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)lt0triangleGge0`
B. `E_(cell)gt0,triangleGlt0`
C. `E_(cell)lt0,triangleG^(@)gt0`
D. `E_(cell)gt0,triangleG^(@)lt0`

Answer

B

Explanation

`M(s)|M^(+)(aq, 0.05M)||M^(+)(aq,1M)|M(s)`
anode: `M(s)toM^(+)(aq)+e^(-)`
Cathode: `underline(M^(+)(aq)+e^(-)toM(s))`
`M^(+)(aq)|_(c)hArrM^(+)(aq)|_(a)`
`E_(Cell)=E_(cell)^(@)-(0.0591)/(1)log((M^(+)(aq)|_(a))/(M^(+)(aq)|_(c)))`
`=0-(0.0591)/(1)log{(0.05)/(1)}`
`=+ve=70mV` and hence `triangleG=-nFE_(cell)=-ve`