Question

At 298 K, if emf of the cell corresponding to the reaction, \( \text{Zn(s)} + 2\text{H}^+ \text{(aq)} \to \text{Zn}^{2+} (0.01 \text{ M}) + \text{H}_2 \text{(g)} \) (1 atm) is 0.28 V, then the pH of the solution at the hydrogen electrode is \( \left( \frac{2.303 RT}{F} = 0.06 \text{ V} \right) \), \( E^0_{\text{Zn}^{2+}|\text{Zn}} = -0.76 \text{ V} \)

• A. 8
• B. 7
• C. 9
• D. 10

Hint:

Be careful with signs in the log term. \( -\log \frac{A}{B} = +\log \frac{B}{A} \). Also, \( \log [H^+]^2 = 2 \log [H^+] = -2 \text{pH} \).

Answer 1

The Correct Option is C

Step 1: Standard Cell Potential:
The cell reaction involves Zinc oxidation (Anode) and Hydrogen reduction (Cathode). \[ E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{H^+/H_2} - E^0_{Zn^{2+}/Zn} \] \[ E^0_{cell} = 0 - (-0.76) = +0.76 \text{ V} \]
Step 2: Applying Nernst Equation:
Reaction: \( \text{Zn} + 2\text{H}^+ \to \text{Zn}^{2+} + \text{H}_2 \). Number of electrons \( n = 2 \). Reaction Quotient \( Q = \frac{[\text{Zn}^{2+}] P_{H_2}}{[\text{H}^+]^2} \). Equation: \[ E_{cell} = E^0_{cell} - \frac{0.06}{n} \log Q \] Substitute values: \[ 0.28 = 0.76 - \frac{0.06}{2} \log \left( \frac{0.01 \times 1}{[\text{H}^+]^2} \right) \]
Step 3: Solving for pH:
Rearrange the equation: \[ 0.28 - 0.76 = -0.03 \log \left( \frac{10^{-2}}{[\text{H}^+]^2} \right) \] \[ -0.48 = -0.03 \left( \log 10^{-2} - \log [\text{H}^+]^2 \right) \] Divide by -0.03: \[ 16 = -2 - 2 \log [\text{H}^+] \] Since \( \text{pH} = -\log [\text{H}^+] \), we can substitute \( \log [\text{H}^+] = -\text{pH} \): \[ 16 = -2 - 2(-\text{pH}) \] \[ 16 = -2 + 2\text{pH} \] \[ 18 = 2\text{pH} \] \[ \text{pH} = \frac{18}{2} = 9 \]
Step 4: Final Answer:
The pH is 9.