Question

need help for half cell potentials pls calculate step by step

(NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) Positive (Ecu) Copper and silver Cu() - Cu2+ + 2e" Ag+ +1e Ag) 0.432 V Zinc + Silver Zn cs + 2n**+ Zé dat + leº nAg (s) 1.484 V Copper & Zinc Zn(s) → Zn²+ +Ze Cutze → Curo 1.119 Copper & Iron Curs) → Cu²+Zé Fest + le Fers 0.340 Lol Bromine & Iron Fe 2+ Festle Br, +2e 2Br 0.427 V non farlo Bromine & Culonine Br 2t 2Br + 2€ Cl₂ +2e=201 -0.312 v) Zine & lodine Znas → Zn²+ + 2€ Izt 2e21 1.456 v pontaneo Calculated Half-Cell Potentials: The measured potential is the sum of the potentials from each half-cell reaction. Ecrit = Ered + Ex The reduction potential of the Ag*, Ag(s) electrode measured with respect to the standard hydrogen electrode (SHE) is 0.799 V. On the next page, calculate the reduction potential for each of the other half- cells measured. Remember that for any half-reaction, Ex = -E

Answer 1

i) Oxidation half-reaction: Cu(s) ----> Cu²⁺(aq) + 2e^-; E^o_anode = 0.34 V

Reduction half-reaction: 2Ag⁺(aq) + 2e^- ----> 2Ag(s); E^o_cathode = 0.80 V

i.e. E^o_cell = E^o_cathode - E^o_anode = 0.80 - 0.34 = 0.46 V (Spontaneous)

ii) Oxidation half-reaction: Zn(s) ----> Zn²⁺(aq) + 2e^-; E^o_anode = -0.76 V

Reduction half-reaction: 2Ag⁺(aq) + 2e^- ----> 2Ag(s); E^o_cathode = 0.80 V

i.e. E^o_cell = E^o_cathode - E^o_anode = 0.80 - (-0.76) = 1.56 V (Spontaneous)

iii) Oxidation half-reaction: Cu(s) ----> Cu²⁺(aq) + 2e^-; E^o_anode = 0.34 V

Reduction half-reaction: Zn²⁺(aq) + 2e^- ----> Ag(s); E^o_cathode = -0.76 V

i.e. E^o_cell = E^o_cathode - E^o_anode = -0.76 - 0.34 = -1.10 V (Nonspontaneous)

iv) Oxidation half-reaction: Zn(s) ----> Zn²⁺(aq) + 2e^-; E^o_anode = -0.76 V

Reduction half-reaction: 2Fe³⁺(aq) + 2e^- ----> 2Fe²⁺(aq); E^o_cathode = 0.77 V

i.e. E^o_cell = E^o_cathode - E^o_anode = 0.77 - (-0.76) = 1.53 V (Spontaneous)