Welcome to Chemca.in's ultimate numerical repository. While the Nernst equation dictates the thermodynamics of a cell, the kinetic and transport phenomena are governed by Conductivity and Faraday's Laws of Electrolysis. The problems below represent the highest tier of conceptual testing found in JEE Advanced and NEET.
Problem 1: Cell Constant and Specific Conductivity
Question: The resistance of a conductivity cell containing $0.001 \text{ M } KCl$ solution at $298 \text{ K}$ is $1500 \ \Omega$. What is the cell constant if conductivity of $0.001 \text{ M } KCl$ solution at $298 \text{ K}$ is $0.146 \times 10^{-3} \text{ S cm}^{-1}$?
Strategy: Direct application of the formula relating cell constant ($G^{\ast}$), resistance ($R$), and specific conductivity ($\kappa$).
$\kappa = \frac{1}{R} \times G^{\ast}$
$G^{\ast} = \kappa \times R = (0.146 \times 10^{-3} \text{ S cm}^{-1}) \times (1500 \ \Omega)$
Final Answer: Cell Constant $G^{\ast} = 0.219 \text{ cm}^{-1}$
Problem 2: Molar Conductivity Calculation
Question: The conductivity of a $0.20 \text{ M}$ solution of $KCl$ at $298 \text{ K}$ is $0.0248 \text{ S cm}^{-1}$. Calculate its molar conductivity.
Strategy: Use the standard conversion formula: $\Lambda_m = \frac{\kappa \times 1000}{M}$. Ensure $\kappa$ is in $S \text{ cm}^{-1}$.
$\Lambda_m = \frac{0.0248 \times 1000}{0.20}$
$\Lambda_m = \frac{24.8}{0.20} = 124$
Final Answer: $\Lambda_m = 124 \text{ S cm}^2 \text{ mol}^{-1}$
Problem 3: Kohlrausch's Law for Weak Electrolytes
Question: Calculate $\Lambda_m^{\circ}$ for Acetic Acid ($CH_3COOH$). Given that $\Lambda_m^{\circ}$ values for $HCl$, $NaCl$, and $CH_3COONa$ are $425.9$, $126.4$, and $91.0 \text{ S cm}^2 \text{ mol}^{-1}$ respectively.
Strategy: Use Kohlrausch's law of independent migration of ions to algebraically isolate the ions of the weak acid using strong electrolytes.
$\Lambda_m^{\circ}(CH_3COOH) = \lambda^{\circ}(CH_3COO^-) + \lambda^{\circ}(H^+)$
$\Lambda_m^{\circ}(CH_3COOH) = \Lambda_m^{\circ}(CH_3COONa) + \Lambda_m^{\circ}(HCl) - \Lambda_m^{\circ}(NaCl)$
$\Lambda_m^{\circ}(CH_3COOH) = 91.0 + 425.9 - 126.4$
Final Answer: $\Lambda_m^{\circ}(CH_3COOH) = 390.5 \text{ S cm}^2 \text{ mol}^{-1}$
Problem 4: Degree of Dissociation ($\alpha$)
Question: The conductivity of a $0.001028 \text{ M}$ acetic acid solution is $4.95 \times 10^{-5} \text{ S cm}^{-1}$. Calculate its degree of dissociation if $\Lambda_m^{\circ}$ for acetic acid is $390.5 \text{ S cm}^2 \text{ mol}^{-1}$.
Strategy: First calculate the molar conductivity at the given concentration ($\Lambda_m^c$), then divide by the limiting molar conductivity ($\Lambda_m^{\circ}$).
Step 1: $\Lambda_m^c = \frac{\kappa \times 1000}{M} = \frac{4.95 \times 10^{-5} \times 1000}{0.001028} = 48.15 \text{ S cm}^2 \text{ mol}^{-1}$
Step 2: $\alpha = \frac{\Lambda_m^c}{\Lambda_m^{\circ}} = \frac{48.15}{390.5}$
Final Answer: $\alpha = 0.1233$ (or $12.33\%$)
Problem 5: Dissociation Constant ($K_a$)
Question: Using the data from Problem 4, calculate the acid dissociation constant ($K_a$) for acetic acid at this concentration.
Strategy: Apply Ostwald's dilution law: $K_a = \frac{c\alpha^2}{1 - \alpha}$.
