A precipitate of copper(II) hydroxide is formed in a reaction. What change will help to dissolve the copper(II) hydroxide?
Cu2+(aq) + 20H*(aq) = Cu(OH)2(s)
A. Adding NaOH
B. Removing CuCl2
C. Adding HCI
D. Removing H20

The answer is letter c

Idon’t know what the answer is i wish i could

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.