You are titrating 100.0 mL of 0.0200 M Fe3 in 1 M HClO4 with 0.100 M Cu to give

Question

You are titrating 100.0 mL of 0.0200 M Fe3 in 1 M HClO4 with 0.100 M Cu to give Fe2 and Cu2 using Pt and saturated Ag | AgCl electrodes to find the endpoint. The two Nernst equations for the cell voltage. (Each applying at different points in the titration.) Eo of the Ag |AgCl electrode is 0.197 V. are E = 0.767- 0.05916log([Fe^2+]/[Fe^3+]-0.197 and E = 0.161- 0.05916log([Cu^+]/[Cu^2+]-0.197. (PLEASE ANSWER ALL PARTS OF THE QUESTION ASAP [parts a, b, and c])

(a) Write the balanced titration reaction.

(b) Complete the two half reactions for the Pt electrode.

(c) Calculate the values of E for the cell when at the volume 21.0 mL of the Cu titrant has been added: (Activity coefficients may be ignored as they tend to cancel when calculating concentration ratios.)

Explanation / Answer

a) The balanced titration reaction is:

Fe3+ + Cu+ -----> Fe2+ + Cu2+ (Fe3+ is an oxidising agent here)

b) Half reactions for the Pt electrode are as follows:

(for Fe3+/Fe ) : Pt/AgCl(s)/Ag(s)/Cl-/Fe3+/Fe2+/Pt

(for Cu+/Cu2+): Pt/Cu+/Cu2+/Cl-/Ag(s)/AgCl(s)/Pt

c) E0 of Ag/AgCl electrode is given 0.197 so, its clear E0 for the Fe3+/Fe2+ is 0.767. When 21 ml of Cu+ is added we have 2.1 millimole of Cu+ is added. Meanwhile, as total only 2 millimole of Fe3+ was there, this leaves with 0.1 millimoles of Cu+ and 2 millimoles of Cu2+. So, Ecell will be

Ecell = 0.161 - 0.05916 * log(0.1/2)-0.197

= 0.161- 0.05916(-1.30)-0.197

= -0.036+0.077

= 0.041 V

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