Question

Electrochemistry

Voltage (V) Expected Voltar 495 Electrochemical Cells 2 -105 M F 05 MF 2017-05M) "05 M) , 22 (0,5 M)005 M . FeFe (0.5 MH0.5 M). FeFe (0.5 M (0.5 M Cucu (0.5 MH 10.5 MM ZnZn (0.10 M ZR 1.0 M) 930 760 1022 1100 404 440 + 55 720 Sede 7143 – 340 29 Post-Lab questions 1 Explain how you can predict which of the electrode in the electrochemical pair should be the cathode or anode. 2 Based on your results, arrange the ion (ZnFeCut and H) in order of decreasing potentials duct Using the electrochemical cell ZI -05 MC 0.5 MC what you have observed Why is it different relate the expected E

Answer 1

Standard reduction potentials: Pencara reducing orang tew - 0.78 v → It oxidizes —>Anode Zn2+ + 20- (aq) + 2e Zn E = -0.76 v -> It oxidizes — Anode 2H*ag + 2€ - H2(g) Ered = 0.00 v Zn/Zn2+(0.5M)//H*(0.5M)/Cu(3) Anode Oxidation Cathode Reduction

Standard reduction potentials: Zn2+29+ 2e → Zn Ered =- 0.76 v → It oxidizes Anode 2H*aq+ 2€ — H2(g) Ered = 0.00 v Zn/Zn2+(0.5M)//H*(0.5M)/Cus) Anode Oxidation Cathode Reduction

Standard reduction potentials: Zn2+ E = -0.76 V (aq) + 2e red → Lower potential value, the greater the tendency to oxidize. Anode cu tag + 2e → Cu, Fed = 0.34 v Zn/Zn2+(0.5M)//Cu²+(0.5M)/Cus) Anode Cathode Oxidation Reduction

Fes/Fe²+(0.5M)//H*(0.5M)/Cus) Standard reduction potentials: Lower potential value, Fe2+29+ 2€ → Fes Eºred = -0.44 v the greater the tendency to 2H*(aq) + 2€ - H29) Ered = 0.00 v oxidize - Anode Anode Fee - Fetag + € Cathode 25*29 +36 %> Haq) Overal Reaction: Fe + 2H* Fe2+an + Halo Fes/Fe2+(0.5M)//H (0.5M)/Cus) Anode Cathode Oxidation Reduction

Fes/Fe2+(0.5M)//Cu²+(0.5M)/Cus) Standard reduction potentials: Fe2+ay+ 2€ — Fe) Ered =- 0.44 v = Lower potential value, the greater the tendency to oxidize Anode Cu?+(aq) + 2€ — Cue) Ered = 0.34 v Anode Fe2+ (aq) Fey) — Cathode Cu?q+ 2€ Cus) Overal Reaction: Fe + Cu2+ (aq) - >Fe2+ (aq) + Cue) Fey/Fe²+(0.5M)//Cu2+(0.5M)/Cue) Anode Cathode Oxidation Reduction

Cus/Cu2+(0.5M)//H*(0.5M)/Cu(s) Standard reduction potentials: Cu2+ (aq) + 2e Cu(s) = 0.34 V - The more positive potential, the greater the tendency to reduce 2H(aq) + 2€ — H2(g) E = 0.00 v Cathode Cus) Cathode Cu?*an + Zé Anode Hage 24 ) + 2 Overal Reaction: Cu2+an + H2(g) - Cus) + 2H*(aq) Cus/Cu2+(0.5M)//H*(0.5M)/Cue) Cathode Anode

Zn/s/Zn2+(0.10M)//Zn2+(1.0M)/ZN) Lower concentration Higher concentration Increase the concentration of Zn2+ Zn2+ concentration decreases Cathode Anode Zn) — Zn²+(aq) + 2€ Zn2+29+ 2 — Zns) Oxidation Reduction

2.- of Decreasing order reduction potential