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Cr(s)|Cr^(3+)||Fe^(2+)|Fe(s) In above ...

`Cr(s)|Cr^(3+)||Fe^(2+)|Fe(s)`
In above cell, the value of n in the Nernst equation :
i.e., `E=E^(@)-(0.059)/n log_(10) Q` will be :

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Cr(s)| Cr^(3+)||Fe^(2+)|Fe(s) In the above cell, the value of n in the Nernst equation i.e., E=E^(@) - (0.059)/(n) log_(10)Q will be :

For the cell : Cr(s)|Cr^(3+)||Fe^(2+)|Fe(s) DeltaG^(@)=-nFE^(@) , the value of n will be :

Assertion A :Increase in the concentration of copper half cell in Daniel cell increases the emf of the cell. Reason R: According to Nernst equation. E_(cell)=E_(cell)^(0)+(0.059)/(2)log.([Cu^(++)])/([Zn^(++)])

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 :