$c = 0.001028 \text{ M}$, $\alpha = 0.1233$
$K_a = \frac{0.001028 \times (0.1233)^2}{1 - 0.1233}$
$K_a = \frac{0.001028 \times 0.0152}{0.8767}$
Final Answer: $K_a = 1.78 \times 10^{-5} \text{ mol L}^{-1}$
Problem 6: Solubility Product ($K_{sp}$) from Conductivity
Question: The specific conductivity of a saturated solution of $AgCl$ at $298 \text{ K}$ is $3.40 \times 10^{-6} \text{ S cm}^{-1}$ (after subtracting the conductivity of water). If $\lambda^{\circ}_{Ag^+} = 61.9$ and $\lambda^{\circ}_{Cl^-} = 76.3 \text{ S cm}^2 \text{ mol}^{-1}$, calculate the $K_{sp}$ of $AgCl$.
Strategy: A saturated solution of a sparingly soluble salt is infinitely dilute. Therefore, $\Lambda_m \approx \Lambda_m^{\circ}$. Find solubility ($S$) from $S = \frac{\kappa \times 1000}{\Lambda_m^{\circ}}$, then $K_{sp} = S^2$.
Step 1: $\Lambda_m^{\circ}(AgCl) = 61.9 + 76.3 = 138.2 \text{ S cm}^2 \text{ mol}^{-1}$
Step 2: $S = \frac{3.40 \times 10^{-6} \times 1000}{138.2} = \frac{3.40 \times 10^{-3}}{138.2} = 2.46 \times 10^{-5} \text{ M}$
Step 3: $K_{sp} = [Ag^+][Cl^-] = S^2 = (2.46 \times 10^{-5})^2$
Final Answer: $K_{sp} = 6.05 \times 10^{-10}$
Problem 7: Equivalent Conductivity and Valency
Question: The specific conductivity of a $0.1 \text{ N}$ solution of $CuSO_4$ is $1.2 \times 10^{-2} \text{ S cm}^{-1}$. Calculate its equivalent conductivity ($\Lambda_{eq}$) and molar conductivity ($\Lambda_m$).
Strategy: Use $N$ for equivalent conductivity. Then, relate $\Lambda_m$ and $\Lambda_{eq}$ using the n-factor (total positive charge on cation). For $CuSO_4$, n-factor = 2.
$\Lambda_{eq} = \frac{\kappa \times 1000}{N} = \frac{1.2 \times 10^{-2} \times 1000}{0.1} = 120 \text{ S cm}^2 \text{ eq}^{-1}$
$\Lambda_m = \Lambda_{eq} \times \text{n-factor} = 120 \times 2$
Final Answer: $\Lambda_{eq} = 120 \text{ S cm}^2 \text{ eq}^{-1}$ and $\Lambda_m = 240 \text{ S cm}^2 \text{ mol}^{-1}$
Problem 8: Mixing Two Strong Electrolytes
Question: $50 \text{ mL}$ of $0.1 \text{ M } HCl$ ($\Lambda_m = 400 \text{ S cm}^2 \text{ mol}^{-1}$) is mixed with $50 \text{ mL}$ of $0.1 \text{ M } NaCl$ ($\Lambda_m = 100 \text{ S cm}^2 \text{ mol}^{-1}$). Assuming $\Lambda_m$ remains constant, what is the specific conductivity ($\kappa$) of the resulting mixture?
Strategy: Find the specific conductivity ($\kappa$) of each solution, calculate their individual conductances in the new volume, and add them (conductance is additive).
New volume = $100 \text{ mL}$. New concentrations: $[HCl] = 0.05 \text{ M}$, $[NaCl] = 0.05 \text{ M}$.
$\kappa_{HCl} = \frac{\Lambda_m \times M}{1000} = \frac{400 \times 0.05}{1000} = 0.02 \text{ S cm}^{-1}$
$\kappa_{NaCl} = \frac{100 \times 0.05}{1000} = 0.005 \text{ S cm}^{-1}$
$\kappa_{\text{mix}} = \kappa_{HCl} + \kappa_{NaCl} = 0.02 + 0.005$
Final Answer: $\kappa_{\text{mix}} = 0.025 \text{ S cm}^{-1}$
Problem 9: Basic Faraday's First Law
Question: A solution of $CuSO_4$ is electrolyzed for 10 minutes with a current of $1.5 \text{ Amperes}$. What is the mass of copper deposited at the cathode? (Atomic mass of $Cu = 63.5 \text{ g/mol}$).
