Consider the titration of 100.0 mL of 0.010 M Ce4+ in 1 M HClO4 by 0.0400 M Cu+
ID: 992580 • Letter: C
Question
Consider the titration of 100.0 mL of 0.010 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.
a) Select a balanced titration reaction.
Ce4+ + Cu2+ Ce3+ + Cu+
Ce4+ + Cu2+ Ce3+ + Cu+
Ce3+ + Cu2+ Ce4+ + Cu+
Ce4+ + Cu+ Ce3+ + Cu2+
(b) Select two different half-reactions for the indicator electrode.
c) Select two different Nernst equations for the cell voltage.
(d) Calculate E at the following volumes of Cu+.
V (three decimal places)
1.17 mL V (two decimal places) 13.9 mL V (two decimal places) 22.4 mL V (two decimal places) 25.0 mL V (three decimal places) 27.0 mL V (three decimal places) 30.7 mL V (three decimal places) 48.8 mLV (three decimal places)
Explanation / Answer
a. Balance chemical Equation
Ce4+ + Cu+ Ce3+ + Cu2+
b. Half Reactions
Ce4+ + e- Ce3+
Cu+ e- + Cu2+
c. Nernst Equation
E = Eo - (0.05916) log( [Ce3+][Cu2+]/[Ce4+][Cu+]