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Question #45029

Difficulty: 8/10 | 🟢 Free

The Question

An electrochemical cell is constructed with two silver electrodes immersed in aqueous \ce{AgNO3} solutions of concentrations $0.1 \text{ M}$ (Left compartment) and $0.001 \text{ M}$ (Right compartment) at $298 \text{ K}$ . The two compartments are separated by a porous plug that allows ion diffusion. The transport numbers of \ce{Ag+} and \ce{NO3-} in this concentration range are $0.4$ and $0.6$ , respectively.

An external DC voltage source of $1.144 \text{ V}$ is connected across the electrodes to force the oxidation of Ag in the Left compartment and the reduction of \ce{Ag+} in the Right compartment. The total internal and external resistance of the circuit is $5 \ \Omega$ .

Assuming the overpotential is negligible, activity coefficients are unity, $\frac{2.303 RT}{F} = 0.06 \text{ V}$ , and the bulk concentrations remain essentially constant, calculate the mass of silver deposited at the cathode after $9650 \text{ s}$ of electrolysis. (Molar mass of Ag = $108 \text{ g/mol}$ , $1 \text{ F} = 96500 \text{ C/mol}$ )

Options

A.
$2.16 \text{ g}$
B.
$2.21 \text{ g}$
C.
$2.26 \text{ g}$
D.
$2.47 \text{ g}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Equivalent Concept in Redox|Electrolysis Product Prediction|Liquid Junction Potential

Why This Question Is Tricky

Common Mistakes

The problem tests the integration of Liquid Junction Potential (LJP) into Faraday's Laws of Electrolysis.

Standard problems assume ideal Nernstian behavior, but here the LJP creates an additional potential difference.

Because the transport number of the anion is greater than the cation ( $t_- > t_+$ ), the LJP increases the spontaneous back-emf of the concentration cell from $0.12 \text{ V}$ to $0.144 \text{ V}$ .

Students who ignore the LJP will calculate a higher net driving voltage ( $1.024 \text{ V}$ ) and incorrectly choose $2.21 \text{ g}$ .

Students who misapply the LJP polarity will subtract it, leading to $2.26 \text{ g}$ .

Recognizing that the thermodynamic work done by the external source must overcome the sum of the half-cell potentials AND the junction potential is the key to finding the true current.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Equivalent Concept in Redox
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free