Consider the following electrochemical cell at standard condition. mathrmAu(s) vert mathrmQH_2, mathrmQ vert mathrmNH_4mathrmX (0.01 mathrmM) vert vert mathrmAg^+ (1 mathrmM) vert mathrmAg(s) mathrmE_textcell = +0.4 mathrmV The couple mathrmQH_2 / mathrmQ represents quinhydrone electrode, the half cell reaction is given below:
Quinhydrone half cell reduction equation diagram for Q47
The diagram displays the balanced chemical equation for quinhydrone reduction, consuming two electrons and two protons to yield hydroquinone.
left[ textGiven: E_Ag^+ / Ag^o = +0.8 mathrmV text and frac2.303 RTF = 0.06 mathrmV right] The mathrmpK_b value of the ammonium halide salt (mathrmNH_4mathrmX) used here is _____.

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

Solution & Explanation

### Related Formula Nernst equation for the net combined redox cell expression: E = E^circ - frac0.062logleft(frac[mathrmH^+]^2[mathrmAg^+]^2right) Hydrolysis equation for a salt composed of a weak base and strong acid: mathrmpH = 7 - frac12mathrmpK_b - frac12logmathrmC ### Core Logic Let's compute the operational values line-by-row: * Combined redox process: mathrmQH_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. * Apply Nernst adjustments using known concentrations ([mathrmAg^+] = 1mathrm~M): 0.4 = 0.1 - 0.06 log [mathrmH^+] 0.3 = 0.06 times mathrmpH implies mathrmpH = 5 ### Step 1: Salt Hydrolysis Substitution Substitute the determined mathrmpH along with salt molarity (C = 0.01mathrm~M = 10^-2mathrm~M) into the hydrolysis equation: 5 = 7 - frac12mathrmpK_b - frac12log(10^-2) 5 = 7 - frac12mathrmpK_b - frac12(-2) 5 = 7 - frac12mathrmpK_b + 1 5 = 8 - frac12mathrmpK_b implies frac12mathrmpK_b = 3 implies mathrmpK_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 at standard ambient conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

More Electrochemistry Previous-Year Questions — Page 6

Q65 jee_main_2024_31_jan_morning Batteries
The metals that are employed in the battery industries are A. Fe B. Mn C. Ni D. Cr E. Cd Choose the correct answer from the options given below:
  • A. textB, C and E only
  • B. textA, B, C, D and E
  • C. textA, B, C and D only
  • D. textB, D and E only

Solution

### Core Logic Mn, Ni, and Cd metals are predominantly used in battery industries. - Mn is used in dry cells (Leclanche cell). - Ni and Cd are used in Nickel-Cadmium (Ni-Cd) rechargeable batteries. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q67 jee_main_2024_31_jan_morning Electrolytic Conductance
Identify the factor from the following that does not affect electrolytic conductance of a solution.
  • A. textThe nature of the electrolyte added.
  • B. textThe nature of the electrode used.
  • C. textConcentration of the electrolyte.
  • D. textThe nature of solvent used.

Solution

### Core Logic Conductivity of an electrolytic cell is affected by the concentration of the electrolyte, the nature of the electrolyte, and the nature of the solvent. It does not depend on the nature of the electrode used. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q90 jee_main_2024_31_jan_morning Faraday's Laws of Electrolysis
One Faraday of electricity liberates x times 10^-1 gram atom of copper from copper sulphate, x is
Numerical Answer. Answer: 5 to 5

Solution

### Core Logic The reduction reaction for copper is: Cu^2+ + 2e^- rightarrow Cu This shows that 2 moles of electrons (2 Faraday) are required to deposit 1 mole (or 1 gram atom) of Cu. Therefore, 1 Faraday of electricity will deposit: frac12 = 0.5 text moles of Cu ### Step 1: Finding x 0.5 text mole = 0.5 text gram atom = 5 times 10^-1 text gram atom Hence, x = 5. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)