For the galvanic cell, Zn(s) + Cu (0.02 M) Zn (0.04 M) + Cu(s), E =_______ × 10 V. (Nearest integer) — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the cell notation and standard cell potential** The reaction is: Zn(s) + Cu²⁺(0.02 M) → Zn²⁺(0.04 M) + Cu(s) Standard potentials: E°(Cu²⁺/Cu) = +0.34 V; E°(Zn²⁺/Zn) = -0.76 V E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 1.10 V **Step 2: Determine the number of electrons transferred** For Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu n = 2 electrons **Step 3: Apply the Nernst equation** $$E_{cell} = E°_{cell} - \frac{0.0592}{n} \log Q$$ **Step 4: Calculate the reaction quotient (Q)** $$Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{0.04}{0.02} = 2$$ **Step 5: Calculate Ecell** $$E_{cell} = 1.10 - \frac{0.0592}{2} \log(2)$$ $$E_{cell} = 1.10 - 0.0296 × 0.301$$ $$E_{cell} = 1.10 - 0.0089 = 1.091 \text{ V}$$ **Step 6: Convert to the required format** 1.091 V = 109.1 × 10⁻² V ≈ 109 × 10⁻² V Therefore, the answer is 109.00.