The pH and conductance of a weak acid (HX) was found to be 5 and 4 times 10^-5 S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1text cm^2 were at a distance of 15text cm apart. The value of the limiting molar conductivity is ..... textS m^2text mol^-1. (nearest integer) (Given: degree of dissociation of the weak acid (alpha) ll 1)

Numerical Answer Type:
Enter a numerical value Answer: 6 to 6 +4 marks

Solution & Explanation

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

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Electrochemistry

More Electrochemistry Previous-Year Questions — Page 5

Q82 jee_main_2024_01_february_morning Nernst Equation
The potential for the given half cell at 298K is (-)dotsdotsdotsdots times 10^-2 mathrm~V. 2mathrmH^+_text(aq) + 2e^- rightarrow mathrmH_2mathrm(g) [mathrmH^+] = 1 mathrmM, P_mathrmH_2 = 2 mathrm~atm Given: 2.303mathrmRT/F = 0.06mathrmV, log 2 = 0.3
Numerical Answer. Answer: 0.9 to 1

Solution

### Related Formula E = E^circ - frac2.303RTnF log Q For the Standard Hydrogen Electrode half-reaction: 2H^+ + 2e^- rightarrow H_2 E_H^+/H_2 = E^circ_H^+/H_2 - frac0.062 log fracP_H_2[H^+]^2 ### Step 1: Substitute the given values E^circ_H^+/H_2 = 0.00 mathrm~V (by definition) [H^+] = 1 mathrm~M P_H_2 = 2 mathrm~atm n = 2 electrons E = 0.00 - frac0.062 log left( frac21^2 right) ### Step 2: Solve the calculation E = -0.03 log 2 Given log 2 = 0.3 E = -0.03 times 0.3 E = -0.009 mathrm~V E = -0.9 times 10^-2 mathrm~V ### Step 3: Match the requested format The question asks for (-) dots times 10^-2 mathrm~V. This gives exactly 0.9. For NAT type with integer expected, 0.9 can be rounded to 1. However, exact calculation yields 0.9. According to official JEE rounding, 0.9 approx 1. ### Pattern Recognition Hydrogen electrode non-standard potential depends strictly on pressure of H_2 and concentration of H^+. If [H^+]=1, increasing H_2 pressure lowers the potential below zero (makes it negative). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q89 jee_main_2024_29_january_evening Faraday's Laws of Electrolysis
A constant current was passed through a solution of mathrmAuCl_4^- ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was 1.314mathrmg. The total charge passed through the solution is ________ times 10^-2mathrmF. (Given atomic mass of mathrmAu = 197)
Numerical Answer. Answer: 2 to 2

Solution

### Related Formula textNumber of equivalents deposited = fracW, E = fracQ, F textEquivalent Weight (E) = fractextAtomic Mass, ntext-factor ### Core Logic In the reduction of gold from the tetrachloroaurate(III) complex anion: mathrmAuCl_4^- + 3e^- rightarrow mathrmAu(s) + 4mathrmCl^- implies ntext-factor = 3 Calculate the equivalent weight (E) of Gold: E = frac197, 3 Set up the Faraday equivalence relation to solve for charge (Q in Faradays): frac1.314, left(frac197, 3right) = Q ### Step 1: Arithmetic Resolution Q = frac1.314 times 3, 197 = frac3.942, 197 = 0.02text F = 2 times 10^-2text F Thus, the required integer value is **2**. ### Pattern Recognition Always determine the correct change in oxidation state (+3 to 0) to establish the proper n-factor value for calculations using Faraday's laws. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q81 jee_main_2024_27_jan_morning Faraday's Laws of Electrolysis
The mass of silver (Molar mass of textAg: 108text g mol^-1) displaced by a quantity of electricity which displaces 5600text mL of O_2 at S.T.P. will be textquadquad g.
Numerical Answer. Answer: 107 to 108

Solution

### Related Formula By Faraday's Second Law of Electrolysis: textEquivalents of Ag = textEquivalents of O_2 textEquivalents = fractextMasstextEquivalent Mass = textMoles times ntext-factor ### Step 1: Calculate equivalents using standard metrics Let x grams of Silver be displaced. Using the older STP molar volume baseline (22.4text L or 22400text mL): textMoles of O_2 = frac560022400 = 0.25text moles Since the n-factor of O_2 is 4 (2textO^2- rightarrow textO_2 + 4texte^-): textEquivalents of O_2 = 0.25 times 4 = 1 ### Step 2: Equating equivalents for silver mass textEquivalents of Ag = fracx108 times 1 = 1 implies x = 108text g ### Step 3: Alternative calculation using current STP metric Using modern STP volume metrics (22.7text L): fracx times 1108 = frac5.622.7 times 4 implies x approx 106.57text g rightarrow 107text g ### Pattern Recognition Equivalents equations bypass complex current/time measurements. Always link volume fractions directly to n-factor equivalents. ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Some Basic Concepts of Chemistry
Q82 jee_main_2024_29_jan_morning Faradays Laws of Electrolysis
The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is \_\_\_\_\_\_ times 10^-4 g. (Atomic mass of zinc = 65.4 amu)
Numerical Answer. Answer: 45.75 to 46

Solution

### Related Formula W = Z cdot I cdot t = fracMn cdot F cdot I cdot t where, W = mass deposited Z = electrochemical equivalent I = current in amperes t = time in seconds M = molar mass n = n-factor (electrons exchanged) F = Faraday's constant (96500 text C/mol) ### Core Logic The electrolysis of zinc sulphate (ZnSO_4) involves the reduction of zinc ions at the cathode: Zn^+2 + 2e^- rightarrow Zn Here, the n-factor (n) is 2. ### Step 1: Calculation Given values: I = 0.015text A t = 15text minutes = 15 times 60text seconds = 900text s M = 65.4text g/mol F approx 96500text C Plugging the values into Faraday's First Law: W = frac65.42 times 96500 times 0.015 times 15 times 60 W = frac65.4193000 times 13.5 W = 3.3886 times 10^-4 times 13.5 W = 45.746 times 10^-4text g Rounding to two decimal places (or nearest integer depending on convention), we get 45.75 times 10^-4text g. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q80 jee_main_2024_30_january_evening Standard Electrode Potential
Reduction potential of ions are given below: mathrmClO_4^- quad E^circ = 1.19mathrmV mathrmIO_4^- quad E^circ = 1.65mathrmV mathrmBrO_4^- quad E^circ = 1.74mathrmV The correct order of their oxidising power is:
  • A. mathrmClO_4^- > mathrmIO_4^- > mathrmBrO_4^-
  • B. mathrmBrO_4^- > mathrmIO_4^- > mathrmClO_4^-
  • C. mathrmBrO_4^- > mathrmClO_4^- > mathrmIO_4^-
  • D. mathrmIO_4^- > mathrmBrO_4^- > mathrmClO_4^-

Solution

### Core Logic The Standard Reduction Potential (E^circ) measures a species' tendency to undergo reduction (gain electrons). A higher, more positive E^circ value means the species has a stronger tendency to be reduced, which in turn makes it a stronger oxidizing agent. Comparing the given E^circ values: mathrmBrO_4^-: 1.74mathrmV mathrmIO_4^-: 1.65mathrmV mathrmClO_4^-: 1.19mathrmV The order of oxidizing power follows the magnitude of the reduction potential: mathrmBrO_4^- > mathrmIO_4^- > mathrmClO_4^- ### Pattern Recognition Higher +ve Standard Reduction Potential (SRP) = Stronger Oxidising Agent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 12 Chemistry: The p Block Elements

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