Redox Titrations?

Question is here:





GivenAnswer: The question states that iodide (I–) is the titrant. The half-reaction shows the reduction of the triiodide anion to the iodide ion. Thus this half reaction occurs in reverse, and in the titration, iodide is oxidized to the triiodide ion (eliminate choices B and D). At the equivalence point, the potential will be equal to the average of the Eo ox and Eo red, so for this titration, it will be (–0.53 + 0.89)/2 = +0.36/2 = +0.18 V, or choice A.

Why does the potential equal the average at the equivalence point? Redox titration is really confusing

 

from tpr