Strategy: Use $m = \frac{M \times I \times t}{n \times F}$. Time must be in seconds. For $Cu^{2+}$, $n = 2$.
$t = 10 \times 60 = 600 \text{ s}$
$m = \frac{63.5 \times 1.5 \times 600}{2 \times 96500} = \frac{57150}{193000}$
Final Answer: Mass of $Cu$ deposited = $0.296 \text{ g}$
Problem 10: Faraday's Second Law (Cells in Series)
Question: Three electrolytic cells A, B, C containing solutions of $ZnSO_4$, $AgNO_3$, and $CuSO_4$ respectively are connected in series. A steady current of $1.5 \text{ A}$ was passed until $1.45 \text{ g}$ of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Strategy: Use the Silver data to find the total charge ($Q$) and time ($t$). Then use Faraday's second law: ratio of masses equals the ratio of equivalent weights.
Step 1: $Ag^+ + e^- \rightarrow Ag$ ($n=1$). Equivalent weight of $Ag = 108 / 1 = 108$.
Moles of $Ag$ = $1.45 / 108 = 0.0134 \text{ mol}$.
Charge $Q = n_e \times F = 0.0134 \times 96500 = 1295.6 \text{ C}$.
Step 2: $t = Q / I = 1295.6 / 1.5 = 863.7 \text{ s}$ (or $\approx 14.4 \text{ min}$).
Step 3: Equivalents of all metals are equal. Eq. weight of $Cu = 63.5/2 = 31.75$. Eq. weight of $Zn = 65.4/2 = 32.7$.
Mass of $Cu = 0.0134 \text{ equivalents} \times 31.75 \text{ g/eq} = 0.425 \text{ g}$.
Mass of $Zn = 0.0134 \text{ equivalents} \times 32.7 \text{ g/eq} = 0.438 \text{ g}$.
Final Answer: $t = 863.7 \text{ s}$; Mass $Cu = 0.425 \text{ g}$; Mass $Zn = 0.438 \text{ g}$
Problem 11: Gas Volume at STP
Question: An electric current is passed through dilute Sulfuric acid for 2 hours. A total of $2.24 \text{ L}$ of mixed gases ($H_2$ and $O_2$) are collected at STP. Calculate the current passed.
Strategy: Water electrolysis yields $2 \text{ volumes of } H_2$ for every $1 \text{ volume of } O_2$. So total volume = $3V$. Find moles of $H_2$, relate to Faradays, and calculate current.
Total volume = $2.24 \text{ L}$. Ratio $H_2:O_2 = 2:1$. Volume of $H_2 = \frac{2}{3} \times 2.24 = 1.493 \text{ L}$.
Moles of $H_2$ at STP = $1.493 / 22.4 = 0.0667 \text{ mol}$.
Reaction: $2H^+ + 2e^- \rightarrow H_2$. $1 \text{ mole } H_2$ requires $2 \text{ Faradays}$.
Total charge $Q = 0.0667 \times 2 \times 96500 = 12873 \text{ Coulombs}$.
$I = Q / t = 12873 / (2 \times 3600 \text{ s}) = 12873 / 7200$
Final Answer: Current $I = 1.788 \text{ Amperes}$
Problem 12: Current Efficiency
Question: A current of $2.0 \text{ A}$ is passed for 5 hours through a molten tin salt, depositing $11.9 \text{ g}$ of tin. What is the oxidation state of tin in this salt? Assuming $100\%$ current efficiency, calculate $n$. What if actual current efficiency was $80\%$?
Strategy: Find total charge, calculate theoretical moles of electrons, and relate it to the moles of tin deposited to find $n$ (valency).
Charge $Q = 2 \times (5 \times 3600) = 36000 \text{ C}$.
Faradays = $36000 / 96500 = 0.373 \text{ F}$ (moles of electrons).
Moles of $Sn$ (atomic mass $118.7$) = $11.9 / 118.7 = 0.1 \text{ mol}$.
Since $0.1 \text{ mol}$ of $Sn$ requires $0.373 \text{ mol}$ of electrons, $n = 0.373 / 0.1 \approx 4$. Oxidation state is $+4$ (e.g., $SnCl_4$).
If efficiency was $80\%$, useful charge = $0.8 \times 36000 = 28800 \text{ C}$ ($0.298 \text{ F}$).
