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For a cell reaction involving a two-ele...

For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be `0.295V` at `25^(@)C`. The equilibrium constant of the reaction at `25^(@)C` will be:

A

`10`

B

`1 xx 10^(10)`

C

`1 xx 10^(-10)`

D

`29.5 xx 10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(cell) = E_(cell)^(@) - (0.0591)/(2) log Q`,
were `Q =` reaction quotient.
At equilibrium
`E_(cell) = 0, Q = K`
`E_(cell) = 0.295 -0.0295 log K`
`log K = 10`
`:. K = 10^(-10)`
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