The cell representations of the above reaction and are: — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Electrode Reactions and the Net Cell Reaction We are given the net equilibrium reaction: $$\ce{Ag^+(aq) + Cl^-(aq) <=> AgCl(s)}$$ To construct an electrochemical cell based on this process, we can split it into two separate half-reactions: * **Anode Half-Reaction (Oxidation):** A silver metal electrode reacts with chloride ions in solution to form insoluble silver chloride, releasing one electron: $$\ce{Ag(s) + Cl^-(aq) -> AgCl(s) + e^-}$$ * **Cathode Half-Reaction (Reduction):** Silver ions in solution are reduced on the surface of a silver electrode, accepting one electron: $$\ce{Ag^+(aq) + e^- -> Ag(s)}$$ Adding these two half-reactions together yields the net overall cell reaction: $$\ce{Ag^+(aq) + Cl^-(aq) -> AgCl(s)}$$ The number of moles of electrons transferred in this balanced reaction is: $$n = 1$$ Step 2 - Calculate the Standard Gibbs Free Energy Change ($\Delta G^\circ$) of the Reaction Using the given standard Gibbs free energy of formation ($\Delta G^\circ_f$) values for the reactants and products: $$\Delta G^\circ_f(\ce{Ag^+(aq)}) = +77\text{ kJ/mol}$$ $$\Delta G^\circ_f(\ce{Cl^-(aq)}) = -129\text{ kJ/mol}$$ $$\Delta G^\circ_f(\ce{AgCl(s)}) = -109\text{ kJ/mol}$$ The standard Gibbs free energy change ($\Delta G^\circ$) for the overall reaction is calculated as: $$\Delta G^\circ = \Delta G^\circ_f(\ce{AgCl(s)}) - \left[ \Delta G^\circ_f(\ce{Ag^+(aq)}) + \Delta G^\circ_f(\ce{Cl^-(aq)}) \right]$$ Substitute the given values into the equation: $$\Delta G^\circ = -109\text{ kJ/mol} - \left[ 77\text{ kJ/mol} + (-129\text{ kJ/mol}) \right]$$ $$\Delta G^\circ = -109\text{ kJ/mol} - (-52\text{ kJ/mol})$$ $$\Delta G^\circ = -57\text{ kJ/mol} = -57,000\text{ J/mol}$$ Step 3 - Calculate the Standard Cell Potential ($E^\circ_{\text{cell}}$) The relationship between the standard Gibbs free energy change and the standard cell potential is given by: $$\Delta G^\circ = -n F E^\circ_{\text{cell}}$$ Where: * $n = 1$ mole of electrons * $F \approx 96,500\text{ C/mol}$ (Faraday's constant) Rearranging the equation to solve for $E^\circ_{\text{cell}}$: $$E^\circ_{\text{cell}} = -\frac{\Delta G^\circ}{n F}$$ Substitute the calculated values into the formula: $$E^\circ_{\text{cell}} = -\frac{-57,000\text{ J/mol}}{1 \times 96,500\text{ C/mol}}$$ $$E^\circ_{\text{cell}} = \frac{57,000}{96,500}\text{ V} \approx 0.5907\text{ V} \approx \boxed{0.59\text{ V}}$$ Step 4 - Formulate the IUPAC Cell Representation According to IUPAC conventions, the anode (oxidation half-cell) is written on the left, and the cathode (reduction half-cell) is written on the right. 1. **Anode Representation:** The anode is a metal-metal insoluble salt electrode. Silver metal ($\ce{Ag}$) is in contact with solid silver chloride ($\ce{AgCl}$), which is immersed in an aqueous solution containing chloride ions ($\ce{Cl^-}$): $$\ce{Ag(s) | AgCl(s) | Cl^-(aq)}$$ 2. **Cathode Representation:** The cathode is a standard metal-ion electrode. Aqueous silver ions ($\ce{Ag^+}$) are in contact with silver metal ($\ce{Ag}$): $$\ce{Ag^+(aq) | Ag(s)}$$ Combining both half-cells with a double vertical line representing the salt bridge yields the final cell notation: $$\ce{Ag(s) | AgCl(s) | Cl^-(aq) || Ag^+(aq) | Ag(s)}$$ This representation along with $E^\circ_{\text{cell}} = 0.59\text{ V}$ corresponds to option A. $$\text{Correct Option: } \boxed{A}$$