A simple model for a concentration cell involving a metal `M` is
`M(s)|M^(o+)(aq,0.05 ` molar`)||M^(o+)(aq,1` molar`)|M(s)`
For the abov electrolytic cell, the magnitude of the cell potential is `|E_(cell)|=70mV.`
For the above cell
(a)`E_(cell)lt0,DeltaGgt0`
(b)`E_(cell)gt0,DeltaGlt0`
(c)`E_(cell)lt0,DeltaG^(c-)gt0`
(d)`E_(cell)gt0,DeltaG^(c-)lt0`

Options

A. `E_(cell) lt 0, DeltaG gt 0`
B. `E_(cell) gt 0, DeltaG lt 0`
C. `E_(cell) lt 0, DeltaG^(@) gt 0`
D. `E_(cell) gt 0, DeltaG^(@) lt 0`

Answer

B

Explanation

`M(s) rarr M_(L)^(+)(aq) +e^(-)` at anode
`M_(R)^(+)(aq) +e^(-) rarr M(s)-` at cathode
`bar("Net cell reaction" : M_(R)^(+)(aq) rarr M_(L)^(+)(aq))`
`Q_(cell) = ([M_(L)^(+)])/([M_(R)^(+)])`
From Nernst equation:
`E_(cell) = E_(cell)^(@) - (0.59)/(n) log Q_(Cell)`
`= 0 -(0.59)/(1) log.([M_(L)^(+)])/([M_(R)^(+)])`
`[ :' E_(cell)^(@) = 0` for concentration cell]
`= +0.059 log.([M_(L)^(+)])/([M_(R)^(+)])`
`= 0.059 log.(1)/(0.05)`
so, `E_(cell) = +ve` and `DeltaG =- nFE_(cell) =- ve`