Question

Consider the titration of 25.0 mL of o.0500 M Sn2+ with 0.100 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+, using Pt and calomel electrodes. (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 E at the following volumes of Fe3+: 1.0, 12.5, 24.0, 25.0, 26.0, and 30.0 mL. Sketch the titration curve.

Answer 1

(a) Balanced equation,

Sn2+ + 2Fe3+ <==> Sn4+ + 2Fe2+

(b) Healf equations,

Sn2+ <===> Sn4+ + 2e-

and,

Fe3+ + e- <==> Fe2+

(c) Nernst equations for,

before equivalence point

E = Eo[Sn] - 0.0592/2 log([Sn2+]/[Sn4+])

and, after the equivalence point

E = Eo[Fe] - 0.0592/2 log([Fe2+]/[Fe3+])

(d) E calculation

1) when 1.0 ml Fe3+ added

moles of Sn2+ = 0.05 M x 25 ml = 1.25 mmol

moles of Fe3+ = 0.1 M x 1 ml = 0.1 mmol

remaining [Sn2+] = 1.15 mmol/26 ml = 0.044 M

formed [Sn4+] = 0.1 mmol/26 ml = 0.0038 M

E = 0.15 - 0.0592/2 log(0.044/0.0038) = 0.118 V

2) when 12.5.0 ml Fe3+ added

moles of Sn2+ = 0.05 M x 25 ml = 1.25 mmol

moles of Fe3+ = 0.1 M x 12.5 ml = 1.25 mmol

This is equivalence point

E = (0.15 + 0.771)/4 = 0.230 V

3) when 24.0 ml Fe3+ added

moles of Sn2+ = 0.05 M x 25 ml = 1.25 mmol

moles of Fe3+ = 0.1 M x 24 ml = 2.4 mmol

remaining [Fe3+] = 1.15 mmol/49 ml = 0.023 M

formed [Fe2+] = 1.25 mmol/49 ml = 0.025 M

E = 0.771 - 0.0592/2 log(0.025/0.0023) = 0.770 V

4) when 25.0 ml Fe3+ added

moles of Sn2+ = 0.05 M x 25 ml = 1.25 mmol

moles of Fe3+ = 0.1 M x 25 ml = 2.5 mmol

E = 0.771 V

5) when 26 ml Fe3+ added

moles of Sn2+ = 0.05 M x 25 ml = 1.25 mmol

moles of Fe3+ = 0.1 M x 26 ml = 2.6 mmol

remaining [Fe3+] = 1.35 mmol/51 ml = 0.026 M

formed [Fe2+] = 1.25 mmol/51 ml = 0.024 M

E = 0.771 - 0.0592/2 log(0.024/0.026) = 0.772 V

6) when 30.0 ml Fe3+ added

moles of Sn2+ = 0.05 M x 25 ml = 1.25 mmol

moles of Fe3+ = 0.1 M x 30 ml = 3.0 mmol

remaining [Fe3+] = 1.75 mmol/55 ml = 0.032 M

formed [Fe2+] = 1.25 mmol/55 ml = 0.023 M

E = 0.771 - 0.0592/2 log(0.023/0.032) = 0.775 V

the plot of E vs ml of Fe3+ is below