[rycharles]

Consider the titration of 100.0 mL of 0.0100 M Ce4+ in 1 M HClO4 by 0.0400 M Cu+ to give Ce3+ and Cu2+, using Pt and saturated Ag / AgCl electrodes to find the end point.  Calculate E at the following volumes of Cu+: 1.00, 25.0, 25.5, and 50.0 mL.

my attempt...

V_e = 25.0 mL

balanced equation: Cu⁺ + Ce⁴⁺  Cu²⁺ + Ce³⁺

Ce⁴⁺ + e^-  Ce³⁺   E^o = 1.70
Cu²⁺ + e^-  Cu⁺   E^o = 0.161
AgCl(s) + e^-  Ag(s) + Cl^-   E^o = 0.222

E = E₊ - E_-

calculating for Cu+ 1.00mL

E₊ = 1.70 - 0.05916*log((25.0-1.00)/1.00) = 1.618

E = 1.618 - 0.222 = 1.40 V

The book says the answer is 1.58 volts, am i plugging in the wrong numbers or did I miss a step?

[Borek]

E₊ = 1.70 - 0.05916*log((25.0-1.00)/1.00) = 1.618

Why 0.170 if you are calculating for copper?

E = 1.618 - 0.222 = 1.40 V

You are not asked to calculate potential against hydrogen electrode, so I don't think this is necessary.

[rycharles]

E_cell = E₊ + E_-

E₊ = 1.70 - 0.05916*log(1.00/24.0) = 1.7816

My mistake, the calomel electrode is saturated in chlorine so its voltage is 0.197

E_cell = 1.7816 - 0.197 = 1.58 V