At 298 K the solubility product constant for PbC2O4 is 8.5e-10 and the standard

Question

At 298 K the solubility product constant for PbC2O4 is 8.5e-10 and the standard reduction potential of the Pb^2+ (aq) to Pb (s) is -0.126 V. Find the standard potential of the half reaction. PbC2O4 (s) + 2e^- -> Pb(s)?+ C2O4^2- (aq)
B. Calculate delta G^0 to the potential of the desired half reaction.
Calculate the potential of the half reaction in a 0.025 M solution of Na_2C_2O_4 At 298 K the solubility product constant for PbC2O4 is 8.5e-10 and the standard reduction potential of the Pb^2+ (aq) to Pb (s) is -0.126 V. Find the standard potential of the half reaction. PbC2O4 (s) + 2e^- -> Pb(s)?+ C2O4^2- (aq)
B. Calculate delta G^0 to the potential of the desired half reaction.
Calculate the potential of the half reaction in a 0.025 M solution of Na_2C_2O_4 PbC2O4 (s) + 2e^- -> Pb(s)?+ C2O4^2- (aq)
B. Calculate delta G^0 to the potential of the desired half reaction.
Calculate the potential of the half reaction in a 0.025 M solution of Na_2C_2O_4

Explanation / Answer

Pb^+2(aq) + 2e-   -----> Pb(s)

Given that; E^0 = -0.126 V

E = E^0 - (RT/nF) log Q

n= 2

R = 8.314

Temperature = 298 K

F= 96485

Q = [products]/[reactants]

E = -0.126 - (8.314 x 298)/(2 x 96485) ln[(C2O4^-2]

We could get the value of C2O4^-2 from Ksp

since, Ksp = [Pb+2] [C2O4^-2]

and the solubility S will calculate from Ksp:

Ksp = (S)^2

S = square root of Ksp

S=(8.5e-10)^1/2

s = 2.92 x 10^-5

E = -0.126 +0.134

E = 0.008 V

Now calculate the delta G^0 as follows:

delta G^0 = - nFE^0

delta G^0 = - 2*96485*(-0.126)

delta G^0 = +24314.22 J or 24.3 kJ

part C here we use common ion concept of PbCrO4

C2O4^-2 is the common ion

Ksp = (Pb+2) (0.25 + S)

S could be neglected as compared to 0.25

Ksp = 0.25S

S =Ksp/0.25

S = 3.4 x 10^-9

Now, use the Nernst equation where Q = 3.4 x 10^-9

E = -0.126 - (8.314 x 298)/(2 x 96485) ln 3.4 x 10^-9

E = -0.126 + 0.250

E = +0.124 V

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