For a cell involving one electron $E^\circ_{cell} = 0.59$ V at 298 K, the equilibrium constant for the cell reaction is : [Given that $\frac{2.303RT}{F} = 0.059$ V at T = 298 K]

Question

Dr. Abubacker · Accepted Answer

[Solution] $\log K = \frac{nE^\circ}{0.059} = \frac{0.59}{0.059} = 10$, so $K = 10^{10}$. Answer (3).

Q. For a cell involving one electron $E^\circ_{cell} = 0.59$ V at 298 K, the equilibrium constant for the cell reaction is : [Given that $\frac{2.303RT}{F} = 0.059$ V at T = 298 K]

Explanation

Reviewed by Dr. Abubacker

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