AirplaneFruit

5+ Year Member
Jul 16, 2012
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Pre-Medical
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
 

bobeanie95

2+ Year Member
Apr 19, 2016
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I ran into the same problem, does anyone understand the reasoning behind this?
 
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theonlytycrane

5+ Year Member
Mar 23, 2014
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As we add I- (to be oxidized), we start out with the full cell potential of .36V. At the end of the titration, the potential will be 0V. The equivalence point (halfway) will be the point where the full cell potential is cut in half.
 
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