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can we write Nernst eequation as E(cell)=emf + 0.0592/n log oxidised state /redused state?
Yes.For Example Lets consider copper/zinc cell,

(anode)        Cu2+(aq)  +  2 e-  = Cu(s)              E0 = +0.337 volt

(cathode)      Zn(s)  =  Zn2+(aq)  +  2 e-              E0 = +0.763 volt
(overall)     Cu2+(aq)  + Zn(s)  = Cu(s) + Zn2+(aq)    E0 = +1.100 volt


Cu2+(aq)  + Zn(s)  = Cu(s) + Zn2+(aq)    E0 = +1.100 volt

 

(1)  Eeff          = 1.100 - 0.0592/2 · log ({Zn++}/{Cu++} )    (Nernst equation for the whole cell)

(2)  Ecathode = +0.337 - 0.0592/2 · log{Cu++}                     (Nernst equation for the cathode)

(3)  Eanode    = -0.763 - 0.0592/2 · log{Zn++}                      (Nernst equation for the anode)

It can be easily shown that eq(2) - eq(3)  yields eq(4), so that      Eeff =  Ecathode  -  Eanode 


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