$n = 0.298 / 0.1 \approx 3$.
Final Answer: Oxidation state = $+4$ (at 100% efficiency)
Problem 13: Electrolysis of Aqueous NaCl & pH
Question: $500 \text{ mL}$ of a $1 \text{ M}$ aqueous $NaCl$ solution is electrolyzed with a current of $2 \text{ A}$ for $965 \text{ seconds}$. Assuming no volume change, calculate the final pH of the solution. (Assume complete dissociation of products).
Strategy: Aqueous NaCl electrolysis produces $NaOH$ at the cathode because $H_2O$ is reduced instead of $Na^+$. Use Faraday's law to find moles of $OH^-$ generated, find molarity, then pOH, then pH.
Charge $Q = 2 \times 965 = 1930 \text{ C}$.
Moles of $e^-$ ($F$) = $1930 / 96500 = 0.02 \text{ mol}$.
Cathode reaction: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$.
Moles of $OH^-$ produced = moles of $e^-$ = $0.02 \text{ mol}$.
$[OH^-] = 0.02 \text{ mol} / 0.500 \text{ L} = 0.04 \text{ M}$.
$\text{pOH} = -\log_{10}(0.04) = 1.4$.
$\text{pH} = 14 - 1.4 = 12.6$.
Final Answer: Final $\text{pH} = 12.6$
Problem 14: Electroplating Thickness
Question: How long must a current of $5.0 \text{ A}$ be applied to a solution of $AgNO_3$ to coat a metal surface of $100 \text{ cm}^2$ with a $0.01 \text{ mm}$ thick layer of silver? (Density of $Ag = 10.5 \text{ g cm}^{-3}$).
Strategy: Calculate the volume of the silver layer, use density to find mass, and then apply Faraday's First Law.
Thickness = $0.01 \text{ mm} = 0.001 \text{ cm}$.
Volume = Area $\times$ Thickness = $100 \text{ cm}^2 \times 0.001 \text{ cm} = 0.1 \text{ cm}^3$.
Mass = Volume $\times$ Density = $0.1 \text{ cm}^3 \times 10.5 \text{ g cm}^{-3} = 1.05 \text{ g}$.
$t = \frac{m \times n \times F}{M \times I} = \frac{1.05 \times 1 \times 96500}{108 \times 5.0} = \frac{101325}{540}$
Final Answer: Time $t = 187.6 \text{ seconds}$
Problem 15: Charge for Water Oxidation
Question: How much charge is required for the complete oxidation of 1 mole of $H_2O$ to $O_2$?
Strategy: Write the balanced oxidation half-reaction for water. Relate moles of water to moles of electrons.
Reaction: $2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+ + 4e^-$
From stoichiometry, 2 moles of $H_2O$ produce 4 moles of electrons.
Therefore, 1 mole of $H_2O$ yields 2 moles of electrons.
Charge = $2 \text{ Faradays} = 2 \times 96500 \text{ C}$.
Final Answer: Charge = $193,000 \text{ Coulombs}$
Problem 16: Active vs Inert Electrodes (CuSO₄)
Question: $CuSO_4$ solution is electrolyzed for 20 minutes with a current of $2 \text{ A}$ using copper electrodes. What is the change in mass of the anode and the cathode?
Strategy: Because the electrodes are active (Copper), the anode dissolves (oxidizes) and the cathode plates (reduces). The mass lost at the anode exactly equals the mass gained at the cathode.
Charge = $2 \text{ A} \times (20 \times 60) \text{ s} = 2400 \text{ C}$.
Mass $m = \frac{63.5 \times 2400}{2 \times 96500} = 0.79 \text{ g}$.
Final Answer: Cathode increases by $0.79 \text{ g}$; Anode decreases by $0.79 \text{ g}$. Solution concentration remains unchanged.
Problem 17: Inert Electrodes and pH (CuSO₄)
Question: If the electrolysis in Problem 16 was done using Platinum (inert) electrodes in $1 \text{ L}$ of solution, what would be the pH of the resulting solution?
Strategy: With inert electrodes, water oxidizes at the anode instead of copper: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. We must calculate the $H^+$ generated.
Total Faradays passed = $2400 / 96500 = 0.02487 \text{ F}$.
