Nernst Equation Calculation - Find a Molarity

Question

A voltaic cell consists of an Mn / Mn2 electrode and a Pb / Pb2 electrode. Calculate [Pb2 ] when [Mn2 ] is 1.5 M and Ecell is 0.27 V

Answer 1

First to find the standard reduction potentials from your book or some other reference/s:

Mn(2+) + 2e- ==> Mn(s), E° = -1.18V, . . . (1)

Pb(2+) + 2e- ==> Pb(s), E° = -0.13V, . . . .(2)

Performing (2) - (1):

Pb(2+) + Mn(s) ==> Mn(2+) + Pb(s), E° = 1.05V

Applying Nernst Equation, we have:

E = E° - (RT/nF) Ln([Mn(2+)]/[Pb(2+)]), or:

0.27 = 1.05-8.314*298/(2*96485)*Ln(1.5/[Pb(2+)]...

[Pb(2+)] = 1.5*exp((0.27-1.05)*2*96485/(8.314*298)) = 6.2x10^-27 (M)

Answer 2

This question took me a few tried, but I got it!

Mn(s) + Pb^2+(aq) => Mn^2+(aq) + Pb(s)

Mn(s) => Mn^2+(aq) + 2e

oxidation occurs - anode

Pb^2+(aq) + 2e => Pb(s)

reduction occurs - cathode

Q = [Mn2+]/[Pb2+], n=2, Ecell=0.27V

Eocell=Eocathode - Eoanode => -0.126 - (-1.18) => 1.054V

E=Eo - 0.0257/n lnQ

0.27 = 1.054 - 0.0257/2 ln[1.5]/[Pb2+]

[1.5]/[Pb2+]=3.1407e26

[Pb2+]=4.7759e-27 M

For the answer, you need the units too.