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) 2Fe^{3+} + Sn^{2+} -> 2Fe^{2+} + Sn^{4+}

b)Fe^{3+} + e^{-} <-> Fe^{2+} ... E_{o} =.732

Sn^{2+} <-> Sn^{4+} + 2e^{-} ... E_{o} = .139

c)E_{-} = .732 -(.05916/2)log([Fe^{2+}]/[Fe^{3+}])

E_{+} = .139 -(.05916/2)log([Sn^{4+}]/[Sn^{2+}])

d)at 1.0 ml Fe^{3+}...

1 mol Fe^{3+} = 1 mol Fe^{2+} + .5 mol Sn^{4+}

(.1 M Fe^{3+})(.001 L) = .0001 mol Fe^{3+}... so there is .0001 mol Fe^{2+} and .00005 mol Sn^{4+}

(.050M Sn^{2+})(.025 L) = .00125 mol Sn^{2+}

E = .139 -(.05916/2)log([Sn^{4+}]/[Sn^{2+}) - .268 (calomel electrode)

E = .139 -(.05916/2)log(.00005/.00125) - .268 (calomel electrode) = -.0876 V

Yet the correct answer is -.143...