From the anode reaction, 4 moles of electrons ($4\text{F}$) generate 4 moles of $H^+$. Thus, moles of $H^+$ = moles of $e^-$ = $0.02487 \text{ mol}$.
$[H^+] = 0.02487 \text{ M}$ (since volume is $1\text{L}$).
$\text{pH} = -\log_{10}(0.02487) \approx 1.6$.
Final Answer: Final $\text{pH} \approx 1.6$
Problem 18: Electro-refining (Anode Mud)
Question: A $100 \text{ g}$ impure block of copper ($90\%$ pure) is used as an anode for electro-refining. A current of $50 \text{ A}$ is passed. How long will it take to completely dissolve all the pure copper? Assuming impurities fall as anode mud without oxidizing.
Strategy: Calculate the mass of pure copper to be dissolved, then apply Faraday's law to find the time.
Mass of pure $Cu$ = $90\% \text{ of } 100 \text{ g} = 90 \text{ g}$.
$t = \frac{m \times n \times F}{M \times I} = \frac{90 \times 2 \times 96500}{63.5 \times 50} = \frac{17370000}{3175}$
$t = 5470.8 \text{ seconds} = 1.52 \text{ hours}$.
Final Answer: Time required = $1.52 \text{ hours}$.
Problem 19: Complete Depletion of Electrolyte
Question: How many hours does it take to completely reduce $500 \text{ mL}$ of a $0.2 \text{ M } AgNO_3$ solution using a current of $2 \text{ A}$?
Strategy: Calculate total moles of $Ag^+$ in the solution, then find the charge required to reduce all of it to $Ag$ metal.
Moles of $Ag^+$ = $0.2 \text{ mol/L} \times 0.5 \text{ L} = 0.1 \text{ mol}$.
Since $Ag^+ + e^- \rightarrow Ag$ ($n=1$), it requires $0.1 \text{ Faradays}$ of charge.
Charge = $0.1 \times 96500 = 9650 \text{ Coulombs}$.
$t = Q / I = 9650 / 2 = 4825 \text{ seconds}$.
Hours = $4825 / 3600 \approx 1.34 \text{ hours}$.
Final Answer: $1.34 \text{ hours}$
Problem 20: Overpotential in Aqueous NaCl
Question: Theoretically, in the electrolysis of aqueous $NaCl$, $O_2$ should evolve at the anode ($E^{\circ}_{\text{ox}} = -1.23 \text{ V}$) instead of $Cl_2$ ($E^{\circ}_{\text{ox}} = -1.36 \text{ V}$). If $Cl_2$ actually evolves, what must be the minimum oxygen overpotential ($\eta$) on that specific electrode?
Strategy: For Chlorine to evolve preferentially, the actual voltage required for Oxygen evolution (Standard + Overpotential) must exceed the standard voltage for Chlorine oxidation.
$|E_{\text{actual, } O_2}| > |E_{\text{std, } Cl_2}|$
$|(-1.23) - \eta| > |-1.36|$
$1.23 + \eta > 1.36$
$\eta > 0.13 \text{ V}$.
Final Answer: Minimum Oxygen Overpotential > $0.13 \text{ V}$
Problem 21: Determining Avogadro's Number
Question: If the charge on a single electron is measured as $1.602 \times 10^{-19} \text{ C}$ by Millikan's oil drop experiment, and the electrochemical equivalent (Faraday's constant) for depositing 1 equivalent of silver is determined as $96485 \text{ C/eq}$, calculate Avogadro's number ($N_A$).
Strategy: Faraday's constant ($F$) is the charge of exactly one mole of electrons. Therefore, $F = N_A \times (\text{charge of one } e^-)$.
$N_A = \frac{F}{e} = \frac{96485 \text{ C mol}^{-1}}{1.602 \times 10^{-19} \text{ C}}$
$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$
Final Answer: $N_A = 6.022 \times 10^{23}$
Problem 22: Integration of Variable Current
Question: A variable current $I = (2 + 0.5t) \text{ Amperes}$ (where $t$ is in seconds) flows through an $AgNO_3$ solution for 10 seconds. Calculate the mass of silver deposited.
Strategy: Since current is not constant, we cannot use $Q = It$. We must integrate the current over time to find the total charge: $Q = \int I dt$.
$Q = \int_0^{10} (2 + 0.5t) dt = [2t + 0.25t^2]_0^{10}$
$Q = (2(10) + 0.25(100)) - 0 = 20 + 25 = 45 \text{ Coulombs}$.
