The standard cell potential left(E_mathrmcell^ominusright)$\left(E_{\mathrm{cell}}^{\ominus}\right)$ of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21mathrm~V$1.21\mathrm{~V}$. The standard half cell reduction potential for mathrmO_2$\mathrm{O}_2$ left(E_mathrmO_2/mathrmH_2mathrmO^circright)$\left(E_{\mathrm{O}_2/\mathrm{H}_2\mathrm{O}}^{\circ}\right)$ is 1.229mathrm~V$1.229\mathrm{~V}$. Choose the correct statement:

### Related Formula kappa = G times fraclA$$\kappa = G \times \frac{l}{A}$$ Lambda_m = frackappa times 1000C$$\Lambda_m = \frac{\kappa \times 1000}{C}$$ alpha = fracLambda_mLambda_m^infty$$\alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}}$$ [H^+] = Calpha$[H^+] = C\alpha$ ### Core Logic Given: pH = 5$pH = 5$, so [H^+] = 10^-5text M$[H^+] = 10^{-5}\text{ M}$. Since [H^+] = C cdot alpha = C cdot fracLambda_mLambda_m^infty$[H^+] = C \cdot \alpha = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$, we have 10^-5 = C cdot fracLambda_mLambda_m^infty$10^{-5} = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$. First, calculate conductivity (kappa$\kappa$): Conductance G = 4 times 10^-5text S$G = 4 \times 10^{-5}\text{ S}$. Cell constant G^* = fraclA = frac15text cm1text cm^2 = 15text cm^-1$G^* = \frac{l}{A} = \frac{15\text{ cm}}{1\text{ cm}^2} = 15\text{ cm}^{-1}$. kappa = G cdot G^* = (4 times 10^-5) times 15 = 6 times 10^-4text S cm^-1$$\kappa = G \cdot G^* = (4 \times 10^{-5}) \times 15 = 6 \times 10^{-4}\text{ S cm}^{-1}$$ Molar conductivity (Lambda_m$\Lambda_m$): Lambda_m = frackappa times 1000C = frac6 times 10^-4 times 1000C = frac0.6C$$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{6 \times 10^{-4} \times 1000}{C} = \frac{0.6}{C}$$ Substitute Lambda_m$\Lambda_m$ into the proton concentration formula: [H^+] = 10^-5 = C cdot frac0.6 / CLambda_m^infty$$[H^+] = 10^{-5} = C \cdot \frac{0.6 / C}{\Lambda_m^{\infty}}$$ 10^-5 = frac0.6Lambda_m^infty$$10^{-5} = \frac{0.6}{\Lambda_m^{\infty}}$$ Lambda_m^infty = frac0.610^-5 = 60000text S cm^2text mol^-1$$\Lambda_m^{\infty} = \frac{0.6}{10^{-5}} = 60000\text{ S cm}^2\text{ mol}^{-1}$$ Convert units to textS m^2text mol^-1$\text{S m}^2\text{ mol}^{-1}$: Since 1text m^2 = 10^4text cm^2$1\text{ m}^2 = 10^4\text{ cm}^2$, Lambda_m^infty = 60000 times 10^-4text S m^2text mol^-1 = 6text S m^2text mol^-1$$\Lambda_m^{\infty} = 60000 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1} = 6\text{ S m}^2\text{ mol}^{-1}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

### 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