Question

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)`

Answer 1

Correct Answer - 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)`