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Question #44755

Difficulty: 8/10 | 🟢 Free

The Question

A solution is prepared by dissolving 15.0 g15.0\text{ g} of a salt \ce{MCl2} (Molar mass = 150 g/mol150\text{ g/mol}) in 500 g500\text{ g} of a solvent LL (Molar mass = 50 g/mol50\text{ g/mol}). The vapor pressure of pure solvent LL at 298 K298\text{ K} is 50.51 torr50.51\text{ torr}, and the vapor pressure of the solution is 50.00 torr50.00\text{ torr}. Assume the volume of the solution is exactly 500 mL500\text{ mL}.

The standard reduction potential E∘(\ceM2+/M)E^\circ(\ce{M^{2+}/M}) is −0.75 V-0.75\text{ V}. The solvent LL undergoes reduction at the cathode at a constant potential of −0.82 V-0.82\text{ V} to evolve \ce{H2} gas.

If this solution is electrolyzed at 298 K298\text{ K} using inert electrodes, what is the primary cathodic product and the actual reduction potential of the \ce{M^{2+}/M} couple in this solution?

(Given: 2.303RTF=0.06 V\frac{2.303 RT}{F} = 0.06\text{ V}, log⁡2=0.3\log 2 = 0.3)

Options

  • A.

    \ce{H2} gas is evolved; E(\ceM2+/M)=−0.831 VE(\ce{M^{2+}/M}) = -0.831\text{ V}

  • B.

    Metal \ce{M} is deposited; E(\ceM2+/M)=−0.771 VE(\ce{M^{2+}/M}) = -0.771\text{ V}

  • C.

    Metal \ce{M} is deposited; E(\ceM2+/M)=−0.810 VE(\ce{M^{2+}/M}) = -0.810\text{ V}

  • D.

    \ce{H2} gas is evolved; E(\ceM2+/M)=−0.822 VE(\ce{M^{2+}/M}) = -0.822\text{ V}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Electrolysis Product Prediction|Degree of Dissociation/Association|Vapor Pressure Lowering (Mole Fraction)

Why This Question Is Tricky

Common Mistakes

This problem tests the integration of Colligative Properties with Electrochemistry, specifically targeting the 'Degree of Dissociation' trap.

Students typically assume complete dissociation (α=1\alpha = 1) for salts like \ce{MCl2}, which would yield [\ceM2+]=0.2 M[\ce{M^{2+}}] = 0.2\text{ M} and E=−0.771 VE = -0.771\text{ V}.

Since −0.771 V>−0.82 V-0.771\text{ V} > -0.82\text{ V}, this leads to the incorrect prediction that Metal \ce{M} deposits (Option B).

By forcing the derivation of ii from vapor pressure data, the true [\ceM2+][\ce{M^{2+}}] is found to be much lower due to solvent dielectric effects or ion pairing.

This drops the metal's reduction potential below that of the solvent, fundamentally changing the major electrolysis product.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterElectrochemistry - Class 12
ConceptElectrolysis Product Prediction
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree