If a chemist titrates 300.0ml of h2so4 with a 3.0m solution of naoh and requires only 3.4ml of the base to reach endpoint, what is the concentration of the sulfuric acid

The balanced equation for the acid-base reaction is as follows;
2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
this means that for complete neutralisation 2 mol of NaOH reacts with 1 mol H₂SO₄
number of NaOH moles reacted = 3.0 M /1000 x 3.4 = 0.0102 mol
Number of H₂SO₄ moles reacted = 0.0102 /2 = 0.0051 mol
Number of H₂SO₄ moles in 300.0 mL - 0.0051 mol
therefore in 1 L solution -0.0051/300 x 1000= 0.017 mol
concentration is number of moles in 1 L solution
Concentration of H₂SO₄ is 0.017 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.