The Iodine "Clock Reaction" Involves the Following Sequence of Reactions $\to$

The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker.
1) IO₃^-(aq) + 5I^-(aq) + 6H⁺(aq) $\to$ 3I₂(aq) + 3H₂O(l)
2) I₂(aq) + 2S₂O₃^2-(aq) $\to$ 2I^-(aq) + S₄O₆^2-(aq)
The molecular iodine (I₂) formed in reaction 1 is immediately used up in reaction 2, so that no iodine accumulates. What is the overall reaction occurring in this experiment?

A) IO₃^-(aq) + 3I^-(aq) + 2S₂O₃^2-(aq) + 6H⁺(aq) $\to$ 2I₂(aq) + S₄O₆^2-(aq) + 3H₂O(l)
B) IO₃^-(aq) + 4S₂O₃^2-(aq) + 6H⁺(aq) $\to$ I^-(aq) + 2S₄O₆^2-(aq) + 3H₂O(l)
C) IO₃^-(aq) + 6S₂O₃^2-(aq) + 6H⁺(aq) $\to$ I^-(aq) + 3S₄O₆^2-(aq) + 3H₂O(l)
D) IO₃^-(aq) + I₂(aq) + 8S₂O₃^2-(aq) + 6H⁺(aq) $\to$ 3I^-(aq) + 4S₄O₆^2-(aq) + 3H₂O(l)
E) IO₃^-(aq) + 2I₂(aq) + 6S₂O₃^2-(aq) + 6H⁺(aq) $\to$ 5I^-(aq) + 3S₄O₆^2-(aq) + 3H₂O(l)