Given below are two statements: Statement I: Mohr's salt is composed of only three types of ions-ferrous, ammonium and sulphate. Statement II: If the molar conductance at infinite dilution of ferrous, ammonium and sulphate ions are mathbfx_1$\mathbf{x}_1$, mathbfx_2$\mathbf{x}_2$ and mathbfx_3$\mathbf{x}_3$ mathrmS\ cm^2\ mathrmmol^-1$\mathrm{S\ cm}^2\ \mathrm{mol}^{-1}$, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by mathbfx_1 + mathbfx_2 + 2mathbfx_3$\mathbf{x}_1 + \mathbf{x}_2 + 2\mathbf{x}_3$. In the light of the given statements, choose the correct answer from the options given below:

### 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

### Related Formula Faraday's law of electrolysis equation: textMoles of electrons (equivalents) = fracI cdot tF$$\text{Moles of electrons (equivalents)} = \frac{I \cdot t}{F}$$ Water ion product relation: textpH + textpOH = 14$$\text{pH} + \text{pOH} = 14$$ ### Core Logic During the electrolysis of brine (NaCl(aq)$NaCl(aq)$), hydroxide ions (OH^-$OH^-$) are generated at the cathode: 2H_2O + 2e^- rightarrow H_2 + 2OH^-$$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$$ Given metrics: - Final textpH = 12 implies textpOH = 14 - 12 = 2$\text{pH} = 12 \implies \text{pOH} = 14 - 12 = 2$ - [OH^-] = 10^-2mathrm\ M$[OH^-] = 10^{-2}\mathrm{\ M}$ - textVolume = 600mathrm\ mL = 0.6mathrm\ L$\text{Volume} = 600\mathrm{\ mL} = 0.6\mathrm{\ L}$ - textTime = 5mathrm\ min = 300mathrm\ s$\text{Time} = 5\mathrm{\ \min} = 300\mathrm{\ s}$ ### Step 1: Calculate Moles of Hydroxide Produced Find the absolute moles of OH^-$OH^-$ ions generated: textMoles = textMolarity times textVolume (L) = 10^-2 times 0.6 = 6 times 10^-3text moles$$\text{Moles} = \text{Molarity} \times \text{Volume (L)} = 10^{-2} \times 0.6 = 6 \times 10^{-3}\text{ moles}$$ ### Step 2: Relate to Electrical Current Since 1$1$ mole of electrons produces 1$1$ mole of OH^-$OH^-$, the moles of charge equals 6 times 10^-3$6 \times 10^{-3}$. Applying Faraday's equation: 6 times 10^-3 = fracI times 30096500$$6 \times 10^{-3} = \frac{I \times 300}{96500}$$ I = frac6 times 10^-3 times 96500300 = 1.93mathrm\ A$$I = \frac{6 \times 10^{-3} \times 96500}{300} = 1.93\mathrm{\ A}$$ Rounding to the nearest integer gives 2$2$. ### Pattern Recognition Always convert a given textpH$\text{pH}$ value into [OH^-]$[OH^-]$ concentration when dealing with cathodic water reduction. Tracking the relationship where 1\ e^- equiv 1\ OH^-$1\ e^- \equiv 1\ OH^-$ provides a direct shortcut to link textpH$\text{pH}$ changes to current flow. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula textOxidising Capacity propto textStandard Reduction Potential (E^ominus)$$\text{Oxidising Capacity} \propto \text{Standard Reduction Potential } (E^{\ominus})$$ ### Core Logic A higher positive value of standard reduction potential (E^ominus$E^{\ominus}$) indicates a stronger tendency to undergo reduction, hence behaving as a stronger oxidising agent. Comparing the given values: * mathrmE_mathrmSn^4+/mathrmSn^2+^ominus = +1.15mathrmV$\mathrm{E}_{\mathrm{Sn}^{4+}/\mathrm{Sn}^{2+}}^{\ominus} = +1.15\mathrm{V}$ * mathrmE_mathrmTl^3+/mathrmTl^ominus = +1.26mathrmV$\mathrm{E}_{\mathrm{T}l^{3+}/\mathrm{T}l}^{\ominus} = +1.26\mathrm{V}$ * mathrmE_mathrmAl^3+/mathrmAl^ominus = -1.66mathrmV$\mathrm{E}_{\mathrm{Al}^{3+}/\mathrm{Al}}^{\ominus} = -1.66\mathrm{V}$ * mathrmE_mathrmPb^4+/mathrmPb^2+^ominus = +1.67mathrmV$\mathrm{E}_{\mathrm{Pb}^{4+}/\mathrm{Pb}^{2+}}^{\ominus} = +1.67\mathrm{V}$ Since +1.67mathrmV$+1.67\mathrm{V}$ is the highest value, mathrmPb^4+$\mathrm{Pb}^{4+}$ possesses the strongest oxidising capacity. ### Pattern Recognition Strongest oxidising agent = Most positive reduction potential. Weakest oxidising agent = Most negative reduction potential. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 12 Chemistry: p-Block Elements

