From the following E° values for the half-cells: (i) D^2+ + 2e^- -> D; E° = -1.5 V, (ii) B^+ + e^- - — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Given Standard Potentials for Each Half-Cell We are given the standard electrode potentials ($E^\circ$) of four different half-cells: 1. **Half-cell (i):** $$\ce{D^2+(aq) + 2e^- -> D(s)} \quad E^\circ = -1.5\text{ V}$$ This is written as a reduction half-reaction. Therefore, its standard reduction potential is: $$E^\circ_{\text{red}}(\text{i}) = -1.5\text{ V}$$ 2. **Half-cell (ii):** $$\ce{B^+(aq) + e^- -> B(s)} \quad E^\circ = -0.5\text{ V}$$ This is written as a reduction half-reaction. Therefore, its standard reduction potential is: $$E^\circ_{\text{red}}(\text{ii}) = -0.5\text{ V}$$ 3. **Half-cell (iii):** $$\ce{A^3-(aq) -> A^2-(aq) + e^-} \quad E^\circ = 1.5\text{ V}$$ Since the electron is released as a product, this is written as an oxidation half-reaction. Therefore, its standard oxidation potential is: $$E^\circ_{\text{ox}}(\text{iii}) = +1.5\text{ V}$$ The corresponding standard reduction potential of the conjugate reduction couple ($\ce{A^2-/A^3-}$) is: $$E^\circ_{\text{red}}(\text{iii}) = -1.5\text{ V}$$ 4. **Half-cell (iv):** $$\ce{C^2+(aq) + e^- -> C^+(aq)} \quad E^\circ = +0.5\text{ V}$$ This is written as a reduction half-reaction. Therefore, its standard reduction potential is: $$E^\circ_{\text{red}}(\text{iv}) = +0.5\text{ V}$$ Step 2 - Analyze the Formula for Cell Potential ($E^\circ_{\text{cell}}$) The standard electromotive force of a cell ($E^\circ_{\text{cell}}$) is determined by subtracting the standard potential of the anode from that of the cathode. To find the combination of half-cells that results in the largest possible potential difference, we evaluate the potential difference between any two given half-cells: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ By pairing the half-cell with the highest potential value with the half-cell with the lowest potential value, we obtain the maximum mathematical potential difference: $$E^\circ_{\text{cell, max}} = E^\circ_{\text{max}} - E^\circ_{\text{min}}$$ Step 3 - Calculate the Potential for Each Option Combination Let us compute the cell potentials for the combinations listed in the options to verify which one gives the largest value: * **For Option (A) - Combination of i and iii:** Here, half-cell (iii) acts as the oxidation anode ($E^\circ_{\text{ox}} = +1.5\text{ V}$) and half-cell (i) acts as the reduction cathode ($E^\circ_{\text{red}} = -1.5\text{ V}$). Subtracting the standard reduction potential of half-cell (i) from the standard oxidation potential of half-cell (iii) to find the total difference: $$E^\circ_{\text{cell}} = E^\circ_{\text{ox}}(\text{iii}) - E^\circ_{\text{red}}(\text{i})$$ $$E^\circ_{\text{cell}} = 1.5\text{ V} - (-1.5\text{ V}) = \mathbf{3.0\text{ V}}$$ * **For Option (B) - Combination of i and iv:** Here, half-cell (iv) acts as the cathode ($E^\circ_{\text{red}} = +0.5\text{ V}$) and half-cell (i) acts as the anode ($E^\circ_{\text{red}} = -1.5\text{ V}$): $$E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{iv}) - E^\circ_{\text{red}}(\text{i})$$ $$E^\circ_{\text{cell}} = 0.5\text{ V} - (-1.5\text{ V}) = \mathbf{2.0\text{ V}}$$ * **For Option (C) - Combination of iii and iv:** Here, half-cell (iv) acts as the cathode ($E^\circ_{\text{red}} = +0.5\text{ V}$) and half-cell (iii) acts as the anode ($E^\circ_{\text{red}} = -1.5\text{ V}$): $$E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{iv}) - E^\circ_{\text{red}}(\text{iii})$$ $$E^\circ_{\text{cell}} = 0.5\text{ V} - (-1.5\text{ V}) = \mathbf{2.0\text{ V}}$$ * **For Option (D) - Combination of ii and iv:** Here, half-cell (iv) acts as the cathode ($E^\circ_{\text{red}} = +0.5\text{ V}$) and half-cell (ii) acts as the anode ($E^\circ_{\text{red}} = -0.5\text{ V}$): $$E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{iv}) - E^\circ_{\text{red}}(\text{ii})$$ $$E^\circ_{\text{cell}} = 0.5\text{ V} - (-0.5\text{ V}) = \mathbf{1.0\text{ V}}$$ Step 4 - Evaluate and Explain the Options * **Option (A) is correct:** Combining half-cell (i) and half-cell (iii) maximizes the difference between the given electrode potentials, producing the largest total cell potential of $3.0\text{ V}$. * **Option (B) is incorrect:** This combination yields a potential of $2.0\text{ V}$, which is smaller than $3.0\text{ V}$. * **Option (C) is incorrect:** This combination yields a potential of $2.0\text{ V}$, which is smaller than $3.0\text{ V}$. * **Option (D) is incorrect:** This combination yields a potential of $1.0\text{ V}$, which is the smallest potential among the choices. $$\text{Correct Option: } \boxed{\text{A}}$$