Answer:
Step-by-step explanation:
From the table of standard reduction potentials we get:
Ag⁺ + e⁻ ⇒ Ag Eº = 0.80 V
Pb²⁺ + 2 e⁻ ⇒ Pb Eº = - 0.13 V
Ag⁺ has a greater reduction potential ( more positive) than Pb²⁺, therefore it wil be reduced while Pb will be oxidized.
In the cathode is where reduction occurs, therefore the Ag metal will act as the cathode. Likewise, the lead metal will act as the anode.
The overall reaction is
Pb ⇒ Pb²⁺ + 2e⁻ Eº (oxidation) = + 0.13 V
2 Ag⁺ + 2e⁻ ⇒ 2 Ag Eº (reduction) = +0.80 V
______________________
Pb + 2 Ag⁺ ⇒ Pb²⁺ + Ag Eºcell = 0.13 V + 0.80 V = 0.93 V
Follow the conventions ( oxidation in the right side, junction, reduction ) to represent the cell diagram which in this case is:
Pb(s) ∣ Pb²⁺ (1 M) ∥ Ag+(1 M) ∣ Ag(s)
For the non standard conditions , [Ag^+] = 0.001 M and [Pb2^+] = 0.01 M , we are going to make use of the Nernst equation:
E cell = Eº cell - 2.303 RT/nF log Q
where,
R = 8.314 J/Kmol
T = 298 K (25ºC)
n = number of electrons transferred
F = Faraday constant ,96500 coulomb/mol e⁻
Q = reaction quotient = molar concentration Reducing agent/ molar concentration Oxidizing agent
Plugging our values, we have
Q = [Pb2^+] / [Ag^+] = 0.01 M/0.001 = 10
E cell = 0.93 V - ( 2.303 x 8.314 J/Kmol x 298 K ) / ( 2 x 96,500 coul/mol) x log 10
E cell = 0.90 V
Note= we could have used the form for the Nernst equation:
Ecell = Eº -0.0592/n log Q
which is simpler but valid only at 298 K ( 25 ºC ), the form we used is more general.