Question

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=96,500 C mol^(-1), R = 8.314 JK^(-1) mol^(-1)`):
A. `2.0 xx 10^(11)`
B. `4.0 xx 10^(12)`
C. `1.0 xx 10^(2)`
D. `1.0 xx 10^(10)`

Answer 1

Correct Answer - D
According to Nernst equation
`E_("cell")^(@) = (2.303 RT)/(nF)log Q`
At equilibrium
`E_("cell") = 0`
`Q = K_(eq.)` Thus,
`E_("cell")^(@) = (2.303 RT)/(nF) log K_(eq.)`
or `log K_(eq.) = (nFE_("cell")^(@))/(2.303 RT)` Substituting the value, we get
`= ((2 mol)(96500 C mol^(-1))(0.295 V))/((2.303)(8.314 JK^(-1) mol^(-1)(298 K))`
= 10
or `K^(eq.) = "antilog" 10`
`= 1.0 xx 10^(10)`