Consider the titration of 100.0 mL of 0.010 M Ce 4+ in 1 M HClO 4 by 0.0400 M Cu

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+ + Cu+ Ce3+ + Cu2+    Ce3+ + Cu2+ Ce4+ + Cu+Ce4+ + Cu2+ Ce3+ + Cu+

(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+.

1.19 mL   V (two decimal places) 13.8 mL   V (two decimal places) 24.7 mL   V (two decimal places) 25.0 mL   V (three decimal places) 26.2 mL   V (three decimal places) 28.7 mL   V (three decimal places) 49.4 mL   V (three decimal places)

Explanation / Answer

a.) Ce4+ + Cu+ --> Cu2+ + Ce3+

b.) Ce4+ + e- ---> Ce3+

Cu+ ---> Cu2+ + e-

c.) E= 0.161-0.05916 log(Cu2+/Cu+) and E= 1.70-0.05916 log(Ce3+/Ce4+)

d.)

moles Cu+ Moles Ce4+=moles Ce4+i-moles Cu+ moles Cu2+ Moles Ce3+ E Ce4+ E Cu+ 1,19 0 0,9524 0,0476 0,0476 1,778 13,8 0 0,448 0,552 0,552 1,715 24,7 0 0,012 0,988 0,988 1,700 25 0 0 1 1 1,700 26,2 0,048 0 1 1 1,700 0.083 28,7 0,148 0 1 1 1,503 0.112 49,4 0,976 0 1 1 1,503 0.160

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