Using tne following half- cell values from the two equations Fe*(aq) + 2e Fe(s); Eo =-0.44 V Cu*(aq) + 2e → Cu(s); Eo = +0.34 V %3D Calculate the following cell potential Fe(s) Fe(C,O4); (0.5M), C,0.(0.02M)|| Cu(NH3),"(0.5M), NH:(IM) Cu(s) Given that Kcomplex formation Cu(NH3),2*: Kof= 1012 Kcomplex formation Fe(C204)3: Kg=2.0*1020

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