The equilibrium constant of the reaction : `A_((s)) + 2 B_((aq))^(+) hArr A_((aq))^(2+) + 2 B_((s)) , E_("cell")^(@) = 0.0295V` is `[(2.303 RT)/(F) = 0.059]`

Options

A. 10
B. `2 xx 10^(2)`
C. `3 xx 10^(2)`
D. `2 xx 10^(5)`

Answer

A

Explanation

`E_("cell")^(@) = (0.0591)/(n) log_(10) K`
`n = 2 , E_("cell")^(@) = 0.0295` V
`0.0295 = (0.0591)/(2) log_(10) K`
K = antilog 0.998
K = 9.96 = 10