What is the [pb2+] in a solution made by adding 100 g of pbcl2(s) (mm = 278.1 g/mol) to a 5.4 m solution of nacl and allowing it to come to equilibrium? ignore any changes to solution volume due to the addition of pbcl2(s). once solution reaches equilibrium, it is noticed that some solid pbcl2 remains undissolved. the ksp for pbcl2 is 1.6 × 10−5?

                    PbCl₂ ⇄ Pb²⁺ + 2 Cl⁻
Initially                         0         5.4 M
Change          -x         + x        + 2x
Equilibrium                  x          5.4 + 2x

Ksp = [Pb²⁺] [Cl⁻]²
1.6 x 10⁻⁵ = x (5.4 + 2x)²
since x <<<< 5.4 so 2x + 5.4 = 5.4
1.6 x 10⁻⁵ = x (5.4)²
x = 5.48 x 10⁻⁷
[Pb²⁺] = 5.48 x 10⁻⁷ M

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

Ithink the answer is d.