Equilibrium & solubility

Psychedelia

Hi, i'd appreciate any help anyone can give me regarding this question. I'm having major problemos with this type of chemistry, mainly in where to start!


Half reaction E°/V (298 K)
Sn(2+) + 2e– → Sn –0.14
Sn(4+) + 2e– → Sn(2+) +0.15
Hg2(2+) + 2e– → 2Hg +0.79
Hg2Cl2 + 2e– → 2Hg + 2Cl– +0.27

Calculate the equilibrium constant, K, for the following reaction at 298 K
Sn(s) + Sn4+(aq) →← 2Sn2+(aq)


Calculate the solubility, S, of Hg2Cl2 in water at 298 K (units for S, mol kg–1). The mercury cation in the aqueous phase is Hg2(2+).

Calculate the voltage, E°, of a fuel cell by using the following reaction involving two
electrons.
H2(g) + 1/2 O2(g) → H2O(l) ΔrG° = –237.1 kJ mol–1

If anyone can give help me understand this it would be greatly appreciated!

Psychedelia

actually i got the first part to work out after some thought!

for the solubility part

i think: S = [Hg2(2+)] = [Cl(-)]/2

and the Ksp is [Hg2(2+)]x[Cl(-)]^2
which means S.4S^2 = 4S^3 = Ksp

so how would one derive the Ksp from the E potentals i have?