`E^(@)` for the electrochemical cell
`Zn(s)|Zn^(2+) 1 M (Aq.)||Cu^(2+) 1 M (aq.)|Cu(s)`
is 1.10 V at `25^(@)C`. The equilibrium constant for the cell reaction,
`Zn(s) +Cu^(2+) (aq.) hArr Zn^(2+) (aq.)+Cu(s)`
Will be :

Options

A. `10^(-37)`
B. `10^(-28)`
C. `10^(18)`
D. `10^(17)`

Answer

A

Explanation

(a) `Zn(s)+Cu^(2+)(aq) iff Cu(s) +Zn^(2+)(aq),E^(@)=+1.10V`
`:. E^(@)=(0.0591)/(n) log_(10) K_(eq)`
because at equilibrium , `E_(cell)=0`
(n= number of electrons exchanged =2)
`1.10=(0.0591)/(2) log_(10)K_(eq)`
`(2.20)/(0.0591)= log_(10) K_(eq)`
`K_(eq)=` antilog 37.225
`K_(eq)=1.66xx10^(-37)`