There is something about these redox titrations that I am just not getting. If anyone has any insight into this problem I would greatly appreciate the assistance
15-5. Consider the titration of 25.0 mL of 0.050 0 M Sn2 with
0.100 M Fe3 in 1 M HCl to give Fe2 and Sn4, using Pt and
calomel electrodes.
(a) Write a balanced titration reaction.
(b) Write two half-reactions for the indicator electrode.
(c) Write two Nernst equations for the cell voltage.
(d) Calculate E at the following volumes of Fe3: 1.0, 12.5, 24.0,
25.0, 26.0, and 30.0 mL. Sketch the titration curve.
I'm stuck on part D. This is what I have so far:
a) 2Fe3+ + Sn2+ -> 2Fe2+ + Sn4+
b)Fe3+ + e- <-> Fe2+ ... Eo =.732
Sn2+ <-> Sn4+ + 2e- ... Eo = .139
c)E- = .732 -(.05916/2)log([Fe2+]/[Fe3+])
E+ = .139 -(.05916/2)log([Sn4+]/[Sn2+])
d)at 1.0 ml Fe3+...
1 mol Fe3+ = 1 mol Fe2+ + .5 mol Sn4+
(.1 M Fe3+)(.001 L) = .0001 mol Fe3+... so there is .0001 mol Fe2+ and .00005 mol Sn4+
(.050M Sn2+)(.025 L) = .00125 mol Sn2+
E = .139 -(.05916/2)log([Sn4+]/[Sn2+) - .268 (calomel electrode)
E = .139 -(.05916/2)log(.00005/.00125) - .268 (calomel electrode) = -.0876 V
Yet the correct answer is -.143...