[Four-digit Integer] The electrode potential (in millivolts) of 2Ag(s) + S^2-(aq) -> Ag2S(s) + 2e^- — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Hydrogen Ion Concentration from pH We are given that the solution is buffered at $\text{pH} = 3$. The concentration of hydrogen ions ($[\ce{H^+}]$) is calculated as: $$[\ce{H^+}] = 10^{-\text{pH}}$$ $$[\ce{H^+}] = 10^{-3}\ \text{M}$$ Step 2 - Calculate the Sulfide Ion Concentration from \ce{H2S} Equilibria Hydrogen sulfide ($\ce{H2S}$) is a weak diprotic acid that undergoes stepwise dissociation in aqueous solution: $$\ce{H2S(aq) <=> H^+(aq) + HS^-(aq)} \quad K_{a1} = 10^{-8}$$ $$\ce{HS^-(aq) <=> H^+(aq) + S^{2-}(aq)} \quad K_{a2} = 10^{-13}$$ The overall dissociation of $\ce{H2S}$ is represented by the sum of these two steps: $$\ce{H2S(aq) <=> 2H^+(aq) + S^{2-}(aq)}$$ The overall equilibrium constant ($K$) is the product of the stepwise constants: $$K = K_{a1} \cdot K_{a2} = 10^{-8} \times 10^{-13} = 10^{-21}$$ The equilibrium expression is: $$\frac{[\ce{H^+}]^2 [\ce{S^{2-}}]}{[\ce{H2S}]} = K_{a1} \cdot K_{a2}$$ Rearranging the formula to solve for the concentration of sulfide ions ($[\ce{S^{2-}}]$): $$[\ce{S^{2-}}] = \frac{K_{a1} \cdot K_{a2} \cdot [\ce{H2S}]}{[\ce{H^+}]^2}$$ Substituting $[\ce{H2S}] = 0.1\text{ M}$ and $[\ce{H^+}] = 10^{-3}\text{ M}$ into the equation: $$[\ce{S^{2-}}] = \frac{10^{-8} \times 10^{-13} \times 0.1\ \text{M}}{(10^{-3}\ \text{M})^2}$$ $$[\ce{S^{2-}}] = \frac{10^{-22}}{10^{-6}}\ \text{M} = 10^{-16}\ \text{M}$$ Step 3 - Calculate the Silver Ion Concentration from solubility product The solubility equilibrium for the sparingly soluble silver sulfide ($\ce{Ag2S}$) in water is: $$\ce{Ag2S(s) <=> 2Ag^+(aq) + S^{2-}(aq)}$$ The solubility product constant ($K_{\text{sp}}$) is given by: $$K_{\text{sp}} = [\ce{Ag^+}]^2 [\ce{S^{2-}}]$$ Substituting the given values $K_{\text{sp}}(\ce{Ag2S}) = 4 \times 10^{-48}$ and our calculated concentration of $[\ce{S^{2-}}] = 10^{-16}\text{ M}$: $$4 \times 10^{-48} = [\ce{Ag^+}]^2 \times 10^{-16}$$ $$[\ce{Ag^+}]^2 = 4 \times 10^{-32}$$ $$[\ce{Ag^+}] = \sqrt{4 \times 10^{-32}} = 2 \times 10^{-16}\ \text{M}$$ Step 4 - Calculate the Reduction Potential of the \ce{Ag^+/Ag} couple The reduction reaction is: $$\ce{Ag^+(aq) + e^- -> Ag(s)}$$ At $298\text{ K}$, we use the Nernst equation for this single-electrode reduction potential: $$E_{\text{red}} = E^\circ_{\ce{Ag^+/Ag}} - \frac{2.303 RT}{F} \log_{10} \left(\frac{1}{[\ce{Ag^+}]}\right)$$ $$E_{\text{red}} = E^\circ_{\ce{Ag^+/Ag}} + 0.06 \log_{10} [\ce{Ag^+}]$$ Substituting the standard potential $E^\circ_{\ce{Ag^+/Ag}} = 0.80\text{ V}$ and $[\ce{Ag^+}] = 2 \times 10^{-16}\text{ M}$: $$E_{\text{red}} = 0.80\text{ V} + 0.06\text{ V} \times \log_{10}(2 \times 10^{-16})$$ $$E_{\text{red}} = 0.80\text{ V} + 0.06\text{ V} \times (\log_{10} 2 - 16)$$ Using the given value $\log_{10} 2 = 0.3$: $$E_{\text{red}} = 0.80\text{ V} + 0.06\text{ V} \times (0.3 - 16)$$ $$E_{\text{red}} = 0.80\text{ V} + 0.06\text{ V} \times (-15.7)$$ $$E_{\text{red}} = 0.80\text{ V} - 0.942\text{ V} = -0.142\text{ V}$$ Step 5 - Calculate the Oxidation Potential in Millivolts The given half-reaction is an oxidation reaction: $$\ce{2Ag(s) + S^{2-}(aq) -> Ag2S(s) + 2e^-}$$ Since the solution contains solid $\ce{Ag2S}$ at equilibrium, the potential of this metal-insoluble salt-anion electrode is thermodynamically coupled with the silver-silver ion reduction system. The oxidation potential ($E_{\text{ox}}$) is equal in magnitude but opposite in sign to the reduction potential ($E_{\text{red}}$): $$E_{\text{ox}} = -E_{\text{red}}$$ $$E_{\text{ox}} = -(-0.142\text{ V}) = +0.142\text{ V}$$ Converting this potential into millivolts ($\text{mV}$): $$E_{\text{ox}} = 0.142\text{ V} \times 1000\text{ mV/V} = 142\text{ mV}$$ Since the question requires a four-digit integer format: $$\boxed{0142}$$