2b of A solution containing the following was prepared: 0.17 M Pb2, 1.5 x10° M P

Question

2b of A solution containing the following was prepared: 0.17 M Pb2, 1.5 x10° M Pb", 1.5 x10° M Mn2 0.17Maph Mn04-. and 0.80 M HNO3. For this solution, the following balanced reduction half-reactions and overall net reaction can occur. E -1690 v E:= 1.507 V IMno; +8H + + 5e- Mn2+ +41,0] A) Determine Eocel, . and K for this reaction. See the lower panel for more information on how to solve part A. E0.183 G"-11-1.065 x 105 | J | K= || 3.74 x 1018 B) Calculate the value for the cell potential, Ecel, and the free energyAG, for the given conditions. &G;= 13068.7 Scroll down for the rest of the question. C) Calculate the value of Ecel for this system at equilibrium. Give Up & View Solution # Try Again Next -1 Exit Explanation E.

Explanation / Answer

For the given reaction,

5Pb4+ + 2Mn2+ + 8H2O <==> 5Pb2+ + 2MnO4- + 16H+

(A) Eo(cell) = Ecathode - Eanode

                   = 1.690 - 1.507 = 0.183 V

dGo = -nFEo(cell)

        = -10 x 96500 x 0.183 = -176595 J

dGo = -RTlnK

[with T = 25oC + 273 = 298 K]

-176595 = -8.314 x 298 lnK

K = 9.02 x 10^30

(B) Using Nernst equation and feeding given concentration values,

Ecell = Eo(cell) - 0.0592/n logK

         = 0.183 - 0.0592/10 log[(0.17)^5.(0.17)^2/(1.5 x 10^-6)^5.(1.5 x 10^-6)^2.(0.80)^16]

         = -0.036 V

dG = dGo + RTlnK

      = -176595 + 8.314 x 298 ln[(0.17)^5.(0.17)^2/(1.5 x 10^-6)^5.(1.5 x 10^-6)^2.(0.80)^16]

      = -175590.11 J

(C) value of Ecell at equilibrium

at equilibrium Ecell = 0 V

(D) pH calculation

Using Nernst equation at equilibrium

Ecell = 0 V

So,

Eo(cell) = 0.0592/n logK

0.183 = 0.0592/10 log[(0.17)^5.(0.17)^2/(1.5 x 10^-6)^5.(1.5 x 10^-6)^2.(H+)^16]

Solving,

[H+] = 1.902 M

pH = -log[H+]

      = -log(1.902) = -0.28

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