Part C and D typed below picture as well Part C) Calculate the value of Ecell fo

Question

Part C and D typed below picture as well

Part C) Calculate the value of Ecell for this system at equlibrium

Part D) Determine the pH at which the given concentrations of Pb2 , Pb4 , Mn2 , and MnO4– would be at equilibrium.

Map Sapling Learning macmillan leaming A solution contain the following was prepared: 0.12 M Pb 1.5 x 10e M Pb 1.5 x 106 M Mn o.12 M Mno4 and 0.87 M HNO3. For this solution, the following balanced reduction half-reactions and overall net reaction can occur. 4 Pb E. 1.690 V 2+ 2 MnO 8H 5e Mn 4H,O 1.507 V 4 2+ 2+ 5Pb 2 Mn 8H,O 5Pb 2Mno 16H A) Determine Eeoell, AG, and K for this reaction Number Number Number A Go cell B) Calculate the value for the cell potential, E and the free energy, AG, for the given conditions Number Number Scroll down for the ce rest of the question. Calri ate the v enf F. fnr this svrstem at en hri Inn A Previous Check Answer Next Exit Hint

Explanation / Answer

Pb+4 +2e------àPb+2,      EO=1.690 V   (1)

MnO4- +8H++5e- ------àMn+2+4H2O , Eo=1.507V   (2)

Multiplying Eq.1 with 5 and Eq.2 with 2 and subtracting Eq.1 and Eq.2 gives

5Pb+4+2Mn+2+8H2O -----à 5Pb+2+2MnO4-+16H+, Eo= 1.690-1.507= 0.183V

deltaG0=-nFE0

n= no of electrons exchanged=5

F= 96500, Eo=0.183V

deltaG=-5*96500*0.183 joules=-88297.5 Joules

but deltaG=-RTlnK, lnK= deltaG/RT= 88297.5/(8.314*298) =35.64

E= EO-(0.0591/n)*logQ

5Pb+4+2Mn+2+8H2O -----à 5Pb+2+2MnO4-+16H+

Since HNO3 is strong acid and ionizes completely [H+]=0.87M

Q= reaction coefficient = [Pb+2]5[H+]16/ [Pb+4]5[ Mn+2]2 = (0.12)5*(0.87)16/(1.5*10-6)5*(1.5*10-8)2=1.57*1039, logQ= 39.2

E= 0.183-(0.0591/5)*39.2 =-0.28V

deltaG=-nFE=5*96500*0.28=135100

deltaG= -RT lnK, K= Equilibrium constant

135100=-8.314*298*lnK

Lnk= -54.53, K= 2.08*10-34

At Equilibrium

K= 2.08*10-34 = [Pb+2]5[H+]16/ [Pb+4]5[ Mn+2]2 = (0.12)5*(H+)16/(1.5*10-6)5*(1.5*10-8)2

[H+]=2.424*10-5

pH= -log[H+]= 4.62

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