Verified Solution • 2025–26 Session moderate

QNEET-2019-Ch-762: NEET 2019 Chemistry - Class 12 NEET-2019-Ch-762 Solution: For a cell involving one electron \(E_{cell}^ = 0.59\ {V}\) at…

For a cell involving one electron Ecell∘=0.59 VE_{cell}^\circ = 0.59\ \text{V} at 298 K the equilibrium constant for the cell reaction is: [Given that 2.303RTF=0.059 V\frac{2.303RT}{F} = 0.059\ \text{V} at T = 298 K] 1.0×10301.0 \times 10^{30} 1.0×10101.0 \times 10^{10} 1.0×1021.0 \times 10^2 1.0×1051.0 \times 10^5

Multiple Choice Options

  • A 1.0 ×\times _ 10 30
  • B 1.0 ×\times _ 10 10
  • C 1.0 ×\times _ 10 2
  • D 1.0 ×\times _ 10 5

Step-by-Step Verified Solution

We know, E cell = E c e l l _ - 2.303 R T n F log _ Q _ E cell = E c e l l _ - 0.059 n log _ Q (At equilibrium, E cell = 0 and Q = K eq ) _ 0 = E c e l l _ - 0.059 1 log _ K e q _ log _ K e q = E c e l l _ 0.059 = 0.59 0.059 = 10 _ K eq = 10 10

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