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NEET CHEMISTRYElectrochemistryEasy

Question

For a cell involving one electron Ecell=0.59 VE^\ominus_{\text{cell}} = 0.59 \text{ V} at 298 K298 \text{ K}. The equilibrium constant for the cell reaction is: [Given that 2.303RTF=0.059 V\frac{2.303 RT}{F} = 0.059 \text{ V} at T=298 KT = 298 \text{ K}]

A

1.0×10301.0 \times 10^{30}

B

1.0×1021.0 \times 10^2

C

1.0×1051.0 \times 10^5

D

1.0×10101.0 \times 10^{10}

Step-by-Step Solution

The relationship between the standard emf of a cell (EcellE^\ominus_{\text{cell}}) and the equilibrium constant (KcK_c) for the cell reaction is given by the Nernst equation at equilibrium:

Ecell=2.303RTnFlogKcE^\ominus_{\text{cell}} = \frac{2.303 RT}{nF} \log K_c

Given: Ecell=0.59 VE^\ominus_{\text{cell}} = 0.59 \text{ V} n=1n = 1 (since the cell involves one electron) 2.303RTF=0.059 V\frac{2.303 RT}{F} = 0.059 \text{ V}

Substituting these values into the equation: 0.59=0.0591logKc0.59 = \frac{0.059}{1} \log K_c logKc=0.590.059=10\log K_c = \frac{0.59}{0.059} = 10

Taking the antilog of both sides: Kc=1010=1.0×1010K_c = 10^{10} = 1.0 \times 10^{10}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Electrochemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYElectrochemistryinvolvingelectroneominustextcellequilibriumconstant

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