electrochemical problem 5. A galvanic cell is constructed in which the overall r

Question

electrochemical problem

5. A galvanic cell is constructed in which the overall reaction is At 25 degree C. the standard reduction potential for each half-cell is given below. (a) Write balanced equations for the half-cell reactions at the cathode and at the anode. (b) Chloride ions are added to the Pb | Pb^2+ (aq) half cell at 25 degree C. The PbCl2(s) is formed when [Cl^-] reaches 0.15 M. and the cell potential is 0.22 V at pH = 0 and PH2 = 1 atm. Calculate [Pb^2+(aq)] under these conditions. (c) Calculate the solubility product constant Ksp of PbC12.

Explanation / Answer

(a)

Anode half cell reaction:

Pb(s) --> Pb2+(aq) + 2e-

Cathode half cell reaction:

2H3O+(aq) + 2e- --> H2(g) + 2H2O(l)

(b)

Overall cell reaction

Pb(s)+2H3O+(aq) --> Pb2+(aq) + H2(g) + 2H2O(l)

E0cell=0-(-.1263)=0.1263V

Ecell=0.22V

No of electrons n=2

From Nerst equation

Ecell=E0cell-0.059/2*log[(Pb2+)*p(H2)/(H3O+^2)]

p(H2)=1

H3O+=10^-pH=1

0.22=0.1263-0.059/2*log(Pb2+)

Pb2+ = 0.67*10^-3M < 0.15 M

So Cl- will not have effect on Pb2+(aq).

(c) PbCl2 --> Pb2+ +2Cl-

Solubility S = 0.15M

Ksp=S*(2*S)^2=4*(.15)^3=0.0135

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