Question

Calculate the cell potential for the reaction below when the concentrations of ions are: [Ag+ ] = 0.010 M and [Cu2+] = 0.750 M, at 25 °C.

Cu(s) + 2Ag+ (aq) --> Cu2+(aq) + 2Ag(s)

Given the standard reduction potentials:

Cu2+(aq) + 2e– → Cu(s) Eϴ = 0.34 V

Ag+ (aq) + e– → Ag(s) Eϴ = 0.80 V

(A) 0.35 V

(B) 0.44 V

(C) 0.46 V

(D) 0.48 V

(E) 0.57 V

Answer 1

Cell reacti on: Cu (s) + 2Ag*(aq) Cu3 (aq) 2Ag (s) 2Ag (s) Cu Reduction half reaction : 2Ag* (aq) + 2e E = 0.80 V cafhod Cu (s) Ед 3D Е — Е" 0.80 V 0.34 V (aq)+2e Oxidation half reaction = 0.34 V 'anode Standard cell potenti al cell 'cathode anode = 0.46V >Cu2* (aq)2Ag (s) Cellreaction Cu (s) + 2Ag' (aq) No of moles of electrons transferredin the cell, » = 2 0.0591V Cul log According to Nemst equation, Ee E) cel cel [Ag] 0.0591 0.750 = 0.46 V -log (0.010) 0.0591V -log (7500) 2 = 0.46 V- 0.0591V -(3.875) 2 = 0.46 V- = 0.46 V - 0.11 V = +0.35 V