If the `E_("cell")` of an equlibrium reaction
`A( s) + 2B^(+ (aq) hArr A^(2+) (aq) + 2B(s )`
at 298 K is 0.59 V , the equlibrium constant K, is

Options

A. `1.0 xx 10^(10)`
B. `1.0 xx 10^(2)`
C. `1.0 xx 10^(-20)`
D. `1.0 xx 10^(20)`

Answer

D

Explanation

For the reaction
`A+ 2B^(2+) (a) hArr A^(2+) (aq) + 2B(s)`
`DeltaG^(@) = RT "in" k_(c) =- n FE^(@)`
where
`Delta G^(@)` Gibbs free energy
n- number of electron involved =(2)
F = Faraday constant = 96 500 C charge
R = gas constant
T = temperature (=298 K )
From Eq . (i) we have
or log `k_(c ) = (n FE^(@))/( 2.303 RT)`
`=(2xx96500 xx 0.59)/( 2.303 xx 8.314 xx 298 ) =-19.96 = 20`
`K_(e ) = 1.0 xx 10^(20)`
Hence option (c ) is the correct answer
Alternativel
At T= 298 K
In `k_(c ) = (nE^(@) )/( 0.059 ) l = (2xx 0.59)/( 0.059) = 20`
`K_(c ) = 1xx 10^(20)`