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The standard emf of galvanic cell involving 3 moles of electrons in its redox reaction is 0.59 V. The equilibrium constant for the reaction of the cell is
- a)1015
- b)1030
- c)1025
- d)1020
Correct answer is option 'B'. Can you explain this answer?
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The standard emf of galvanic cell involving 3 moles of electrons in it...
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The standard emf of galvanic cell involving 3 moles of electrons in it...
Given data:
Standard EMF of cell, E°cell = 0.59 V
Number of electrons involved in the redox reaction, n = 3
We know that the relation between standard EMF and equilibrium constant for a cell reaction is given by the following formula:
E°cell = (0.0591/n) x log K
where,
K = equilibrium constant of the cell reaction
Substituting the given values in the above equation, we get:
0.59 = (0.0591/3) x log K
log K = (3 x 0.59) / 0.0591
log K = 30.303
Taking antilog on both sides, we get:
K = antilog (30.303)
K = 10^30.303
K = 1.995 x 10^30
Therefore, the equilibrium constant for the given cell reaction is 10^30.303, which is approximately equal to 10^30. Hence, the correct option is (b) 1030.
Standard EMF of cell, E°cell = 0.59 V
Number of electrons involved in the redox reaction, n = 3
We know that the relation between standard EMF and equilibrium constant for a cell reaction is given by the following formula:
E°cell = (0.0591/n) x log K
where,
K = equilibrium constant of the cell reaction
Substituting the given values in the above equation, we get:
0.59 = (0.0591/3) x log K
log K = (3 x 0.59) / 0.0591
log K = 30.303
Taking antilog on both sides, we get:
K = antilog (30.303)
K = 10^30.303
K = 1.995 x 10^30
Therefore, the equilibrium constant for the given cell reaction is 10^30.303, which is approximately equal to 10^30. Hence, the correct option is (b) 1030.
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