Question

19. a. For the titration of 50.00 mL of 0.0500 M Fe2+ with 0.1000 M Ce4+ in the presence of 1 M H2SO4, please calculate the system potential after 2.25 mL of Ce4+ is added .

Ce4+ + e ÍÎ Ce3+, E0 = + 1.44 V (in 1M H2SO4)

Fe3+ + e ÍÎ Fe2+, E0 = + 0.68 V (in 1M H2SO4).

They got Esystem = 0.62 V.

b)  For the same titration as in question (a), what’s the system potential after 32.00 mL of Ce4+ is added?

Answer: Esystem = 1.41 V

c) If we set up an electrochemical cell where the cathode is the system that we have in question (b), whereas the anode is a saturated calomel electrode (0.2444 V relative to standard hydrogen electrode). Please determine the potential of this electrochemical cell.

Answer: Ecell = 1.16 V

d) Please determine the system potential at the equivalence point for the titration described in question (a).

Answer: Esystem = 1.06 V

I listed all the answers. I just want to see how they got that. Thanks.

Answer 1

19.

a. moles of Fe2+ = 0.05 M x 50 ml = 2.5 mmol

mols of Ce4+ = 0.1 M x 2.25 ml = 0.225 mmol

[Fe2+] remaining = 2.275 mmol/52.25 ml = 0.043 M

[Fe3+] formed = 0.225 mmol/52.25 ml = 0.0043 M

Using Nernst equation,

E = 0.68 - 0.0592 log([Fe2+]/[Fe3+])

   = 0.68 - 0.0592 log(0.043/0.0043)

   = 0.62 V

b) moles of Fe2+ = 0.05 M x 50 ml = 2.5 mmol

mols of Ce4+ = 0.1 M x 32 ml = 3.2 mmol

[Ce4+] remaining = 0.7 mmol/82 ml = 0.0085 M

[Ce3+] formed = 2.5 mmol/82 ml = 0.03 M

Using Nernst equation,

E = 1.44 - 0.0592 log([Ce3+]/[Ce4+])

   = 1.44 - 0.0592 log(0.03/0.0085)

   = 1.41 V

c) When anode is saturated calomel electrode

moles of Fe2+ = 0.05 M x 50 ml = 2.5 mmol

mols of Ce4+ = 0.1 M x 32 ml = 3.2 mmol

[Ce4+] remaining = 0.7 mmol/82 ml = 0.0085 M

[Ce3+] formed = 2.5 mmol/82 ml = 0.03 M

Using Nernst equation,

E = [1.44 - 0.0592 log([Ce3+]/[Ce4+])] - 0.2444

   = [1.44 - 0.0592 log(0.03/0.0085)] - 0.2444

   = 1.16 V

d) Volume of Ce4+ added at equivalence point = 2.5 mmol/0.1 M = 25 ml

At equivalence point, [Fe2+] = [Ce4+] and [Fe3+] = Ce3+]

E = (1.44 + 0.68)/2 = 1.06 V