### Related Formula kappa = frac1rho$\kappa = \frac{1}{\rho}$ Lambda_m = frackappaM$\Lambda_m = \frac{\kappa}{M}$ ### Core Logic To compute the molar conductivity, we first calculate the molarity of the solution and the conductivity of the electrolyte from the given resistivity. ### Step 1: Calculate Molarity (M) 0.2\%$0.2\%$ (w/v) mathrmNaOH$\mathrm{NaOH}$ means 0.2~mathrmg$0.2~\mathrm{g}$ of mathrmNaOH$\mathrm{NaOH}$ is present in 100~mathrmmL$100~\mathrm{mL}$ of solution. textMolar mass of NaOH = 23 + 16 + 1 = 40~mathrmg~mol^-1$\text{Molar mass of NaOH} = 23 + 16 + 1 = 40~\mathrm{g~mol^{-1}}$ textMolarity M = fractextMass of solutetextMolar mass times frac1000V_mathrmmL = frac0.240 times frac1000100 = 0.05~mathrmmol~L^-1 = 0.05~mathrmmol~dm^-3$\text{Molarity } M = \frac{\text{Mass of solute}}{\text{Molar mass}} \times \frac{1000}{V_{\mathrm{mL}}} = \frac{0.2}{40} \times \frac{1000}{100} = 0.05~\mathrm{mol~L^{-1}} = 0.05~\mathrm{mol~dm^{-3}}$ ### Step 2: Calculate Conductivity (kappa) in dm Units Given resistivity rho = 870.0~mathrmmOmega~m = 870 times 10^-3~Omega~m = 0.87~Omega~m$\rho = 870.0~\mathrm{m\Omega~m} = 870 \times 10^{-3}~\Omega~m = 0.87~\Omega~m$. Since 1~mathrmm = 10~mathrmdm$1~\mathrm{m} = 10~\mathrm{dm}$: rho = 0.87~Omega times (10~mathrmdm) = 8.7~Omega~dm$\rho = 0.87~\Omega \times (10~\mathrm{dm}) = 8.7~\Omega~dm$ Now, conductivity kappa$\kappa$ is: kappa = frac1rho = frac18.7~Omega^-1~dm^-1$\kappa = \frac{1}{\rho} = \frac{1}{8.7}~\Omega^{-1}~dm^{-1}$ ### Step 3: Calculate Molar Conductivity (Lambda_m) Lambda_m = frackappaM = fracfrac18.7~mathrmS~dm^-10.05~mathrmmol~dm^-3 = frac18.7 times 0.05 = frac10.435 approx 2.29885~mathrmS~dm^2~mol^-1$\Lambda_m = \frac{\kappa}{M} = \frac{\frac{1}{8.7}~\mathrm{S~dm^{-1}}}{0.05~\mathrm{mol~dm^{-3}}} = \frac{1}{8.7 \times 0.05} = \frac{1}{0.435} \approx 2.29885~\mathrm{S~dm^2~mol^{-1}}$ Converting mathrmS$\mathrm{S}$ to mathrmmS$\mathrm{mS}$ (1~mathrmS = 10^3~mathrmmS$1~\mathrm{S} = 10^3~\mathrm{mS}$): Lambda_m = 2.29885 times 10^3~mathrmmS~dm^2~mol^-1 = 22.9885 times 10^2~mathrmmS~dm^2~mol^-1$\Lambda_m = 2.29885 \times 10^3~\mathrm{mS~dm^2~mol^{-1}} = 22.9885 \times 10^2~\mathrm{mS~dm^2~mol^{-1}}$ Rounding off to the nearest integer gives **23**. ### Pattern Recognition Ensure careful handling of volumetric conversions. Since concentration is expressed in moles per liter (equivalent to mathrmdm^-3$\mathrm{dm^{-3}}$), expressing conductivity in terms of mathrmdm^-1$\mathrm{dm^{-1}}$ directly eliminates the need for arbitrary 1000$1000$ multiplication factors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula Nernst equation for the net combined redox cell expression: E = E^circ - frac0.062logleft(frac[mathrmH^+]^2[mathrmAg^+]^2right)$E = E^\circ - \frac{0.06}{2}\log\left(\frac{[\mathrm{H}^+]^2}{[\mathrm{Ag}^+]^2}\right)$ Hydrolysis equation for a salt composed of a weak base and strong acid: mathrmpH = 7 - frac12mathrmpK_b - frac12logmathrmC$\mathrm{pH} = 7 - \frac{1}{2}\mathrm{pK_b} - \frac{1}{2}\log\mathrm{C}$ ### Core Logic Let's compute the operational values line-by-row: * Combined redox process: mathrmQH_2 + 2Ag^+ rightarrow Q + 2Ag + 2H^+$\mathrm{QH_2 + 2Ag^+ \rightarrow Q + 2Ag + 2H^+}$. * Standard cell potential difference: E^circ_textcell = E^circ_mathrmAg^+/Ag - E^circ_mathrmQ/QH_2 = 0.8 - 0.7 = +0.1mathrm~V$E^\circ_{\text{cell}} = E^\circ_{\mathrm{Ag^+/Ag}} - E^\circ_{\mathrm{Q/QH_2}} = 0.8 - 0.7 = +0.1\mathrm{~V}$. * Apply