A solution containing the following was prepared at 25degreeC: 0.18 M Pb^2+, 1.5

Question

A solution containing the following was prepared at 25degreeC: 0.18 M Pb^2+, 1.5 Times 10^-6 M Pb^4+, 1.5 Times 10^-6 M Mn^2+, 0.18 M MnO_4-, and 0.88 M HNO_3. For this solution, the following balanced reduction half-reactions and overall net reaction can occur. 5[Pb^4+ + 2_e- Pb^2+] 2[MnO^-_4 + 8H^+ +5e^- Mn^2+ +4H_2O] E degree_+ = 1.690 V E degree_+ = 1.507 V 5Pb^4+ + 2Mn^2+ +8H_2O 5Pb^2+ + 2NnO^-_4 + 16H^+ A) Determine E degree_cell, Deltal G degree, and K for this reaction. E^0_cell = Delta Gdegree = K = B) Calculate the value for the cell potential, E_cell, and the free energy, Delta G, for the given conditions. E_cell = Delta G = Scroll down to answer all parts of this question. (A through D). C) Calculate the value of E_cell for this system at equilibrium. E_cell = D) Determine the pH at which the given concentrations of Pb^2+, Pb^4+, Mn^2+, and MnO_4- would be at equilibrium. pH =

Explanation / Answer

A) E0cell = 1.690 - 1.507 = 0.183 V

del G0 = -nFE0cell = = -10 x 96500 x 0.183 = -176595 J

del G0 = - RTlnK

therefore K = exp(-del G0/ RT) = exp{-(-176595 / 8.314 x 298)} = 9.02 x 1030.

C) At equlibrium Ecell = 0 Volts

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