Pb2+ + 2 e- ? Pb (s) ?o = -0.13 V Ag+ + 1 e- ? Ag (s) ?o = 0.80 V What is the vo

Question

Pb2+ + 2 e- ? Pb (s) ?o = -0.13 V Ag+ + 1 e- ? Ag (s) ?o = 0.80 V What is the voltage, at 298 K, of this voltaic cell starting with the following non-standard concentrations: [Pb2+] (aq) = 0.073 M [Ag+] (aq) = 0.91 M Use the Nernst equation: ? = ?o - (RT/nF) ln Q First calculate the value of Q, and enter it into the first answer box. Q is dimensionless. Then calculate ?, the non-standard cell potential, and enter its value into the second answer box (remember the unit of ? is Volts).

Explanation / Answer

Pb(s) --> Pb2+ + 2 e- ...?o = +0.13 V Ag+ + 1 e- ? Ag (s)... ?o = 0.80 V Q =0.073/0.91^2 = 0.0882 Ecell = 0.93-(0.0591xlogQ)/2 = 0.961 V

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