Using the Nernst equation calculate the cell voltage?

Question

Fe(s) + Cd2+(aq) ---> Fe2+(aq) + Cd(s).
when the [Fe2+] = 0.15 M and [Cd2+] = 1.0 M?

Answer 1

DeltaG = DeltaG0 - RT ln ({reactants}/{products})

And

Electrical energy per mole reaction = potential x charge = E x (-zF)

Hence E = E0 + RT/zF ln ({reactants}/{products})

This IS the dreaded Nernst Equation; nothing else to it

In this case, z = 2; and you can find E0 from standard potentials.

{reactants} = [Fe(s)] x [Cd2+] = 1 (for a solid) x 1.0M (from information given in question); use similar reasoning for {products}, and the rest is just arithmetic.

Answer 2

Lancenigo di Villorba (TV), Italy

Before to apply Nernst's equation, I have to treat about Electrodical Half-Reactions.
Yes, now I have to break the written reaction in two or more half-reactions, one or more losing electrons

i) Fe°(s) ---> Fe++(aq) + 2e

while the second one is catching them

ii) Cd++(aq) + 2e ---> Cd°(s)

The great german scientist W. H. Nernst got the Classical Electrochemistry's Fundaments where he attributed the
well-known Standard Potentials as E°,i = -0.44 V and
E°,ii = - 0.40 V.
Since the written equation tends to oxidize iron instead cadmium, I may define DeltaE° = E°,ii - E°,i = (-0.40) - (-0.44) = +0.04 V.
Now, I apply the formula

DeltaE = DeltaE° + (R * T / (n * F)) * LN[|Fe++| / |Cd++|]
DeltaE = ((-0.40) -(-0.44)) + (8.31 * 298.16 / (2 * 96,500)) *
* LN[0.15 / 1.00] = + 0.016 V

a very small value near zero volts.

I hope this helps you.