V)

• A. \( -0.6 \)
• B. \( -0.06 \)
• C. \( +0.6 \)
• D. \( +0.06 \)

Hint:

For hydrogen electrode potential calculations, use the Nernst equation: \[ E = E^\circ - \frac{2.303 RT}{F} \log [H^+] \] where \( E^\circ = 0 \) V.

Answer 1

The Correct Option is B

Step 1: Nernst Equation The electrode potential of the hydrogen electrode is given by: \[ E = E^\circ - \frac{2.303 RT}{F} \log [H^+] \] where: - \( E^\circ = 0 \) V for the standard hydrogen electrode, - \( [H^+] = 10^{-pH} = 10^{-10} \), - \( \frac{2.303 RT}{F} = 0.06 \) V. Step 2: Compute Electrode Potential \[ E = 0 - (0.06 \log 10^{-10}) \] \[ E = - (0.06 \times -10) \] \[ E = -0.06 \text{ V} \] Conclusion Thus, the correct answer is: \[ -0.06 \text{ V} \]