The standard emf of a galvanic cell involving cell reaction with n=2 is found to be 0.295 v at `25^(@)c` the equilibrium constant of the reaction would be (Given F= 96500 C `"mol"^(-1)m R =8 .314 "jk"^(-1) "mol"^(-1)` )

Options

A. `2.0 xx10^(11)`
B. `4.0 xx10^(12)`
C. `1.0xx10^(2)`
D. `1.0 xx10^(10)`

Answer

B

Explanation

By Nernst equation `E_(Cell) =E_(cell)^(@)-(2.303 RT)/(nF) log_(120) k`
At equilibriuim `E_(cell) =0`
Given that `R= 8.315 JK^(-1) mol^(-1)`
`therefore E_(Cell)^(@) =(0.0591)/(2) log_(10) K`
`because` Given that `E_(cell)^(@) =0.295 V`
`k=1xx10^(10)`