1. 1.0×10² 2. 1.0×10⁵ 3. 1.0×10¹⁰ 4. 1.0×10³⁰

Answer: 3. 1.0×10¹⁰ At 298 K, E^(ominus)_cell = 0.059/nlog K . log K = dfracnE^(ominus)0.059 = 1×0.59/0.059 = 10 , so K = 1.0×10¹⁰ . Option (3).

For a cell involving one electron E^_cell = 0.59 V at 298 K, the equilibrium constant for the cell reaction is :[Given that 2.303 RT/F = 0.059 V at T = 298 K]

1. \(1.0\times10^{2}\)

2. \(1.0\times10^{5}\)

3. \(1.0\times10^{10}\)

4. \(1.0\times10^{30}\)

Doubtly Asked 30 Oct 2026 NEET UG 2019 Practice test neet Save WhatsApp Report

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Answer: 3. \(1.0\times10^{10}\)

At 298 K, \(E^{\ominus}_{cell} = \dfrac{0.059}{n}\log K\).

\(\log K = \dfrac{nE^{\ominus}}{0.059} = \dfrac{1\times0.59}{0.059} = 10\), so \(K = 1.0\times10^{10}\). Option (3).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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