Please help:Electrochemistry?

Question

An electrochemical cell is set up where the overall reaction is : Pb2+ + Sn → Pb + Sn.
Calculate the cell potential Ecell,for this process under the following conditions:
a.Standard conditions at both electrodes
b.[Sn2+ ]=0.01 M; [Pb2+]=0.2 M
c.[Sn2+ ]=0.40 M; [Pb2+]=0.005 M
In each case decide the direction in which the reaction will spontaneously speed.

Answer 1

Use the Nernst equation:
dE = dE0 - 0.0592 V/n ln [C]^c [D]^d /([A]^a [B]^b)

n is the number of electrons involved, 2 in this case.
Therefore we have in all cases:
dE = dE0 - 0.0296 V ln [Sn2+] /([Fe2+]
Furthermore we use the standard electrode potentials to calculate dE0
Sn2+(aq) + 2e- -> Sn(s) -0.14 V
Fe2+(aq) + 2e- -> Fe(s) -0.41 V
Therefore dE0 = 0.41 - 0.14 = 0.27 V
a) The concentration are 1 M/L and we have dE = 0.27 V
b) dE = 0.27 - 0.0296 ln 0.01/0.2 = 0.27 + 0.09 = 0.36 V
c) dE = 0.27 - 0.0296 ln 0.4/0.005 = 0.27 - 0.13 = 0.14 V

Only in the last case the reaction moves to the left: the concentration of Sn2+ is the highest.

You can take a look here for explanations and examples:

http://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html

Answer 2

First of all you have written incorrectly the reaction

Pb(+2) + Sn <=> Pb + Sn(+2)

Pb(+2) +2e ->Pb(0) E01=-0.1262 Volt
Sn(+2) +2e ->Sn(0) E02=-0.1375 Volt

So for the above total reaction E0=E01-E02= -0.1262-(-0.1375) = +0.0113 Volt so it is written in the spontaneous direction (E>0)

For the other conditions you have to find E based on the Nernst equation:
E=E0 - (RT/nF)*ln([Sn+2]/[Pb+2])

R=8.314 J/(mole*K)
n=2
F=96500 Coulomb/mole
T is the absolute temperature. What is your temperature?
If you find that E>0 then the reaction is spontaneous as it is (to the right), if E<0 then the reverse reaction is spontaneous (to the left)