Question

Consider the titration of 20 mL of 0.010 M V2+ with 0.010 M Fe3+ in 1 M HCl using Pt and saturated calomel electrodes. Write a balanced titration reaction and the two half reactions for the indicator electrode. Calculate E at the following volumes of Fe3+ added: 1.00, 10.00, 20.00, and 30.00 mL

Answer 1

Titration

Nernst equation,

E = Eo - 0.0592/n log[Red/Oxd]

Balanced titration reaction,

V^2+(aq) + Fe^3+(aq) <==> V^3+(aq) + Fe^2+(aq)

Half-reactions

Oxidation : V^2+ <==> V^3+ + e-     Eo = 0.26 V

Reduction : Fe3+ + e- <==> Fe2+    Eo = 0.771 V

Calculation of E,

after 1.00 ml Fe3+ added

initial moles V^2+ = 0.01 M x 20 ml = 0.20 mmol

moles Fe3+ added = 0.01 m x 1 ml = 0.01 mmol

V2+ remained = 0.19 mmol

V^3+ formed = 0.01 mmol

Using Nernst equation,

E = -0.26 - 0.0592log(0.19/0.01) - 0.241 = -0.58 V

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after 10.00 ml Fe3+ added

initial moles V^2+ = 0.01 M x 20 ml = 0.20 mmol

moles Fe3+ added = 0.01 m x 10 ml = 0.10 mmol

half-equivalence point

V2+ remained = V^3+ formed = 0.10 mmol

Using Nernst equation,

E = -0.26 - 0.241 = -0.50 V

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after 20.00 ml Fe3+ added

initial moles V^2+ = 0.01 M x 20 ml = 0.20 mmol

moles Fe3+ added = 0.01 m x 20 ml = 0.20 mmol

Equivalence point

Using Nernst equation,

E = (-0.26 + 0.771)/2 - 0.241 = 0.015 V

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after 30.00 ml Fe3+ added

initial moles V^2+ = 0.01 M x 20 ml = 0.20 mmol

moles Fe3+ added = 0.01 m x 30 ml = 0.30 mmol

Fe^3+ remained = 0.10 mmol

Fe^2+ formed = 0.20 mmol

Using Nernst equation,

E = 0.771 - 0.0592log(0.20/0.10) - 0.241 = 0.512 V