The concentrations of NaCl and Kl are at 0.150 M and 0.120 M, respectively, in t

Question

The concentrations of NaCl and Kl are at 0.150 M and 0.120 M, respectively, in the following electrochemical cell: Cu(s)|Cul(s)| (aq)||C (aq)|AgCl(s)||Ag(s) Using the following half-reactions, calculate cell voltage: CuI(s) + e^- Cu(s) + E degree = - 0.185 V AgCI(s) + e^- Ag(s) + CI^- E degree = 0.222 V E = Now, calculate the cell voltage using the following half-reactions. The solubility products (K_sp) 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

E0cell = 0.222 - (-0.185) = 0.407 v

Ecell = E0cell - (0.0591/n)log(Cl-/I-)

     = 0.407 - (0.0591/1)log(0.15/0.12)

   = 0.401 v

b)

E0 = 0.7993- 0.518 = 0.2813 V

Ecell = E0cell - (0.0591/n)log(Cu+/Ag+)

molarsolubility OF Cu+ = s = sqrt(1.8*10^-10) = 1.342*10^-5

molarsolubility OF Ag+ = s = sqrt(1.0*10^-12) = 1.0*10^-6

   = 0.2813 - (0.0591/1)log((1.342*10^-5)/(1*10^-6))

= 0.214

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