For an electrode reaction written as
`M^(n+)+n e^(-)rarr M` ,
`E_("red")=E_("red")^(@)-(RT)/(nf)ln(1)/([M^(n+)])`
`=E_("red")^(@)-(0.059)/(n)"llog"(1)/([M^(n+)])` at 298 k
For the cell reaction `aA+bB rarr xX+yY`
`rArr E_("cell")=E_("cell")^(@)-(RT)/(nf)"log"([X]^(x)[Y]^(y))/([A]^(a)[B]^(b))`
For pure solids, liquids or gases at lampt.
molarconc = 1
Std. free energy change `Delta G^(@)=-nfE^(@)` where f - for faraday = 96,500 c, n = no. of `e^(-)`
`Delta G=-2.303 RT log K_(c )` where `K_(c )` is equilibrium constant `K_(c )` can be calculated by using `E^(@)` of cell.
The emf of the cell
`Zn//zn^(2+)(0.01M)//Fe^(2+)(0.001M)//Fe` at 298 K is `0.2905`. The value of equilibrium constant for cell reaction si

Options

A. `e^((0.32)/(0.0295))`
B. `1-^((0.32)/(0.0295))`
C. `10^(0.26//0.0295)`
D. `10^((0.32)/(0.059))`

Answer

B

Explanation

`Zn+Fe^(2+)rarr Fe+Zn^(2+)`
`E=E^(@)-(0.059)/(n)log Ke` Given E = 0.2905
i.e., `0.2905 = E^(@)-(0.059)/(2)"log"(0.01)/(0.001)`
`E^(@)=0.2905+0.0295 log 10=0.32`
`E^(@)=(0.059)/(n) log K_(eq) =0.32 =(0.059)/(2)log K_(eq)`
`K_(eq)=10^(0.32//0.0295)`