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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Chapter1: Chemical Foundations
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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,04), (0.5M), C,0,(0.02M)|| Cu(NH;),"(0.5M), NH;(1M)| Cu(s)
Given that
Kcomplex formation Cu(NH3)42*: Kef = 1012
Kcomplex formation Fe(C204)3: Kg=2.0*1020
Transcribed Image Text: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,04), (0.5M), C,0,(0.02M)|| Cu(NH;),"(0.5M), NH;(1M)| Cu(s) Given that Kcomplex formation Cu(NH3)42*: Kef = 1012 Kcomplex formation Fe(C204)3: Kg=2.0*1020
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