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14. Consider the titration of 25.00 mL of 0.0500 M Sn2+ with 0.100 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+, using Pt and satu

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Answer #1

(14)

Stoichiometry of the balanced equation:

Number of moles of Fe3+ = 2 x number of the moles of Sn4+

MFe3+ x VFe3+ = 2 x MSn2+ x VSn2+

Ecell = Ecello -2.303RT/nF log [Product]/[Reactant] - Ecalomel

E+ = +0.732 - 0.05916/1 log [Fe2+]/[Fe3+] - 0.241

and

E- = +0.139 - 0.05916/2 log [Sn2+]/[Sn4+] - 0.241

Answer: (a)

(15)

After adding 12.5 mL of Fe3+ i.e. before reaching the equivalence point

[Sn4+] = half the moles of Fe3+ added / total volume = (0.5 x MFe3+ x VFe3+) / (VFe3+ + VSn2+)

[Sn4+] = (0.5 x 0.1 x 12.5) / (25.0 +12.5)

[Sn4+] = 0.0167 M

[Sn2+] = (initial moles of Sn2+ - half the moles of Fe3+ added ) / (total volume)

[Sn2+] = (MSn2+ x VSn2+ - 0.5 x MFe3+ x VFe3+) /  (VFe3+ + VSn2+)

[Sn2+​​​​​​​] = (0.05 x 25.0 - 0.5 x 0.1 x 12.5) / (25.0 +12.5)

[Sn2+​​​​​​​] = 0.0167 M

Ecell = +0.139 - 0.05916/2 log [Sn2+]/[Sn4+] - 0.241

Ecell = +0.139 - 0.05916/2 log 0.0167/0.0167 - 0.241

Ecell = -0.102 V

Answer: (c)

16. After adding 25.0 mL of Fe3* i.e. at equivalence point n1EFes+ n2Esn2+ Ecalomel E = n1n2 [(1 x 0.732 +2 x 0.139) / (1 2

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