1. Consider the titration of 25.00 mL of 0. 0 750 M Sn 2+ with 0.10 00 M Fe 3+ i

Question

1. Consider the titration of 25.00 mL of 0. 0 750 M Sn 2+ with 0.10 00 M Fe 3+ in 1 M HCl to give Fe 2+ and Sn 4+ , using a Pt electrode for the cathode and a Standard Hydrogen Electrode for the anode (connected to negative terminal of potent iometer).

1A. Write a balanced titration reaction:

1B. Write two half - reactions for the indicator electrode

1C. Write two Nernst equations for the cell voltage.

1D. Calculate Etitr after the addition of the following volumes of 0.1800 M Fe3+:

10.0 mL:

E titr______________

18.75 mL:

Etitr_______________

37.50 mL:

Etitr_______________

40.00 mL

Etitr_______________

1E. Sketch the titration curve and indicate the equivalent pt, and the E

titr when ½ of the moles of analyte has been oxidized

Explanation / Answer

1A. Balance titration equation

Sn2+(Aq) + 2Fe3+(Aq) à Sn4+(Aq) + 2Fe2+(Aq)

1B. Half cell reaction at electrodes as follows,

At Anode,   Sn2+ à Sn4+ + 2e-

At Cathode, 2Fe3+ + 2e- à 2Fe2+

Erxn, is the difference between the reduction potentials for each half-reaction.

Erxn, = [EFe(Ox)/ EFe(red) ] - [ESn(Ox)/ ESn(red)]

After each addition of titrant the reaction between the titrand and the titrant reaches a state of equilibrium. Because the potential at equilibrium is zero, the titrand’s and the titrant’s reduction potentials are same.

EFe(Ox)/ EFe(red) = ESn(Ox)/ ESn(red)

The potential, therefore, is easier to calculate if we use the Nernst equation for the titrand’s half-reaction Sn2+ half cell

Erxn = EoSnox/Snred – [RT/nF] × ln[Snred]/[Snox]

Erxn = EoFeox/Fered – [RT/nF] × ln[Fered]/[Feox]

The first we need to calculate the volume of Ce4+ needed to reach the titration’s equivalence point. From the reaction’s stoichiometry we know that

Moles of Sn2+ = 2 × moles of Fe3+

MSn × VSn = 2(MFe × VFe)

At equilibrium , DeltaE° = RT/nF × ln [Sn4+][ Fe2+]2/ [Sn2+][Fe3+]2

Initial moles of Sn2+ = 0.0750M × 0.025L = 0.001875 moles

So to reach equivalence point moles of Fe3+ needed will be = 0.00375 moles

So DeltaE° at equilibrium = RT/nF   = 8.314 Jmol-1K-1 × 298.15/ 2 × 96485 C = 0.0128 V

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