Emf of the following cell at 298 K in V is x × 10 . Zn|Zn (0.1 M)||Ag (0.01 M)|Ag The value of x is — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the cell reaction and standard EMF** Cell: Zn|Zn²⁺(0.1 M)||Ag⁺(0.01 M)|Ag Standard reduction potentials: - Ag⁺ + e⁻ → Ag: E° = +0.80 V - Zn²⁺ + 2e⁻ → Zn: E° = -0.76 V Cell reaction: Zn + 2Ag⁺ → Zn²⁺ + 2Ag E°cell = E°cathode - E°anode = 0.80 - (-0.76) = 1.56 V **Step 2: Apply the Nernst equation** E_cell = E°cell - (0.0592/n) × log Q where n = 2 (electrons transferred) **Step 3: Calculate the reaction quotient (Q)** Q = [Zn²⁺]/[Ag⁺]² = (0.1)/(0.01)² = 0.1/0.0001 = 1000 **Step 4: Substitute into Nernst equation** E_cell = 1.56 - (0.0592/2) × log(1000) E_cell = 1.56 - 0.0296 × 3 E_cell = 1.56 - 0.0888 = 1.4712 V **Step 5: Express in the form x × 10⁻²** 1.4712 V = 147.12 × 10⁻² V Rounded to nearest integer: x = 147 Therefore, the answer is 147.00.