What is the standard cell EMF? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Cell Representation The given Edison storage cell is represented using IUPAC notation as: $$\ce{Fe(s) \mid FeO(s) \mid KOH(aq) \mid Ni2O3(s) \mid Ni(s)}$$ According to standard electrochemical conventions: * The left-hand side represents the **anode** (oxidation half-cell): $$\ce{Fe(s) \mid FeO(s)}$$ * The right-hand side represents the **cathode** (reduction half-cell): $$\ce{Ni2O3(s) \mid Ni(s)}$$ Step 2 - Identify the Standard Reduction Potentials We are given the standard reduction potentials ($E^\circ$) of both systems under standard state conditions: 1. **Cathode reaction (Reduction):** $$\ce{Ni2O3(s) + H2O(l) + 2e^- <=> 2NiO(s) + 2OH^-(aq)} \quad E^\circ_{\text{cathode}} = E^\circ_{\ce{Ni2O3/NiO}} = +0.40\text{ V}$$ 2. **Anode reaction (written as a reduction half-reaction):** $$\ce{FeO(s) + H2O(l) + 2e^- <=> Fe(s) + 2OH^-(aq)} \quad E^\circ_{\text{anode}} = E^\circ_{\ce{FeO/Fe}} = -0.87\text{ V}$$ Although the reaction taking place at the anode is the oxidation of metallic iron ($\ce{Fe}$) to iron(II) oxide ($\ce{FeO}$), the thermodynamic calculations are performed using standard reduction potentials. Step 3 - Calculate the Standard Cell EMF ($E^\circ_{\text{cell}}$) The standard electromotive force ($E^\circ_{\text{cell}}$) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ Substitute the given values into the formula: $$E^\circ_{\text{cell}} = 0.40\text{ V} - (-0.87\text{ V})$$ $$E^\circ_{\text{cell}} = 0.40\text{ V} + 0.87\text{ V}$$ $$E^\circ_{\text{cell}} = \boxed{1.27\text{ V}}$$ Step 4 - Evaluate each Option * **Option (A) is correct:** The calculated standard EMF of the Edison storage cell is $+1.27\text{ V}$. * **Option (B) is incorrect:** This value ($0.47\text{ V}$) is obtained if one incorrectly subtracts the absolute values without taking care of the signs. * **Option (C) is incorrect:** This negative potential ($-1.27\text{ V}$) corresponds to a non-spontaneous process. * **Option (D) is incorrect:** This represents a potential of $-0.47\text{ V}$, which does not arise from correct thermodynamic calculations. $$\text{Correct Option: } \boxed{A}$$