on the following reaction: Sn2-(aq) +Pb(s) Sn(s) +Pb2-(aq) A) (6pts) If the conc

Question

on the following reaction: Sn2-(aq) +Pb(s) Sn(s) +Pb2-(aq) A) (6pts) If the concentration of Sn2 in the cathode half-cell is 2.50 M and the cell generates an emf of +0.39 V, what is the concentration of Pb2+ in the anode half-cell? B)_(5pts) If the anode half-cell contains [SO21- 2.50 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4? Q20 (10pts) Consider the following half-reactions: N2(g) NH3(g) A) (5pts) Balance the following half-reaction B) (5pts) Is the half-reaction an oxidation or a reduction? (basic solution)

Explanation / Answer

19) A) Given reaction is Sn2+ (aq)+ Pb (s) -------------> Sn (s) + Pb2+(aq)

Ecell = Eocell - RT/nF In {[Pb2+] / [Sn2+]} (solids cannot be taken into consideration)

Ecell = Eocell - 0.059/n Iog {[Pb2+] / [Sn2+]}  where n = no of electrons transferred

At cathode : Sn2+ (aq) + 2e- -------------> Sn (s)     Eo = - 0.14 V

At anode :   Pb (s) -----------> Pb2+(aq) + 2e- Eo = + 0.13 V

Eocell = Eoreduction of reaction at cathode + Eooxidation of reaction at anode

= -0.14 V + 0.13 V

= -0.01 V

Eocell = - 0.01 V

no of electrons transferred n = 2

Given that [Sn2+]= 2.5 M

Ecell = + 0.39 V

Ecell = Eocell - 0.059/n Iog {[Pb2+] / [Sn2+]}

0.39 = (0.01) - 0.059/2 log {[Pb2+] /[2.5]}

0.38 = - 0.059/2 log {[Pb2+] /[2.5]}

[Pb2+] = 3.28 x 10-13 M

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B)

Given that [SO42-] = 2.5 M

Ft=rom part A,  [Pb2+] = 3.28 x 10-13 M

PbSO4 <-----------> Pb2+ + SO42-

Ksp = [Pb2+] [SO42-]

= 3.28 x 10-13 M x 2.5 M

= 8.2 x 10-13

Therefore,

Ksp of PbSO4 = 8.2 x 10-13

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