Determine the number of moles of mg2+(aq) ions produced when 2.5 moles of zn 2+aq react completely in this cell

The question is incomplete. Complete question is attached below.

Solution:
The reaction involved in present electrochemical cell is:

Mg(s)     +       Zn2+(aq.)      ↔        Zn(s)      +      Mg2+ (aq.)

Thus, from above reaction it can be seen that 1 mole of Zn2+ produces 1 mole of Mg2+.

Thus, 2.5 moles of Mg2+(aq) ions would be produced when 2.5 moles of Zn2+(aq) will react.

Answer (b) supersaturation;

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.