Astrong monoprotic acid ionizes in water. if the [h3o+] concentration is 1.56 x 10-4 m, what else is true?

If,
              [H⁺]  =  1.56 × 10⁻⁴ M
Then,
              pH  = -log (1.56 × 10⁻⁴)

              pH  =  3.80
Also,
              pH + pOH  =  14
So,
              pOH  =  14- 3.80

              pOH  =  10.2

Taking Anti log of pOH,
            
              [OH⁻]  = 10⁻¹⁰°²

              [OH⁻]  =  6.30 × 10⁻¹¹

Result:
           For [H⁺] = 1.56 × 10⁻⁴ other true statements are,

                        1)  [OH⁻]  =  6.30 × 10⁻¹¹

                        2)  pH  =  3.80

                        3)  pOH  =  10.2

Ithink the answer is d

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.