Question

Pb²⁺ + 2 e^- → Pb (s)      ξ^o = -0.13 V
Ag⁺ + 1 e^- → Ag (s)      ξ^o = 0.80 V

What is the voltage, at 298 K, of this voltaic cell starting with the following non-standard concentrations:

[Pb²⁺] (aq) = 0.109 M
[Ag⁺] (aq) = 1.01 M

Use the Nernst equation:

ξ = ξ^o - (RT/nF) ln Q

First calculate the value of Q, and enter it into the first answer box. Q is dimensionless.
Then calculate ξ, the non-standard cell potential, and enter its value into the second answer box (remember the unit of ξ is Volts).

Q =

Show work/reasoning please, trying to understand the material

Answer 1

Pb2⁺ + 2 e- → Pb (s)      ξo = -0.13 V
Ag⁺ + 1 e- → Ag (s)      ξo = 0.80 V

Since the reduction potential of Pb2+ is less than that of Ag+ so Pb undergoes reduction at cathode

Catode reaction(reduction) : Pb2⁺ + 2 e- → Pb (s)

Anode reaction (oxidation) : Ag(s)  → Ag+ + 1 e-

Overall reaction is  Pb2⁺ + 2 e- +2 Ag(s)  → 2 Ag+ + 2 e- + Pb (s)

OR   Pb2⁺ + 2 Ag(s)  → 2 Ag+ Pb (s)

0.109 M   1.01 M

reaction Quotient , Q = [Ag+]² [Pb2+]

=1.01 ² / 0.109

= 9.359

Eocell = Eo cathode - Eo anode

= 0.80 -(-0.13)

= 0.93 V

Ecell = Eocell - (0.0591/n) log Q

Where

n = number of electrons transferred = 2 mol

So E cell = 0.93 - (0.0591/2)x log9.359

= 0.901 V