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What is the value of the equilibrium constant when the cell potential is found to be −0.29 v for a transfer of 2 moles of electrons at 298.15 k?
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What is the value of the equilibrium constant when the cell potential is found to be −0.29 v for a transfer of 2 moles of electrons at 298.15 k?

User Kalatabe
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7 votes

Answer:

Step-by-step explanation:

The formula relating equilibrium constant and cell potential is given below

E = RT / nF ln K , E is cell potential , T is temperature , n is no of moles , F is

one farad .

.29 = (8.31 x 298.15 / 2 x 96485) ln K

lnK = 22.58

K = e ²²°⁵⁸

= 6.4 x 10⁹

User RichieV
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