A chemist makes 0.410L of mercury (II) iodide (HgI2) working solution by adding solution by adding distilled water to 0.0900L of a 0.0000350 M stock solution of mercury (II) iodide in water. Calculate the concentration of the chemists working solution.

The answer for this is a

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.

The answer is solid copper