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The driving force DeltaG diminishes to z...

The driving force `DeltaG` diminishes to zero on the way to equilibrium, just as in any other spontaneous process. Both `DeltaG` and the corresponding cell potential `(E=-(DeltaG)/(nF))` are zero when the redox reaction comes to equilibrium. The Nernst equation for the redox process of the cell may be given as :
`E=E^(@)-0.059/n log Q`
The key to the relationship is the standard cell potential `E^(@)`, derived from the standard free energy changes as :
`E^(@)=-(DeltaG^(@))/(nF)`
At equilibrium, the Nernst equation is given as :
`E^(@)=0.059/n log K`
the equilibrium constant `K_(c)` will be equal to Q, when :

A

`E=E^(@)`

B

`RT//nF=1`

C

`E=0`

D

`E^(@)=1`

Text Solution

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The correct Answer is:
C
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The driving force DeltaG diminishes to zero on the way to equilibrium, just as in any other spontaneous process. Both DeltaG and the corresponding cell potential (E=-(DeltaG)/(nF)) are zero when the redox reaction comes to equilibrium. The Nernst equation for the redox process of the cell may be given as : E=E^(@)-0.059/n log Q The key to the relationship is the standard cell potential E^(@) , derived from the standard free energy changes as : E^(@)=-(DeltaG^(@))/(nF) At equilibrium, the Nernst equation is given as : E^(@)=0.059/n log K The equilibrium constant K_(c) for the reaction : Cu(s)+2Ag^(+) (aq.)+2Ag(s)" "(E_(cell)^(@)=0.46 V) will be :

The Nernst equation E= E^(@) -RT/nF in Q indicates that the Q will be equal to equilibrium constant K_c when:

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