### Core Logic Limiting molar conductivity relies heavily on the charge and hydrodynamic radius of the hydrated ion. Let us verify the standard experimental limiting molar ionic conductivities (lambda^circ$\lambda^{\circ}$) at 298text K$298\text{ K}$: * H^+: 349.8text S cm^2text mol^-1$H^{+}: 349.8\text{ S cm}^2\text{ mol}^{-1}$ (exhibits Grotthuss proton-hopping conduction mechanism) * Ca^2+: 119.0text S cm^2text mol^-1$Ca^{2+}: 119.0\text{ S cm}^2\text{ mol}^{-1}$ * Mg^2+: 106.1text S cm^2text mol^-1$Mg^{2+}: 106.1\text{ S cm}^2\text{ mol}^{-1}$ * K^+: 73.5text S cm^2text mol^-1$K^{+}: 73.5\text{ S cm}^2\text{ mol}^{-1}$ * Na^+: 50.1text S cm^2text mol^-1$Na^{+}: 50.1\text{ S cm}^2\text{ mol}^{-1}$ ### Step 1: Trend Layout Arranging these values in descending order yields: H^+ > Ca^2+ > Mg^2+ > K^+ > Na^+$H^{+} > Ca^{2+} > Mg^{2+} > K^{+} > Na^{+}$ ### Pattern Recognition Shortcut: H^+$H^{+}$ always has the absolute highest value due to its unique proton-hopping transport system. For metal ions, a higher ionic charge boosts conductivity (M^2+ > M^+$M^{2+} > M^{+}$), and within a group, a smaller hydrated radius (larger bare ion) increases mobility (K^+ > Na^+$K^{+} > Na^{+}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula Lambda_m^circ(NH_4OH) = lambda^circ(NH_4^+) + lambda^circ(OH^-) quad text(Kohlrausch's Law)$\Lambda_m^\circ(NH_4OH) = \lambda^\circ(NH_4^+) + \lambda^\circ(OH^-) \quad \text{(Kohlrausch's Law)}$ alpha = fracLambda_m^cLambda_m^circ$\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}$ ### Core Logic 1. Find the limiting molar conductance of NH_4^+$NH_4^+$ using the NH_4Cl$NH_4Cl$ data: Lambda_m^circ(NH_4Cl) = lambda^circ(NH_4^+) + lambda^circ(Cl^-) = 185$\Lambda_m^\circ(NH_4Cl) = \lambda^\circ(NH_4^+) + \lambda^\circ(Cl^-) = 185$ lambda^circ(NH_4^+) = 185 - 70 = 115 mathrm~S cdot cm^2 cdot mol^-1$\lambda^\circ(NH_4^+) = 185 - 70 = 115 \mathrm{~S \cdot cm^2 \cdot mol^{-1}}$ 2. Calculate Lambda_m^circ$\Lambda_m^\circ$ for the weak electrolyte ammonium hydroxide (NH_4OH$NH_4OH$): Lambda_m^circ(NH_4OH) = 115 + 170 = 285 mathrm~S cdot cm^2 cdot mol^-1$\Lambda_m^\circ(NH_4OH) = 115 + 170 = 285 \mathrm{~S \cdot cm^2 \cdot mol^{-1}}$ 3. Evaluate the degree of dissociation (alpha$\alpha$): alpha = frac85.5285 = 0.3 = 3 times 10^-1$\alpha = \frac{85.5}{285} = 0.3 = 3 \times 10^{-1}$ Comparing with the format mathbfx times 10^-1$\mathbf{x} \times 10^{-1}$, the value of mathbfx$\mathbf{x}$ is **3**. ### Pattern Recognition Kohlrausch's law allows direct algebraic recombination of ion conductances. Always construct the targeted weak base compound value by filtering out the spectator chloride contribution from the initial salt parameters. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula textNet Charging Reaction: 2PbSO_4(s) + 2H_2O(l) rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)$\text{Net Charging Reaction: } 2PbSO_4(s) + 2H_2O(l) \rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)$ ### Core Logic During the **charging** cycle, the discharge chemical reactions are driven in reverse: * **At Anode:** Lead sulfate (PbSO_4$PbSO_4$, where Lead is +2$+2$) is reduced back to metallic lead (Pb$Pb$, oxidation state 0$0$): x_1 = +2 rightarrow y_1 = 0$x_1 = +2 \rightarrow y_1 = 0$ * **At Cathode:** Lead sulfate (PbSO_4$PbSO_4$, where Lead is +2$+2$) is oxidized back into lead dioxide (PbO_2$PbO_2$, where Lead is +4$+4$): x_2 = +2 rightarrow y_2 = +4$x_2 = +2 \rightarrow y_2 = +4$ Thus, the values are +2, 0, +2, +4$+2, 0, +2, +4$. ### Pattern Recognition Be careful with wording! Discharging consumes Pb$Pb$ and PbO_2$PbO_2$ to create PbSO_4$PbSO_4$. **Charging** does the exact opposite, converting PbSO_4$PbSO_4$ (+2$+2$) back into its parent elements (0$0$ and +4$+4$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula textEase of discharge at Cathode propto textStandard Reduction Potential (E^0)$\text{Ease of discharge at Cathode} \propto \text{Standard Reduction Potential } (E^0)$ ### Core Logic - At the cathode, the metal ion with the highest standard reduction potential (E^0$E^0$) gets reduced and deposited first. Arranging the given potentials: E^0_textAg^+/textAg (0.80textV) > E^0_textHg_2^2+/textHg (0.79textV) > E^0_textCu^2+/textCu (0.24textV) $E^0_{\text{Ag}^+/\text{Ag}} (0.80\text{V}) > E^0_{\text{Hg}_2^{2+}/\text{Hg}} (0.79\text{V}) > E^0_{\text{Cu}^{2+}/\text{Cu}} (0.24\text{V}) $ Thus, deposition follows the order textAg ightarrow textHg ightarrow textCu$\text{Ag} ightarrow \text{Hg} ightarrow \text{Cu}$ as voltage is steadily increased, confirming Statement I. - For textMg^2+$\text{Mg}^{2+}$, its reduction potential is highly negative (-2.37text V$-2.37\text{ V}$), much lower than that of water (-0.83text V$-0.83\text{ V}$). Consequently, water undergoes reduction at the cathode instead of magnesium: 2textH_2textO + 2e^- ightarrow textH_2(g) + 2textOH^- $2\text{H}_2\text{O} + 2e^- ightarrow \text{H}_2(g) + 2\text{OH}^- $ This results in the evolution of **Hydrogen gas** at the cathode, not oxygen gas. Oxygen gas is evolved at the *anode* via water oxidation. Thus, Statement II is incorrect. [cite: 1042, 1044] ### Step 1: Conclusion Match Since Statement I is correct and Statement II is incorrect, we select option (2). ### Pattern Recognition Cathode vs Anode Gas Trap: During the aqueous electrolysis of highly reactive metals (Groups 1, 2, and textAl$\text{Al}$), textH_2$\text{H}_2$ gas is always discharged at the cathode due to water's easier reduction profile. Oxygen gas (textO_2$\text{O}_2$) is an anodic product generated by water oxidation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula Delta G^0 = -nFE^0$\Delta G^{0} = -nFE^{0}$ ### Core Logic Using Gibbs free energy changes for individual steps to find the target reduction potential: 1. Ag^+ + e^- rightarrow Ag quad Delta G_1^0 = -1Fx$Ag^{+} + e^{-} \rightarrow Ag \quad \Delta G_1^0 = -1Fx$ 2. Fe^2+ + 2e^- rightarrow Fe quad Delta G_2^0 = -2Fy$Fe^{2+} + 2e^{-} \rightarrow Fe \quad \Delta G_2^0 = -2Fy$ 3. Fe^3+ + 3e^- rightarrow Fe quad Delta G_3^0 = -3Fz$Fe^{3+} + 3e^{-} \rightarrow Fe \quad \Delta G_3^0 = -3Fz$ For the conversion of Fe^2+ rightarrow Fe^3+ + e^-$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$, we compute the free energy change as: Delta G^0 = Delta G_2^0 - Delta G_3^0 = -2Fy - (-3Fz) = 3Fz - 2Fy$\Delta G^0 = \Delta G_2^0 - \Delta G_3^0 = -2Fy - (-3Fz) = 3Fz - 2Fy$ Thus, E^0_Fe^2+/Fe^3+ = 2y - 3z$E^{0}_{Fe^{2+}/Fe^{3+}} = 2y - 3z$ Combining with silver reduction: E^0_cell = E^0_Ag^+/Ag + E^0_Fe^2+/Fe^3+ = x + 2y - 3z$E^{0}_{cell} = E^{0}_{Ag^{+}/Ag} + E^{0}_{Fe^{2+}/Fe^{3+}} = x + 2y - 3z$ ### Step 1: Final Calculation The overall potential equals x + 2y - 3z$x + 2y - 3z$.

### Pattern Recognition Direct application of Delta G^0$\Delta G^0$ summation. Remember that standard cell potentials cannot be added directly unless the number of electrons involved is identical. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry