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Question #44783

Difficulty: 8/10 | 🟢 Free

The Question

A 500 mL500\text{ mL} aqueous solution of 2.0 M2.0\text{ M} \ce{NaCl} is electrolyzed using inert Platinum electrodes at 298 K298\text{ K}. A constant current of 9.65 A9.65\text{ A} is passed for 1000 s1000\text{ s}. Given the standard reduction potentials:\nE\ceCl2/Cl−∘=+1.36 VE^\circ_{\ce{Cl_2/Cl^-}} = +1.36\text{ V}\nE\ceO2/H2O∘=+1.23 VE^\circ_{\ce{O_2/H_2O}} = +1.23\text{ V}\nE\ceH2O/H2∘=−0.83 VE^\circ_{\ce{H_2O/H_2}} = -0.83\text{ V}\nE\ceNa+/Na∘=−2.71 VE^\circ_{\ce{Na^+/Na}} = -2.71\text{ V}\nAssuming standard conditions for gas collection (273 K273\text{ K}, 1 atm1\text{ atm}, molar volume =22.4 L/mol= 22.4\text{ L/mol}) and no change in solution volume, calculate the total volume of gases evolved and the final pH of the solution. (Take log⁡2=0.3\log 2 = 0.3, 1 F=96500 C/mol1\text{ F} = 96500\text{ C/mol})

Options

  • A.

    2.24 L, 13.3

  • B.

    1.68 L, 7.0

  • C.

    1.12 L, 13.3

  • D.

    1.68 L, 13.3

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

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Concepts Behind This Question

Electrode Potential Dependencies|Electrolysis Product Prediction

Why This Question Is Tricky

Common Mistakes

This question tests the critical distinction between thermodynamic prediction and kinetic reality in electrochemistry.

Students typically rely solely on standard electrode potentials (E∘E^\circ) to predict electrolysis products.

Thermodynamically, water should oxidize at the anode to produce \ce{O_2} because it requires a lower potential (1.23 V1.23\text{ V}) than chloride ions (1.36 V1.36\text{ V}).

However, the kinetic activation barrier for \ce{O_2} evolution on platinum creates a significant 'overpotential', making \ce{Cl_2} evolution the actual favored pathway.

Failing to account for this exception leads to predicting \ce{O_2} evolution, which yields incorrect gas volumes (1.68 L1.68\text{ L}) and a false neutralization of the solution (pH 7.0) due to the erroneous assumption that \ce{H^+} is produced at the anode to cancel the \ce{OH^-} from the cathode.

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Question Information

SubjectChemistry
ChapterElectrochemistry - Class 12
ConceptElectrode Potential Dependencies
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree