[Single-digit Integer] If it is desired to construct the following galvanic cell: Ag(s) — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the Type of Electrochemical Cell The given cell representation is: $$\ce{Ag(s) \mid Ag^+(sat. AgI) \parallel Ag^+(sat. AgCl, } x \times 10^{-4}\text{ M }\ce{Cl^-) \mid Ag(s)}$$ This is an **electrode-metal concentration cell** because both the anode (left side) and cathode (right side) consist of the same metal ($\ce{Ag}$) and are in contact with solutions containing the same ionic species ($\ce{Ag^+}$), but at different concentrations. For any concentration cell: * The standard cell potential ($E^\circ_{\text{cell}}$) is exactly: $$E^\circ_{\text{cell}} = 0\text{ V}$$ * The overall cell reaction is the transfer of silver ions from the higher concentration (cathode) to the lower concentration (anode) compartment: $$\ce{Ag^+(cathode) -> Ag^+(anode)}$$ Step 2 - Calculate the Silver Ion Concentration at the Anode The anode compartment contains a pure saturated solution of the sparingly soluble salt, silver iodide ($\ce{AgI}$). The solubility equilibrium is: $$\ce{AgI(s) <=> Ag^+(aq) + I^-(aq)}$$ Since the solution is pure and saturated, the concentration of silver ions at the anode is: $$[\ce{Ag^+}]_{\text{anode}} = \sqrt{K_{\text{sp}}(\ce{AgI})}$$ Substituting the given value $K_{\text{sp}}(\ce{AgI}) = 8.1 \times 10^{-17}$: $$[\ce{Ag^+}]_{\text{anode}} = \sqrt{8.1 \times 10^{-17}\text{ M}^2} = \sqrt{81 \times 10^{-18}\text{ M}^2}$$ $$[\ce{Ag^+}]_{\text{anode}} = 9 \times 10^{-9}\text{ M}$$ Step 3 - Determine the Silver Ion Concentration at the Cathode At $298\text{ K}$, the Nernst equation for this concentration cell (with $n = 1$ electron transferred) is: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{F} \log_{10} \left(\frac{[\ce{Ag^+}]_{\text{anode}}}{[\ce{Ag^+}]_{\text{cathode}}}\right)$$ Substituting $E^\circ_{\text{cell}} = 0\text{ V}$ and the given slope parameter $\frac{2.303 RT}{F} = 0.06\text{ V}$: $$E_{\text{cell}} = 0.06\text{ V} \times \log_{10} \left(\frac{[\ce{Ag^+}]_{\text{cathode}}}{[\ce{Ag^+}]_{\text{anode}}}\right)$$ Now, substitute the given cell EMF ($E_{\text{cell}} = 0.102\text{ V}$) and our calculated value of $[\ce{Ag^+}]_{\text{anode}}$: $$0.102\text{ V} = 0.06\text{ V} \times \log_{10} \left(\frac{[\ce{Ag^+}]_{\text{cathode}}}{9 \times 10^{-9}\text{ M}}\right)$$ Divide both sides by $0.06\text{ V}$: $$\log_{10} \left(\frac{[\ce{Ag^+}]_{\text{cathode}}}{9 \times 10^{-9}\text{ M}}\right) = \frac{0.102}{0.06} = 1.7$$ Taking the antilog (base 10) of both sides: $$\frac{[\ce{Ag^+}]_{\text{cathode}}}{9 \times 10^{-9}\text{ M}} = 10^{1.7}$$ Using logarithmic approximations, we know that: $$10^{1.7} = 10^1 \times 10^{0.7} \approx 10 \times 5 = 50$$ $$\implies [\ce{Ag^+}]_{\text{cathode}} = 9 \times 10^{-9}\text{ M} \times 10^{1.7} = 9 \times 10^{-9}\text{ M} \times 50.119$$ $$[\ce{Ag^+}]_{\text{cathode}} \approx 4.5 \times 10^{-7}\text{ M}$$ Step 4 - Calculate the Chloride Ion Concentration and Solve for $x$ The cathode compartment contains saturated silver chloride ($\ce{AgCl}$) along with additional chloride ions ($\ce{Cl^-}$). The solubility equilibrium for $\ce{AgCl}$ in this half-cell is: $$\ce{AgCl(s) <=> Ag^+(aq) + Cl^-(aq)}$$ The solubility product constant ($K_{\text{sp}}$) of $\ce{AgCl}$ is: $$K_{\text{sp}}(\ce{AgCl}) = [\ce{Ag^+}]_{\text{cathode}} [\ce{Cl^-}]$$ Rearranging the formula to solve for the concentration of chloride ions ($[\ce{Cl^-}]$): $$[\ce{Cl^-}] = \frac{K_{\text{sp}}(\ce{AgCl})}{[\ce{Ag^+}]_{\text{cathode}}}$$ Substituting the given value of $K_{\text{sp}}(\ce{AgCl}) = 1.8 \times 10^{-10}$ and our calculated $[\ce{Ag^+}]_{\text{cathode}} = 4.5 \times 10^{-7}\text{ M}$: $$[\ce{Cl^-}] = \frac{1.8 \times 10^{-10}}{4.5 \times 10^{-7}}\text{ M}$$ $$[\ce{Cl^-}] = 0.4 \times 10^{-3}\text{ M} = 4 \times 10^{-4}\text{ M}$$ Comparing this result with the given expression $[\ce{Cl^-}] = x \times 10^{-4}\text{ M}$, we find: $$x = \boxed{4}$$