The concentrations of NaCl and Kl are at 0.170 M and 0.110 M. respectively, in t

Question

The concentrations of NaCl and Kl are at 0.170 M and 0.110 M. respectively, in the following electrochemical cell: Cu(s) I Cul(s) |1 (aq) || Cl(aq) | AgCl(s) | Ag(s) Using the following half-reactions, calculate cell voltage: Cul(s) + e Cu(s) + I E degree = -0.185 V AgCl(s) + e Ag(s) + Cl E degree = 0.222 V Now, calculate the cell voltage using the following half-reactions. The solubility products (Kap) for AgCI and Cul are 1.8 times 10^-10 and 1.0 times 10^-12, respectively Cu + e Cu(s) E degree = 0.518 V Ag + e Ag(s) E degree = 0.7993 V

Explanation / Answer

Eocell = Eo( right half ) - Eo( left half)                     ( right half inc ell notation shows reduction , left half oxidation)

        = 0.222-(-0.185) = 0.407

Now   reaction is   Cu(s) + I- (aq) + AgCl(s)   =   CuI(s) + Ag(s) + Cl-(aq)

equilibrium constant K = [Cl-]/[I-] = 0.17/0.11 = 1.5454

Ecell = Eo cell - ( 0.059/n) log K   where n = number of electron involver per reaction = 1

E cell = 0.407 - ( 0.059) log ( 1.5454) = 0.396 volts

B) E cell = Eo( Ag+/Ag) - Eo( Cu+/Cu) = 0.7993-0.518 = 0.2813

AgCl (s) <---> Ag+ (aq) + Cl-(aq)   , Ksp = [Ag+][Cl-]

CuI(s) <---> Cu+ (aq) + I- (aq)   , Ksp = [Cu+] [I-]

reaction we have is     Ag+(aq) + Cl -(aq) + CuI (s) =   AgCl (s) + Cu+ (aq) + I- (aq)

K reaction = [Cu+] [I-] / [Ag+] [Cl-] = Ksp of CuI / Ksp of AgCl = 10^ -12 / ( 1.8x10^-10) = 0.00556

E cell = Eo - ( 0.059/n) log K   where n =number of elctrons per reaction = 1

E cell = 0.2813 - ( 0.059) log 0.00556

= 0.414 V

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