Consider the titration of 25.00 mL of 0.0750 M Sn 2+ with 0.1000 M Fe 3+ in 1 M
ID: 965929 • Letter: C
Question
Consider the titration of 25.00 mL of 0.0750 M Sn2+ with 0.1000 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+ , using a Pt electrode for the cathode and a Standard Hydrogen Electrode for the anode (connected to negative terminal of potentiometer).
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 Etitr after the addition of the following volumes of 0.1800 M Fe3+ for 10 mL
e) Sketch the titration curve and indicate the equivalent pt, and the Etitr when ½ of the moles of analyte has been oxidized.
Explanation / Answer
Titration
a) Balanced titration reaction,
Sn2+ + 2Fe3+ ----> Sn4+ + 2Fe2+
b) Anode half reaction,
Sn2+ ---> Sn4+ + 2e-
Cathode half reaction,
Fe3+ + e- ---> Fe2+
c) Nernst equations for,
Anode half cell,
E = 0.15 - 0.0592/2 log([Sn2+]/[Sn4+])
Cathode half cell,
E = 0.771 - 0.0592/2 log([Fe2+]/[Fe3+])
d) Etitr after 10 ml of 0.1800 M of Fe3+ added
moles of Sn2+ present = 0.0750 M x 25 ml = 1.875 mmol
moles of Fe3+ added = 0.18 M x 10 ml = 1.8 mmol
[Sn2+] remaining = 0.075 mmol/35 ml = 0.002 M
[Sn4+] formed = 1.8 mmol/35 ml = 0.051 M
Etitr = 0.15 - 0.0592/2 log(0.002/0.051) = 0.19 V