The concentrations of NaCl and Kl are at 0.150 M and 0.100 M, respectively, in t
ID: 833537 • Letter: T
Question
The concentrations of NaCl and Kl are at 0.150 M and 0.100 M, respectively, in the following electrochemical cell: Cu(s) | CuI(s) | I-(aq) || CI-(aq) | AgCl(s) | Ag(s) Using the following half-reactions, calculate cell voltage: CuI(s) + e- Cu(s) + I- E degree = -0.185 V AgCl(s) + e- Ag(s) + CI- E degree = 0.222 V E = Now, calculate the cell voltage using the following half-reactions. The solubility products (Ksp) for AgCl and CuI are 1.8times 10-10 and 1.0times 10-12, respectively. Cu + + e- Cu(s) E degree = 0.518 V Ag + + e- Ag(s) E degree = 0.7993 V E =Explanation / Answer
A)
Oxidation half-rxn (occurs at anode): Cu(s) + I^-(aq) = CuI(s) + 1e^-
Reduction half-rxn (occurs at cathode): AgCl(s) + 1e^- = Ag(s) + Cl^-(aq)
Combine both half-rxns to produce an overall balanced reaction
Cu(s) + I^-(aq) + AgCl(s) + 1e^- = CuI(s) + Ag(s) + Cl^-(aq) + 1e^- ... electrons MUST cancel
Cu(s) + I^-(aq) + AgCl(s) = CuI(s) + Ag(s) + Cl^-(aq)
Solve for E^o(cell)
E^o(cell) = E^o(cathode) - E^o(anode) = 0.222 V - (-0.185 V) = 0.407 V
Solve for E(cell)
E(cell) = E^o(cell) - (0.0592 / n) log Q ... Q = [Cl^-] / [I^-]
E(cell) = 0.407 V - (0.0592 / 1 mole of electrons) log (0.150 / 0.100)
E(cell) = 0.397 V
B)
Oxidation half-rxn (occurs at anode): Cu(s) = Cu^+(aq) + 1e^-
Reduction half-rxn (occurs at cathode): Ag^+(aq) + 1e^- = Ag(s)
Combine both half-rxns to produce an overall balanced reaction
Cu(s) + Ag^+(aq) + 1e^- = Cu^+(aq) + Ag(s) + 1e^- ... electrons MUST cancel
Cu(s) + Ag^+(aq) = Cu^+(aq) + Ag(s)
Solve for Cu^+ and Ag^+
Ksp = [Cu^+] [I^-] ... let x = [Cu^+] = [I^-]
(1.0 x 10^-12) = x^2
x = square root (1.0 x 10^-12) = 1.0 x 10^-6 M
Ksp = [Ag^+] [Cl^-] ... let x = [Ag^+] = [Cl^-]
(1.8 x 10^-10) = x^2
x = square root (1.8 x 10^-10) = 1.3 x 10^-5 M
Solve for E^o(cell)
E^o(cell) = E^o(cathode) - E^o(anode) = 0.7993 V - 0.518 V = 0.2813 V
Solve for E(cell)
E(cell) = E^o(cell) - (0.0592 / n) log Q ... Q = [Cu^+] / [Ag^+]
E(cell) = 0.2813 V - (0.0592 / 1 mole of electrons) log [(1.0 x 10^-6) / (1.3 x 10^-5)]
E(cell) = 0.347 V
Hope this helps! :)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.