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Answer the question below regarding the reaction, Pb+2(aq) + 2Ag(s) → Pb(s) + 2Ag+(aq) a. Use...

Answer the question below regarding the reaction, Pb+2(aq) + 2Ag(s) → Pb(s) + 2Ag+(aq) a. Use the standard reduction potentials in the textbook to calculate the Eocell and determine whether this reaction would occur in a voltaic cell or an electrolytic cell. b. Use Eocell to calculate Kc for this reaction at 25oC. c. Use Eocell to calculate Go for this reaction (the o symbol denotes standard conditions; what temperature is standard conditions?)

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Answer #1

Pb+2(aq) + 2Ag(s) → Pb(s) + 2Ag+(aq)

oxidation : 2 Ag -----> 2 Ag+ + 2e-

Reduction : Pb+2 + 2e- ------> Pb

E0 cell = E0 cathode - E0 anode

E0 cell = -0.13 - 0.80

E0 cell = -0.93

Since the reduction potensial value of Ag is higher than the Pb , So Ag under go oxidation and Pb under go reduction.

Hence the reaction possible in electrolytic cell not in voltoic cell.

dG0 = -nFE0 cell

dG0 = - 2 * 96500 * (-0.93)

dG0 = 179.49 kJ so from dG value the reaction is endo thermic reaction and it is possible only in electrolytic cell.

dG0 = -2.303RT logKeq

179490 = -2.303 * 8.314 * 298 log K

Keq = 3.4898 * 10^-32

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