The potential of electrode in a saturated solution of at 298 K is: — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Determine the Concentration of Silver Ions in Saturated \ce{AgI} We are given a saturated solution of the sparingly soluble salt, silver iodide ($\ce{AgI}$), at $298\text{ K}$. The solubility equilibrium for $\ce{AgI}$ in water is: $$\ce{AgI(s) <=> Ag^+(aq) + I^-(aq)}$$ The solubility product constant ($K_{\text{sp}}$) of $\ce{AgI}$ is defined as: $$K_{\text{sp}} = [\ce{Ag^+}][\ce{I^-}]$$ Since $\ce{AgI}$ dissociates in a $1:1$ stoichiometric ratio, the concentration of silver ions ($[\ce{Ag^+}]$) is equal to the concentration of iodide ions ($[\ce{I^-}]$) in a pure saturated solution: $$[\ce{Ag^+}] = [\ce{I^-}] = \sqrt{K_{\text{sp}}}$$ Substitute the given value $K_{\text{sp}}(\ce{AgI}) = 6.4 \times 10^{-17}$: $$[\ce{Ag^+}] = \sqrt{6.4 \times 10^{-17}\text{ M}^2} = \sqrt{64 \times 10^{-18}\text{ M}^2}$$ $$[\ce{Ag^+}] = 8 \times 10^{-9}\text{ M}$$ Step 2 - Formulate the Nernst Equation for the \ce{Ag^+ | Ag} Electrode The reduction half-reaction taking place at the silver electrode is: $$\ce{Ag^+(aq) + e^- -> Ag(s)}$$ At $298\text{ K}$, the Nernst equation for this single-electrode reduction potential ($E_{\ce{Ag^+/Ag}}$) is expressed as: $$E_{\ce{Ag^+/Ag}} = E^\circ_{\ce{Ag^+/Ag}} - \frac{2.303 RT}{nF} \log_{10} \left(\frac{1}{[\ce{Ag^+}]}\right)$$ Using the given slope factor parameter $\frac{2.303 RT}{F} = 0.06\text{ V}$ and substituting the number of electrons transferred $n = 1$: $$E_{\ce{Ag^+/Ag}} = E^\circ_{\ce{Ag^+/Ag}} - 0.06\text{ V} \times \log_{10} \left(\frac{1}{[\ce{Ag^+}]}\right)$$ $$E_{\ce{Ag^+/Ag}} = E^\circ_{\ce{Ag^+/Ag}} + 0.06\text{ V} \times \log_{10} [\ce{Ag^+}]$$ Step 3 - Calculate the Non-Standard Electrode Potential We substitute the standard reduction potential $E^\circ_{\ce{Ag^+/Ag}} = 0.80\text{ V}$ and the silver ion concentration $[\ce{Ag^+}] = 8 \times 10^{-9}\text{ M}$ into the equation: $$E_{\ce{Ag^+/Ag}} = 0.80\text{ V} + 0.06\text{ V} \times \log_{10}(8 \times 10^{-9})$$ Let us compute the logarithmic term: $$\log_{10}(8 \times 10^{-9}) = \log_{10}(8) + \log_{10}(10^{-9})$$ $$\log_{10}(8 \times 10^{-9}) = 3\log_{10}(2) - 9$$ Using the standard approximation $\log_{10}(2) \approx 0.3010$: $$\log_{10}(8 \times 10^{-9}) \approx 3(0.3010) - 9 = 0.9030 - 9 = -8.097$$ Now, substitute this back into the potential expression: $$E_{\ce{Ag^+/Ag}} = 0.80\text{ V} + 0.06\text{ V} \times (-8.097)$$ $$E_{\ce{Ag^+/Ag}} = 0.80\text{ V} - 0.48582\text{ V}$$ $$E_{\ce{Ag^+/Ag}} \approx \boxed{+0.314\text{ V}}$$ Step 4 - Evaluate and Explain each Option * **Option (A) is incorrect:** This option represents $-0.314\text{ V}$. This is incorrect because the high initial positive reduction potential of silver ($0.80\text{ V}$) is decreased by the low concentration of silver ions, but it remains positive ($+0.314\text{ V}$). A negative potential would represent a sign error in the calculation. * **Option (B) is correct:** As mathematically proven in Step 3, the potential of the electrode is exactly $+0.314\text{ V}$. * **Option (C) is incorrect:** This option represents $-0.172\text{ V}$. This specific distractor is obtained if a student forgets to take the square root of $K_{\text{sp}}$ and incorrectly assumes $[\ce{Ag^+}] = K_{\text{sp}} = 6.4 \times 10^{-17}\text{ M}$: $$E = 0.80\text{ V} + 0.06\text{ V} \times \log_{10}(6.4 \times 10^{-17})$$ $$E \approx 0.80\text{ V} + 0.06\text{ V} \times (-16.194) = 0.80\text{ V} - 0.972\text{ V} = -0.172\text{ V}$$ * **Option (D) is incorrect:** This option represents $+0.172\text{ V}$. This is the positive counterpart of the conceptual mistake described in option (C). $$\text{Correct Option: } \boxed{B}$$