Nernst adjustments using known concentrations ([mathrmAg^+] = 1mathrm~M$[\mathrm{Ag}^+] = 1\mathrm{~M}$): 0.4 = 0.1 - 0.06 log [mathrmH^+]$0.4 = 0.1 - 0.06 \log [\mathrm{H}^+]$ 0.3 = 0.06 times mathrmpH implies mathrmpH = 5$0.3 = 0.06 \times \mathrm{pH} \implies \mathrm{pH} = 5$ ### Step 1: Salt Hydrolysis Substitution Substitute the determined mathrmpH$\mathrm{pH}$ along with salt molarity (C = 0.01mathrm~M = 10^-2mathrm~M$C = 0.01\mathrm{~M} = 10^{-2}\mathrm{~M}$) into the hydrolysis equation: 5 = 7 - frac12mathrmpK_b - frac12log(10^-2)$5 = 7 - \frac{1}{2}\mathrm{pK_b} - \frac{1}{2}\log(10^{-2})$ 5 = 7 - frac12mathrmpK_b - frac12(-2)$5 = 7 - \frac{1}{2}\mathrm{pK_b} - \frac{1}{2}(-2)$ 5 = 7 - frac12mathrmpK_b + 1$5 = 7 - \frac{1}{2}\mathrm{pK_b} + 1$ 5 = 8 - frac12mathrmpK_b implies frac12mathrmpK_b = 3 implies mathrmpK_b = 6$5 = 8 - \frac{1}{2}\mathrm{pK_b} \implies \frac{1}{2}\mathrm{pK_b} = 3 \implies \mathrm{pK_b} = 6$ ### Pattern Recognition Quinhydrone electrodes act as excellent pH indicators in electrochemical cells. Note that each change of 1 pH unit shifts the cell output potential by exactly 0.06mathrm~V$0.06\mathrm{~V}$ at standard ambient conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

### Related Formula At the equivalence point during titration: M_textacid V_textacid = M_textbase V_textbase$M_{\text{acid}} V_{\text{acid}} = M_{\text{base}} V_{\text{base}}$ ### Core Logic In a mixture of a strong acid (mathrmHCl$\mathrm{HCl}$) and a weak acid (mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$): 1. mathrmHCl$\mathrm{HCl}$ is a strong acid and is completely ionized. When mathrmNaOH$\mathrm{NaOH}$ is added, highly mobile mathrmH^+$\mathrm{H}^+$ ions are replaced by less mobile mathrmNa^+$\mathrm{Na}^+$ ions, causing a sharp drop in conductance (segment AB). 2. At point B (2.0mathrm~mL$2.0\mathrm{~mL}$), mathrmHCl$\mathrm{HCl}$ is completely neutralized. 3. Segment BC represents the neutralization of the weak acid mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ to form highly conducting sodium acetate, causing a moderate rise in conductance up to point C (5.0mathrm~mL$5.0\mathrm{~mL}$). 4. Beyond point C, excess mathrmOH^-$\mathrm{OH}^-$ ions cause a rapid rise in conductance (segment CD). ### Step 1: Calculate concentration of mathrmHCl$\mathrm{HCl}$ Volume of mathrmNaOH$\mathrm{NaOH}$ used to neutralize mathrmHCl$\mathrm{HCl}$ is V_1 = 2.0mathrm~mL$V_1 = 2.0\mathrm{~mL}$: M_mathrmHCl times 40mathrm~mL = 0.1mathrm~M times 2.0mathrm~mL$M_{\mathrm{HCl}} \times 40\mathrm{~mL} = 0.1\mathrm{~M} \times 2.0\mathrm{~mL}$ M_mathrmHCl = frac0.240 = 0.005mathrm~M$M_{\mathrm{HCl}} = \frac{0.2}{40} = 0.005\mathrm{~M}$ ### Step 2: Calculate concentration of mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ Volume of mathrmNaOH$\mathrm{NaOH}$ used to neutralize mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ is V_2 = 5.0mathrm~mL - 2.0mathrm~mL = 3.0mathrm~mL$V_2 = 5.0\mathrm{~mL} - 2.0\mathrm{~mL} = 3.0\mathrm{~mL}$: M_mathrmCH_3mathrmCOOH times 40mathrm~mL = 0.1mathrm~M times 3.0mathrm~mL$M_{\mathrm{CH}_3\mathrm{COOH}} \times 40\mathrm{~mL} = 0.1\mathrm{~M} \times 3.0\mathrm{~mL}$ M_mathrmCH_3mathrmCOOH = frac0.340 = 0.0075mathrm~M$M_{\mathrm{CH}_3\mathrm{COOH}} = \frac{0.3}{40} = 0.0075\mathrm{~M}$ ### Pattern Recognition Conductometric titration curves are analyzed sequentially: the strongest electrolyte is always neutralized first. A steep drop in conductance