Consider the cell: . The cell potential: — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Electrode Types and Half-Cell Reactions The given electrochemical cell is represented using standard IUPAC notation: $$\ce{Ag(s), AgCl(s) \mid KCl(0.1\ M) \parallel Hg2Cl2(s), Hg(l)}$$ This setup is composed of two metal-insoluble salt-anion electrodes sharing a common electrolyte solution of chloride ions ($\ce{Cl^-}$) from $\ce{KCl(aq)}$: 1. **Anode (Left Electrode):** Silver-silver chloride electrode ($\ce{Ag(s), AgCl(s) \mid Cl^-(aq)}$), where oxidation takes place. 2. **Cathode (Right Electrode):** Calomel electrode ($\ce{Cl^-(aq) \mid Hg2Cl2(s), Hg(l)}$), where reduction takes place. Step 2 - Derive the Nernst Equation for the Anode (Silver-Silver Chloride Electrode) At the anode, metallic silver undergoes oxidation in the presence of chloride ions to form solid silver chloride: $$\ce{Ag(s) + Cl^-(aq) -> AgCl(s) + e^-}$$ The reduction potential of this half-cell ($E_{\text{anode}}$) is given by the Nernst equation: $$E_{\text{anode}} = E_{\ce{Ag/AgCl/Cl^-}} = E^\circ_{\ce{Ag/AgCl/Cl^-}} - \frac{RT}{F} \ln a_{\ce{Cl^-}}$$ Assuming ideal behavior where the activity of chloride ions is represented by their molar concentration ($a_{\ce{Cl^-}} \approx [\ce{Cl^-}]$): $$E_{\text{anode}} = E^\circ_{\ce{Ag/AgCl/Cl^-}} - \frac{RT}{F} \ln [\ce{Cl^-}]$$ Step 3 - Derive the Nernst Equation for the Cathode (Calomel Electrode) At the cathode, mercurous chloride (calomel) undergoes reduction to liquid mercury, releasing chloride ions: $$\ce{Hg2Cl2(s) + 2e^- -> 2Hg(l) + 2Cl^-(aq)}$$ Dividing the reaction by $2$ to represent it per mole of electrons transferred ($n = 1$): $$\ce{1/2 Hg2Cl2(s) + e^- -> Hg(l) + Cl^-(aq)}$$ The reduction potential of this half-cell ($E_{\text{cathode}}$) is: $$E_{\text{cathode}} = E_{\text{calomel}} = E^\circ_{\text{calomel}} - \frac{RT}{F} \ln \left( \frac{a_{\ce{Hg}} \cdot a_{\ce{Cl^-}}}{(a_{\ce{Hg2Cl2}})^{1/2}} \right)$$ Since pure liquids and pure solids have thermodynamic activities equal to unity ($a_{\ce{Hg(l)}} = 1$ and $a_{\ce{Hg2Cl2(s)}} = 1$): $$E_{\text{cathode}} = E^\circ_{\text{calomel}} - \frac{RT}{F} \ln [\ce{Cl^-}]$$ Step 4 - Calculate the Overall Cell Potential ($E_{\text{cell}}$) The overall potential of the electrochemical cell is the difference between the reduction potentials of the cathode and the anode: $$E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$$ Substituting the Nernst expressions from Step 2 and Step 3 into this equation: $$E_{\text{cell}} = \left( E^\circ_{\text{calomel}} - \frac{RT}{F} \ln [\ce{Cl^-}] \right) - \left( E^\circ_{\ce{Ag/AgCl/Cl^-}} - \frac{RT}{F} \ln [\ce{Cl^-}] \right)$$ Expanding and simplifying: $$E_{\text{cell}} = E^\circ_{\text{calomel}} - \frac{RT}{F} \ln [\ce{Cl^-}] - E^\circ_{\ce{Ag/AgCl/Cl^-}} + \frac{RT}{F} \ln [\ce{Cl^-}]$$ $$E_{\text{cell}} = E^\circ_{\text{calomel}} - E^\circ_{\ce{Ag/AgCl/Cl^-}}$$ The terms involving the concentration of chloride ions ($\ce{Cl^-}$) cancel out completely. Therefore: $$E_{\text{cell}} = E^\circ_{\text{cell}}$$ Step 5 - Evaluate and Explain Each Option * **Option (A) is incorrect:** The cell potential does not increase on increasing the concentration of $\ce{Cl^-}$ ions because the concentration of $\ce{Cl^-}$ has no net effect on the EMF due to complete cancellation. * **Option (B) is incorrect:** The cell potential does not decrease on decreasing the concentration of $\ce{Cl^-}$ ions for the same reason. * **Option (C) is correct:** As mathematically demonstrated in Step 4, the logarithmic concentration terms of the chloride ions cancel out completely, making the cell potential entirely independent of the concentration of $\ce{Cl^-}$ ions. * **Option (D) is correct:** Pure solids ($\ce{AgCl(s)}$, $\ce{Ag(s)}$, and $\ce{Hg2Cl2(s)}$) and pure liquids ($\ce{Hg(l)}$) have activities equal to $1$, meaning their chemical potentials are intensive and independent of the physical masses or quantities of these substances present in the electrodes. Thus, the cell potential is independent of the amounts of $\ce{AgCl}$ and $\ce{Hg2Cl2}$. $$\text{Correct Options: } \boxed{C, D}$$