Electricity is passed through an acidic solution of Cu^2+ till all the Cu^2+ was exhausted, leading to the deposition of 300 text mg of Cu metal. However, a current of 600 text mA was continued to pass through the same solution for another 28 text minutes by keeping the total volume of the solution fixed at 200 text mL. The total volume of oxygen evolved at STP during the entire process is ____ textmL. (Nearest integer) [Given : Cu^2+(aq)+2e^- rightarrow Cu(s) quad E_red^0 = +0.34 text V O_2(g)+4H^++4e^- rightarrow 2H_2O quad E_red^0 = +1.23 text V Molar mass of Cu = 63.54 text g mol^-1 Molar mass of O_2 = 32 text g mol^-1 Faraday Constant = 96500 text C mol^-1 Molar volume at STP = 22.4 text L ]

Numerical Answer Type:
Enter a numerical value Answer: 111 to 111 +4 marks

Solution & Explanation

### Related Formula textEquivalents of metal deposited = textEquivalents of gas evolved (in Phase 1) n_e^- = fracQF = fracI times t96500 ### Core Logic Phase 1: Deposition of 300 text mg Cu. By Faraday's Laws, the equivalents of copper deposited at the cathode must equal the equivalents of oxygen evolved at the anode during this period. Equivalents of Cu = fracWE = frac300 times 10^-3frac63.542 Equivalents of O_2 = n_O_2 times 4 frac300 times 10^-3 times 263.54 = n_O_2 times 4 2.36 times 10^-3 = n_O_2 (moles of O_2 in Phase 1) Phase 2: Current continued. I = 600 text mA = 0.6 text A, t = 28 text mins = 28 times 60 text seconds. Moles of electrons passed = frac0.6 times 28 times 6096500 = 0.010445 text moles of e^- Equivalents of O_2 evolved in Phase 2 = Moles of e^- passed = 0.010445 n_O_2 text (Phase 2) = frac0.0104454 = 2.611 times 10^-3 text mol ### Step 1: Calculate Total Volume of Oxygen Total moles of O_2 evolved = (2.36 times 10^-3) + (2.611 times 10^-3) = 4.971 times 10^-3 text mol Total volume at STP: V_O_2 = n_total times 22400 text mL V_O_2 = 4.971 times 10^-3 times 22400 text mL = 111.35 text mL Rounding to nearest integer implies 111 text mL. ### Pattern Recognition Equivalents of products at cathode and anode are always equal in any given time span. Valency factor for O_2 evolution from water is 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

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Electrochemistry

More Electrochemistry Previous-Year Questions

Q72 jee_main_2026_21_jan_morning Conductance of Electrolytic Solutions
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. Answer: 6 to 6

Solution

### 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
Q67 jee_main_2026_21_jan_evening Daniell Cell and Cell Potential Variation
For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?
  • A. (1) text Plot 1
  • B. (2) text Plot 2
  • C. (3) text Plot 3
  • D. (4) text Plot 4

Solution

### Core Logic For a closed circuit Daniell cell operating under given conditions, standard cell potential E_textcell^circ remains constant with time during discharge until equilibrium or noticeable concentration shifts occur in a specific manner, or cell potential versus time plots reflect constancy of standard electromotive force terms before polarization effects take over. ### Step 1: Final Conclusion Plot 2 accurately depicts the expected trend where E_textcell^circ remains constant with time. ### Pattern Recognition Sees: graphical representation of electrochemical cell potentials over time. Trap: Confusing standard cell potential E^circ with operating cell potential E_textcell. ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q75 jee_main_2026_21_jan_evening Solubility Product and Electrode Potential
textMX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K: textMX(texts) rightleftharpoons textM^+(textaq) + textX^-(textaq); quad K_textsp = 10^-10 If the standard reduction potential for textM^+(textaq) rightarrow textM(texts) is (E_textM^+/textM^ominus) = 0.79 text V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode E_textX^-/textMX(texts)/textM^ominus is \_\_\_\_ text mV. (nearest integer) [Given: frac2.303RTF = 0.059 text V]
Numerical Answer. Answer: 200 to 200

Solution

### Related Formula E_textX^-/textMX/textM^circ = E_textM^+/textM^circ + frac0.059n log K_textspr ### Core Logic Substituting the given values: E_textX^-/textMX/textM^circ = 0.79 + frac0.0591 log(10^-10) E_textX^-/textMX/textM^circ = 0.79 + 0.059(-10) = 0.79 - 0.59 = 0.20 text V ### Step 1: Converting to Millivolts 0.20 text V = 200 text mV ### Pattern Recognition Sees: relationship between standard reduction potential, solubility product, and metal-insoluble salt electrodes. Trap: Sign or logarithmic base calculation errors. ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q72 jee_main_2026_22_january_morning Nernst Equation
Consider the following electrochemical cell at 298K Pt | HSnO_2^-(aq) | Sn(OH)_6^2-(aq) || Bi_2O_3(s) | Bi(s) If the reaction quotient at a given time is 10^6, then the cell EMF (E_textcell) is x times 10^-1text V (Nearest integer). Given the standard half-cell reduction potential as E_Bi_2O_3/Bi,OH^-^0 = -0.44text V text and E_Sn(OH)_6^2-/HSnO_2^-,OH^-^0 = -0.90text V
Numerical Answer. Answer: 4 to 4

Solution

### Related Formula E_cell = E_cell^circ - frac0.0591n log Q (Using 0.06 for approximation if needed, as per standard practice unless specified). ### Core Logic First, calculate E^circ_cell: E^circ_cell = E^circ_cathode - E^circ_anode E^circ_cell = (-0.44text V) - (-0.90text V) = +0.46text V Identify the number of transferred electrons (n) for the overall reaction. Anode reaction: Sn^2+ rightarrow Sn^4+ + 2e^- (Tin goes from +2 in HSnO_2^- to +4 in Sn(OH)_6^2-). Cathode reaction: Bi^3+ + 3e^- rightarrow Bi (Bismuth goes from +3 in Bi_2O_3 to 0). To balance the electrons, we multiply anode by 3 and cathode by 2, yielding n = 6 total electrons transferred. ### Step 1: Apply Nernst Equation Applying the Nernst equation (approximating 0.0591 to 0.06 for simplicity): E_cell = E^circ_cell - frac0.06n log_10 Q E_cell = 0.46 - frac0.066 log_10(10^6) E_cell = 0.46 - 0.01 times 6 E_cell = 0.46 - 0.06 = 0.40text V ### Step 2: Format Output The problem asks for the answer in the form x times 10^-1text V. 0.40 = 4 times 10^-1 So, x = 4. ### Pattern Recognition Be extremely careful finding 'n'. You must balance the half-reactions. A +2 to +4 change (n=2) and a +3 to 0 change (n=3) resolves to a common multiple of 6. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q67 jee_main_2026_22_january_evening Standard Reduction Potentials and Reducing Strength
Consider the following reduction processes: textAl^3+ + 3e^- rightarrow textAl(texts), E^0 = -1.66text V textFe^3+ + e^- rightarrow textFe^2+, E^0 = +0.77text V textCo^3+ + e^- rightarrow textCo^2+, E^0 = +1.81text V textCr^3+ + 3e^- rightarrow textCr(texts), E^0 = -0.74text V The tendency to act as reducing agent decreases in the order:
  • A. textAl > textCr > textFe^2+ > textCo^2+
  • B. textAl > textFe^2+ > textCr > textCo^2+
  • C. textAl > textCr > textCo^2+ > textFe^2+
  • D. textCr > textFe^2+ > textAl > textCo^2+

Solution

### Related Formula textReducing Power propto frac1textStandard Reduction Potential (E^0_red) ### Core Logic Step 1: List standard reduction potential values: - E^0(textAl^3+/textAl) = -1.66text V - E^0(textCr^3+/textCr) = -0.74text V - E^0(textFe^3+/textFe^2+) = +0.77text V - E^0(textCo^3+/textCo^2+) = +1.81text V Step 2: Arrange species in increasing order of E^0_red (decreasing reducing power): textAl > textCr > textFe^2+ > textCo^2+ ### Pattern Recognition Sees: Reduction potential values for reducing strength ordering. Shortcut: More negative standard reduction potential implies stronger reducing agent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

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