always signals the neutralization of a strong acid (mathrmH^+$\mathrm{H}^+$ depletion). A weak acid titration shows a gentle upward slope due to salt formation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

### Related Formula Standard cell EMF is related to standard reduction potentials: E_mathrmcell^circ = E_mathrmcathode^circ - E_mathrmanode^circ$E_{\mathrm{cell}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ}$ ### Core Logic In a methanol-oxygen fuel cell: - Anode reaction (Oxidation): Methanol is oxidized to carbon dioxide: mathrmCH_3mathrmOH + mathrmH_2mathrmO rightarrow mathrmCO_2 + 6mathrmH^+ + 6e^-$\mathrm{CH}_3\mathrm{OH} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{CO}_2 + 6\mathrm{H}^+ + 6e^-$ - Cathode reaction (Reduction): Oxygen is reduced to water: mathrmO_2 + 4mathrmH^+ + 4e^- rightarrow 2mathrmH_2mathrmO$\mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- \rightarrow 2\mathrm{H}_2\mathrm{O}$ Hence, cathode is the oxygen electrode, and anode is the methanol electrode. ### Step 1: Calculate Standard Reduction Potential of Anode Using the EMF equation: 1.21mathrm~V = 1.229mathrm~V - E_mathrmanode^circ$1.21\mathrm{~V} = 1.229\mathrm{~V} - E_{\mathrm{anode}}^{\circ}$ E_mathrmanode^circ = 1.229 - 1.21 = 0.019mathrm~V = 19mathrm~mV$E_{\mathrm{anode}}^{\circ} = 1.229 - 1.21 = 0.019\mathrm{~V} = 19\mathrm{~mV}$ The standard half-cell reduction potential for the mathrmCO_2/mathrmCH_3mathrmOH$\mathrm{CO}_2/\mathrm{CH}_3\mathrm{OH}$ couple is 19mathrm~mV$19\mathrm{~mV}$, matching Option (1). ### Pattern Recognition Fuel cells are galvanic cells where reactants (like fuels and oxidants) are fed continuously to the electrodes, not at one go. Oxidation always occurs at the anode (methanol) and reduction at the cathode (oxygen). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula lambda_m^infty = nu_+ lambda_+^infty + nu_- lambda_-^infty$\lambda_m^{\infty} = \nu_+ \lambda_+^{\infty} + \nu_- \lambda_-^{\infty}$ ### Core Logic Statement I: Mohr's salt is a double salt with chemical formula: mathrmFeSO_4 cdot (NH_4)_2SO_4 cdot 6H_2O$\mathrm{FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O}$ When dissolved in water, it completely dissociates into three distinct ionic species: mathrmFe^2+ text (ferrous), quad mathrmNH_4^+ text (ammonium), quad textand mathrmSO_4^2- text (sulphate)$\mathrm{Fe}^{2+} \text{ (ferrous)}, \quad \mathrm{NH}_4^+ \text{ (ammonium)}, \quad \text{and } \mathrm{SO}_4^{2-} \text{ (sulphate)}$ Thus, Statement I is true. Statement II: According to Kohlrausch's law of independent migration of ions: lambda_m^infty(textMohr's Salt) = 1 cdot lambda_m^infty(mathrmFe^2+) + 2 cdot lambda_m^infty(mathrmNH_4^+) + 2 cdot lambda_m^infty(mathrmSO_4^2-)$\lambda_m^{\infty}(\text{Mohr's Salt}) = 1 \cdot \lambda_m^{\infty}(\mathrm{Fe}^{2+}) + 2 \cdot \lambda_m^{\infty}(\mathrm{NH}_4^+) + 2 \cdot \lambda_m^{\infty}(\mathrm{SO}_4^{2-})$ lambda_m^infty = x_1 + 2x_2 + 2x_3$\lambda_m^{\infty} = x_1 + 2x_2 + 2x_3$ Statement II claims the expression is x_1 + x_2 + 2x_3$x_1 + x_2 + 2x_3$ (missing the coefficient 2$2$ for ammonium). Thus, Statement II is false. ### Pattern Recognition Kohlrausch's law matches stoichiometric coefficients directly to the ion quantities released. Mohr's salt formula contains (NH_4)_2$(NH_4)_2$, requiring a multiplier of 2$2$ for ammonium ion conductance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 12 Chemistry: d- and f-Block Elements