### Related Formula For weak electrolytes, the degree of dissociation alpha$\alpha$ increases sharply near infinite dilution according to Ostwald's Dilution Law: alpha = sqrtfracK_aC$$\alpha = \sqrt{\frac{K_a}{C}}$$ ### Core Logic When a weak electrolyte is diluted (concentration C rightarrow 0$C \rightarrow 0$), its molar conductivity increases steeply. Conversely, when plotted against sqrtC$\sqrt{C}$, as concentration increases, the degree of dissociation drops rapidly, causing a sharp decrease in molar conductivity. This matches the curve given below:

### Pattern Recognition Weak electrolyte plots feature a steep asymptotic exponential-like rise towards the y-axis as C rightarrow 0$C \rightarrow 0$, meaning a sharp decrease occurs with increasing concentration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula E_textcell = E_textcell^circ - fracRTnF ln Q$$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{RT}{nF} \ln Q$$ ### Core Logic Let us explicitly formulate the complete chemical oxidation-reduction equations : Anode oxidation: mathrmMg_(s) rightarrow mathrmMg^2+_(aq) + 2e^-$\mathrm{Mg}_{(s)} \rightarrow \mathrm{Mg}^{2+}_{(aq)} + 2e^-$ Cathode reduction: 2mathrmAg^+_(aq) + 2e^- rightarrow 2mathrmAg_(s)$2\mathrm{Ag}^{+}_{(aq)} + 2e^- \rightarrow 2\mathrm{Ag}_{(s)}$ Net total equation : mathrmMg_(s) + 2mathrmAg^+_(aq) rightleftharpoons mathrmMg^2+_(aq) + 2mathrmAg_(s)$$\mathrm{Mg}_{(s)} + 2\mathrm{Ag}^{+}_{(aq)} \rightleftharpoons \mathrm{Mg}^{2+}_{(aq)} + 2\mathrm{Ag}_{(s)}$$ Total transferred moles of electrons n = 2$n = 2$ . Reaction quotient : Q = frac[mathrmMg^2+][mathrmAg^+]^2$$Q = \frac{[\mathrm{Mg}^{2+}]}{[\mathrm{Ag}^{+}]^2}$$ Substituting into Nernst form : E_textcell = E_textcell^circ - fracRT2F lnleft( frac[mathrmMg^2+][mathrmAg^+]^2 right)$$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{RT}{2F} \ln\left( \frac{[\mathrm{Mg}^{2+}]}{[\mathrm{Ag}^{+}]^2} \right)$$ Inverting the inside quotient changes the sign of the logarithm term from negative to positive: E_textcell = E_textcell^circ + fracRT2F lnleft( frac[mathrmAg^+]^2[mathrmMg^2+] right)$$E_{\text{cell}} = E_{\text{cell}}^{\circ} + \frac{RT}{2F} \ln\left( \frac{[\mathrm{Ag}^{+}]^2}{[\mathrm{Mg}^{2+}]} \right)$$ ### Pattern Recognition A standard negative logarithmic quotient can always toggle into an addition configuration by inverting the products/reactants variables concentration ratio.