19. a. For the titration of 50.00 mL of 0.0500 M Fe2+ with 0.1000 M Ce4+ in the
ID: 953001 • Letter: 1
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.
Explanation / Answer
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
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