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For the cell reaction Cu^(2+)(C(1),aq....

For the cell reaction
`Cu^(2+)(C_(1),aq.)+Zn(s)hArrZn^(2+)(C_(2),aq)+Cu(s)`
of an electrochemical cell, the change in free energy `(DeltaG)` of a given temperature is a function of

A

ln `(C_(1))`

B

ln `(C_(2))`

C

ln `(C_(1)+C_(2))`

D

ln `(C_(2)//C_(1))`

Text Solution

Verified by Experts

The correct Answer is:
D
`E_(cell)=E_(cell)^(@)-(RT)/(nF)` In `(C_(2))/(C_(1)) and DeltaG=-nFE_(cell)`
Hence `DeltaG` is the function of `ln ((C_(2))/(C_(1)))`
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