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) = 0.065 M
[Ag⁺] (aq) = 0.93 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).

Answer 1

1)

from data table:

Eo(Pb2+/Pb(s)) = -0.13 V

Eo(Ag+/Ag(s)) = 0.80 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Ag+/Ag(s))

anode is (Pb2+/Pb(s))

The chemical reaction taking place is

2Ag+(aq) + Pb(s) --> 2Ag(s) + Pb2+(aq)

Q = [Pb2+]/[Ag+]^2

= 0.065/ (0.93)^2

= 0.0752

Answer: 0.0752

2)

Eocell = Eocathode - Eoanode

= (0.7996) - (-0.126)

= 0.9256 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

use:

E = Eo - (2.303*RT/nF) log {[Pb2+]^1/[Ag+]^2}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Pb2+]^1/[Ag+]^2}

E = 0.9256 - (0.0591/2) log (0.0752)

E = 0.9256-(-3.323*10^-2)

E = 0.9588 V

Answer: 0.959 V