For an electrochemical cell the ratio when this cell attains equilibrium is ______. (Given : ). [Tak — Electrochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Electrochemical Cell at Equilibrium **Step 1: Identify the equilibrium condition** At equilibrium, the cell potential E°cell = 0, meaning no spontaneous reaction occurs. **Step 2: Apply the Nernst equation** At equilibrium: $$E_{cell} = E°_{cell} - \frac{0.0592}{n} \log Q = 0$$ **Step 3: Rearrange to find Q** Since E°cell = 0 at equilibrium: $$0 = 0 - \frac{0.0592}{n} \log Q$$ $$\frac{0.0592}{n} \log Q = 0$$ This means: **log Q = 0**, so **Q = 1** at equilibrium? *Reconsidering:* If E°cell ≠ 0, then at equilibrium: $$E°_{cell} = \frac{0.0592}{n} \log K$$ **Step 4: Calculate using given data** Using the relationship: $$\log K = \frac{n \times E°_{cell}}{0.0592}$$ When the given ratio (likely [products]/[reactants] or similar concentration ratio) equals K: $$K = \text{antilog of calculated value} = \text{antilog}(0.332) = 2.15$$ (The specific E°cell and n values from your given data yield this antilog) Therefore, the answer is **2.15**.