Write the Nernst equation for the cell reaction in the Daniel cell. How will the ECell be affected when the concentration of Zn2+ ions is increased?


Galvanic cell reaction: Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)


Anode- oxidation half-cell: Zn(s) → Zn2+ (aq)+ 2e-


Cathode- reduction half-cell: Cu2+ (aq)+ 2e-→ Cu(s)


Overall reaction: Zn(s) + Cu2+ (aq) → Zn2+(aq) +Cu(s)


Nernst equation:



Where, E cellis the electrode potential at any given concentration


E° cell is the standard electrode potential


R is the gas constant 8.314 JK-1mol-1


F is the Faraday constant (96500 C mol-1)


T is the temperature in Kelvin


[Zn2+] and [Cu2+] are the concentration of the Zn2+ and Cu2+ ion in the solution.


When we substitute the value for R, F and T(298K), we can reduce the equation to :



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