$m = \frac{108 \times 45}{1 \times 96500} = 0.0503 \text{ g}$
Final Answer: Mass = $0.0503 \text{ g}$
Problem 23: Energy Consumption in kWh
Question: In an electrolytic cell producing Aluminum by the Hall-Héroult process, a voltage of $5.0 \text{ V}$ is applied across the cell. If a current of $100,000 \text{ A}$ flows for 24 hours, how much electrical energy is consumed in Kilowatt-hours (kWh)?
Strategy: Power ($P$) = Voltage $\times$ Current. Energy = Power $\times$ Time. Ensure units are correctly converted to kilo-Watts and hours.
$P = V \times I = 5.0 \text{ V} \times 100,000 \text{ A} = 500,000 \text{ Watts} = 500 \text{ kW}$.
Energy ($E$) = $P \times t = 500 \text{ kW} \times 24 \text{ h} = 12,000 \text{ kWh}$.
Final Answer: Energy consumed = $12,000 \text{ kWh}$
Problem 24: Chlor-Alkali Production Rates
Question: An industrial chlor-alkali cell operates at $50,000 \text{ A}$ with $90\%$ current efficiency. Calculate the mass of $Cl_2$ gas produced per hour.
Strategy: Apply efficiency to the current, find Faradays passed per hour, and relate to the anode reaction ($2Cl^- \rightarrow Cl_2 + 2e^-$).
Effective current $I = 50,000 \times 0.90 = 45,000 \text{ A}$.
Charge per hour $Q = 45,000 \times 3600 = 1.62 \times 10^8 \text{ C}$.
Faradays = $(1.62 \times 10^8) / 96500 = 1678.7 \text{ F}$.
2 Faradays produce 1 mole of $Cl_2$ ($71 \text{ g}$).
Mass of $Cl_2$ = $(1678.7 / 2) \times 71 = 59593 \text{ g} \approx 59.6 \text{ kg}$.
Final Answer: Mass of $Cl_2$ per hour = $59.6 \text{ kg}$
Problem 25: Master Challenge - Conductivity Altered by Electrolysis
Question: $1.0 \text{ L}$ of a $0.1 \text{ M } CuCl_2$ solution is electrolyzed with inert electrodes for 2 hours at $2.68 \text{ A}$. Assuming complete dissociation and ignoring volume changes, calculate the new molar conductivity of the resulting solution if the specific conductivity drops by $40\%$. (Initial $\Lambda_m = 200 \text{ S cm}^2 \text{ mol}^{-1}$).
Strategy: Electrolysis removes $Cu^{2+}$ and $Cl^-$ from the solution. Find the new Molarity ($M_{\text{new}}$) using Faraday's law, find the new specific conductivity ($\kappa_{\text{new}}$), and recalculate $\Lambda_m$.
Step 1: Initial $\kappa = \frac{\Lambda_m \times M}{1000} = \frac{200 \times 0.1}{1000} = 0.02 \text{ S cm}^{-1}$.
New $\kappa = 0.02 \times (1 - 0.40) = 0.012 \text{ S cm}^{-1}$.
Step 2: Faradays passed = $\frac{2.68 \times 7200}{96500} = 0.2 \text{ F}$.
$Cu^{2+} + 2e^- \rightarrow Cu$. $0.2\text{F}$ consumes $0.1 \text{ moles of } Cu^{2+}$.
Step 3: Initial moles of $Cu^{2+} = 0.1 \text{ mol/L} \times 1 \text{ L} = 0.1 \text{ mol}$.
Wait! All $Cu^{2+}$ is consumed! The solution is now essentially depleted of copper salt, leaving only water (if all $Cl^-$ is also gone) or creating a massive $\text{pH}$ shift if water oxidizes. The premise shows total consumption!
Final Answer: The salt is completely electrolyzed. Conductivity relies solely on auto-ionization of water and produced ${H^+}$.
Mastering the Kinetics of Electrochemistry
Congratulations on navigating these 25 grueling, JEE Advanced-tier numericals. While the thermodynamic Nernst equation tells you if a reaction will happen, Conductivity and Faraday's Laws tell you how fast and how much. Remember: always account for the n-factor carefully, watch out for active vs. inert electrodes, and never forget that kinetic overpotential can completely flip theoretical predictions!
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