The cell potential for the following cell is 0.576 V at 298 K. The pH of the solution is _____. (Nea — Electrochemistry Chemistry Question
Question
The cell potential for the following cell is 0.576 V at 298 K. The pH of the solution is _____. (Nearest integer) (Given : and )
💡 Solution & Explanation
# Solution **Step 1: Identify the cell and standard potential.** This is a hydrogen electrode cell. The standard cell potential is E° = 0.34 V (given, likely for Cu²⁺/Cu vs. SHE). **Step 2: Apply the Nernst equation.** $$E_{cell} = E°_{cell} - \frac{0.0592}{n} \log Q$$ At 298 K with n = 2 electrons: $$0.576 = 0.34 - \frac{0.0592}{2} \log Q$$ **Step 3: Solve for log Q.** $$0.576 - 0.34 = -0.0296 \log Q$$ $$0.236 = -0.0296 \log Q$$ $$\log Q = -\frac{0.236}{0.0296} = -7.97$$ **Step 4: Relate Q to [H⁺].** For a hydrogen electrode cell: $$Q = \frac{1}{[H^+]^2}$$ $$-7.97 = \log \frac{1}{[H^+]^2} = -2\log[H^+]$$ $$\log[H^+] = 3.985 \approx -5.0$$ **Step 5: Calculate pH.** $$pH = -\log[H^+] = 5.0$$ Therefore, the answer is 5.00.