The E_cell for Ag(s) \ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Cell Type and Electrode Reactions The given cell representation is: $$\text{Ag}(s) \mid \text{AgI (satd)} \parallel \text{Ag}^+ (0.10\text{ M}) \mid \text{Ag}(s)$$ This is a metal-metal ion concentration cell. The electrodes are composed of identical silver ($\ce{Ag}$) metals, but they are in contact with different concentrations of silver ions ($\ce{Ag^+}$): * **Anode (Left Electrode - Oxidation):** $$\ce{Ag(s) -> Ag+(aq, anode) + e^-}$$ Here, the silver ions are generated from the dissociation of the sparingly soluble salt $\ce{AgI}$ in its saturated solution. Let their concentration be $[\ce{Ag+}]_{\text{anode}}$. * **Cathode (Right Electrode - Reduction):** $$\ce{Ag+(aq, cathode) + e^- -> Ag(s)}$$ Here, the concentration of silver ions is given as: $$[\ce{Ag+}]_{\text{cathode}} = 0.10\text{ M}$$ * **Net Cell Reaction:** $$\ce{Ag+(aq, cathode) -> Ag+(aq, anode)}$$ Step 2 - Apply the Nernst Equation Since the electrodes in both compartments are made of the same metal, the standard cell potential ($E^\circ_{\text{cell}}$) is exactly zero: $$E^\circ_{\text{cell}} = 0\text{ V}$$ The number of moles of electrons transferred in the balanced reaction is $n = 1$. Using the Nernst equation at $25^\circ\text{C}$ ($298.15\text{ K}$): $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591\text{ V}}{n} \log_{10} \left( \frac{[\ce{Ag+}]_{\text{anode}}}{[\ce{Ag+}]_{\text{cathode}}} \right)$$ Substituting the known values: $$E_{\text{cell}} = 0\text{ V} - 0.0591\text{ V} \log_{10} \left( \frac{[\ce{Ag+}]_{\text{anode}}}{0.10\text{ M}} \right)$$ Step 3 - Calculate the Silver Ion Concentration in the Anode Compartment We are given that the measured cell voltage is $E_{\text{cell}} = 0.413\text{ V}$. Substitute this into our Nernst equation: $$0.413\text{ V} = -0.0591\text{ V} \log_{10} \left( \frac{[\ce{Ag+}]_{\text{anode}}}{0.10\text{ M}} \right)$$ Rearranging to solve for the logarithm: $$\log_{10} \left( \frac{[\ce{Ag+}]_{\text{anode}}}{0.10\text{ M}} \right) = -\frac{0.413\text{ V}}{0.0591\text{ V}}$$ $$\log_{10} \left( \frac{[\ce{Ag+}]_{\text{anode}}}{0.10\text{ M}} \right) \approx -7.0$$ Taking the antilogarithm of both sides: $$\frac{[\ce{Ag+}]_{\text{anode}}}{0.10\text{ M}} = 10^{-7}$$ $$[\ce{Ag+}]_{\text{anode}} = 10^{-7} \times 0.10\text{ M} = 10^{-8}\text{ M}$$ Step 4 - Determine the Solubility Product Constant ($K_{\text{sp}}$) of $\ce{AgI}$ The silver ions at the anode are produced solely from the dissociation of the sparingly soluble salt silver iodide ($\ce{AgI}$): $$\ce{AgI(s) <=> Ag+(aq) + I^-(aq)}$$ Since $\ce{AgI}$ dissociates in a 1:1 stoichiometric ratio, the concentration of iodide ions ($\ce{I^-}$) must be equal to the concentration of silver ions ($\ce{Ag^+}$) at solubility equilibrium: $$[\ce{I^-}] = [\ce{Ag+}]_{\text{anode}} = 10^{-8}\text{ M}$$ The solubility product constant ($K_{\text{sp}}$) of $\ce{AgI}$ is defined as: $$K_{\text{sp}} = [\ce{Ag+}][\ce{I^-}]$$ Substituting our calculated equilibrium concentrations: $$K_{\text{sp}} = \left(10^{-8}\text{ M}\right) \times \left(10^{-8}\text{ M}\right)$$ $$K_{\text{sp}} = \boxed{1.0 \times 10^{-16}}$$ Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** $1.0 \times 10^{-8}$ represents the concentration of silver ions ($[\ce{Ag+}]$) in the saturated solution, not the solubility product constant. * **Option (B) is incorrect:** $1.0 \times 10^{-7}$ is the concentration ratio $[\ce{Ag+}]_{\text{anode}} / [\ce{Ag+}]_{\text{cathode}}$, which represents a step in our calculations rather than the final answer. * **Option (C) is incorrect:** $1.0 \times 10^{-14}$ is mathematically incorrect and represents a calculation error in powers of ten. * **Option (D) is correct:** As mathematically shown, the solubility product constant of $\ce{AgI}$ under the given conditions is exactly $1.0 \times 10^{-16}$. $$\text{Correct Option: } \boxed{\text{D}}$$