A solution containing the following was prepared: o.18 M Pb2 1.5 x 106 M Pb4 1.5

Question

A solution containing the following was prepared: o.18 M Pb2 1.5 x 106 M Pb4 1.5 x 106 M Mn? o.18 M Mnos and 0.92 M HNO3. For this solution, the following balanced reduction half-reactions and overall net reaction can occur. 2+ Pb E. 1.690 V 2 MnO, 8H 5e Mn 4H,O E- 1.507 V 2 5Pb 2Mn 8H,O 5Pb 2Mno 16H A) Determine Eee AG and K for this reaction Number Number Number cel B) calculate the value for the cell potential, Eoel, and the free energy, AG, for the given conditions Number Number V AG cell Scroll down for the rest of the question

Explanation / Answer

For thr overall reaction 5Pb+4 +2Mn+2 + 6H2Oß-> 5Pb+2 + 2MnO4- + 16H+

Is obtained from subsreaction reaction 1 from reaction-2. Hence Eo= 1.690-1.507= 0.183 V

deltaGo=-nFE, n= no of electrons exchanged =10, F=96500 and Eo=0.183V

deltaGo= -10*96500*0.183=-176595 Joules /mole=-176.595 KJ/mole

deltaGo=-RTlnK, lnK= -deltaG/RT= 176595/(8.314*298), K= 9.02*1030,

E= Eo-(0.0591/n)*logQ =0.183- (0.0591/10)*logQ

[H+]= 0.92M ( since HNO3 is strong and suppliments H+]

Q= [Pb+2]5 [MoO4-]2 [H+]16/ [Pb+4]5 [Mn+2]2 = (0.18)5* (0.18)2*(0.92)16/(1.5*10-6)5 (1.5*10-6)2=1.54*1040, logQ= 40.2

E= 0.183-(0.0591/10)*40.2 =-0.0534

deltaG=-nFE= -10*96500*0.0534 j/mole=51531 J/mole

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