The correct cell diagram for the following reaction and E° for the cell is 2AgBr(s) + (g) -> 2Ag(s) — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Half-Cell Reactions To construct the correct cell diagram, we must first break down the given overall spontaneous cell reaction into its corresponding oxidation (anodic) and reduction (cathodic) half-reactions: $$\ce{2AgBr(s) + H2(g) -> 2Ag(s) + 2H^+(aq) + 2Br^-(aq)}$$ 1. **Oxidation Half-Reaction (At the Anode):** Gaseous hydrogen ($\ce{H2}$) is oxidized to hydrogen ions ($\ce{H^+}$): $$\ce{H2(g) -> 2H^+(aq) + 2e^-}$$ Since hydrogen is a non-conducting gas, an inert platinum ($\ce{Pt}$) electrode is required to establish electrical contact with the solution. 2. **Reduction Half-Reaction (At the Cathode):** Insoluble silver bromide ($\ce{AgBr}$) in contact with metallic silver ($\ce{Ag}$) and bromide ions ($\ce{Br^-}$) undergoes reduction: $$\ce{2AgBr(s) + 2e^- -> 2Ag(s) + 2Br^-(aq)}$$ This constitutes a metal-insoluble salt-anion electrode system ($\ce{Ag | AgBr | Br^-}$). Step 2 - Apply IUPAC Rules for Cell Representation According to standard IUPAC conventions: * The anode (oxidation half-cell) is always written on the left side. * The cathode (reduction half-cell) is always written on the right side. * Phase boundaries are represented by single vertical lines ($\mid$). * The salt bridge connecting the two half-cells is represented by a double vertical line ($\parallel$). Let us represent each electrode compartment: 1. **Anode representation (left side):** The inert platinum terminal is in contact with hydrogen gas, which is in contact with the aqueous acid solution: $$\text{(Pt)} \ce{H2} \mid \ce{H^+}$$ 2. **Cathode representation (right side):** The aqueous bromide ions are in contact with the solid silver bromide phase, which is in contact with the solid silver metal electrode: $$\ce{Br^-} \mid \ce{AgBr} \mid \ce{Ag}$$ Combining both half-cells with the salt bridge yields the complete cell notation: $$\text{(Pt)} \ce{H2} \mid \ce{H^+} \parallel \ce{Br^-} \mid \ce{AgBr} \mid \ce{Ag}$$ Step 3 - Calculate the Standard Cell Potential ($E^\circ_{\text{cell}}$) The standard electromotive force ($E^\circ_{\text{cell}}$) of the cell is calculated using the formula: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ Substitute the given values: * Standard reduction potential of the cathode ($E^\circ_{\ce{AgBr/Ag,Br^-}}$) = $+0.10\text{ V}$ * Standard reduction potential of the standard hydrogen electrode anode ($E^\circ_{\ce{H^+/H2}}$) = $0.00\text{ V}$ $$E^\circ_{\text{cell}} = +0.10\text{ V} - 0.00\text{ V}$$ $$E^\circ_{\text{cell}} = \boxed{+0.10\text{ V}}$$ Step 4 - Evaluate the Options * **Option (A) is correct:** It correctly places the hydrogen oxidation anode on the left and the silver-silver bromide reduction cathode on the right in accordance with IUPAC rules. * **Option (B) is incorrect:** This is the reverse representation of the cell, which would represent a non-spontaneous cell reaction with a negative cell potential. * **Option (C) is incorrect:** This representation is physically incorrect and fails to follow the correct phase boundaries for the anode and cathode. * **Option (D) is incorrect:** This is a structurally incorrect cell diagram that misplaces the electrode materials and phase sequences. $$\text{Correct Option: } \boxed{\text{A}}$$