x10 M Mn2, 0.20M solution containing the following was prepared: 0.20 M Pb2. 1.5

Question

x10 M Mn2, 0.20M solution containing the following was prepared: 0.20 M Pb2. 1.5 x 10 M Pb", 1.5 no4, and 0.80 M HNO3. For this solution, the following balanced reduction half-reactions and overall ne action can occur. E+= 1.690 V 2+ E-=1.507 V ) Determine Ecell, G, and K for this reaction Number Number Number Ecell = K= 3) Calculate the value for the cell potential, Ecell, and the free energy, AG, for the given conditions Number Number EA cell Scroll down to answer all parts of this question.(A through D). C) Caleulate the value of F.for thie evetem at enuilihrium Previous Check Answer Next Ex

Explanation / Answer

1. Mn2+ ---------> Mn+7

Eoxid = -1.507 x 2 = -3.014 V

Pb+4 --------> Pb+2

Ered= 1.690 x 5 = 8.45

Ecell = Ered + Eoxid = 8.45 -3.014 = 5.436 V

G = -nFEcell

= -20x 96500 x 5.436 = -10539.73 kJ

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