What will be the magnitude of electrode potential for the given half cell reaction at pH = 5? (R = 8 — Electrochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Electrode Potential at pH = 5 **Step 1: Identify the half-reaction** O₂ + 4H⁺ + 4e⁻ → 2H₂O Standard electrode potential: E° = +1.23 V **Step 2: Apply the Nernst equation** $$E = E° - \frac{2.303RT}{nF}\log Q$$ Where: - R = 8.314 J mol⁻¹ K⁻¹ - T = 298 K - n = 4 (electrons transferred) - F = 96,500 C mol⁻¹ - At 298 K: $\frac{2.303RT}{F}$ = 0.0592 V **Step 3: Determine the reaction quotient (Q)** At pH = 5: [H⁺] = 10⁻⁵ M At 1 bar O₂: P(O₂) = 1 bar $$Q = \frac{1}{[H⁺]^4 \cdot P(O_2)} = \frac{1}{(10^{-5})^4 \times 1} = 10^{20}$$ **Step 4: Calculate the potential** $$E = 1.23 - \frac{0.0592}{4}\log(10^{20})$$ $$E = 1.23 - 0.0148 \times 20$$ $$E = 1.23 - 0.296 = 0.934 \text{ V}$$ **Step 5: Round to appropriate significant figures** E ≈ 0.93 V Therefore, the answer is 0.93.