Question

Redox Titrations Redox Titrations Like Monoprotic Titrations . Three (3) Regions Practice Extra Credit Titrate 100.0mL of 0.0500M Fe? with 0.100M Ce using the cell I drew at the beginning of the lecture. The equivalence point is when volume of Ce added is 50.0mL. Find the voltage of the cell at the follawing points: Bafore the equivalence point At the equivalence paint After the equivalence poi Take a reaction with Fe with a Ce4-, Cerium(IV), standard 20.0mL, 36.0mL, 50.0ml, 51.0mL and 63.0mL Use a Pt working electrode and a S.C.E reference Electroanalytical Techniques Typically two (2) types of Electroanalytical Electroanalytical Techniques Control Current-; Vary Potential - E Instrument Used: Galvanostat

  

Hello, this is in the area of electrochemistry. I posted the picture to the extra credit problem. the little drawing correlates to the problem. I need help solving this problem. I would appreciate the help tremendously since I really would like the points. Thank you on advance.

Answer 1

For the reaction,

Fe2+ + Ce4+ ----> Fe3+ + Ce3+

Titration

(a) 20 ml of 0.1 M Ce4+ added

initial moles Fe2+ = 0.05 m x 100 ml = 5 mmol

moles of Ce4+ added = 0.1 M x 20 ml = 2 mmol

[Fe2+] formed = 2 mmol/120 ml = 0.0167 M

[Fe3+] remain = 3 mmol/120 ml = 0.025 M

Ecell = Eo - 0.0592 log([Fe2+]/[Fe3+])

         = 0.767 - 0.0592 log(0.0167/0.025)

         = 0.780 V

(b) 36 ml of 0.1 M Ce4+ added

initial moles Fe2+ = 0.05 m x 100 ml = 5 mmol

moles of Ce4+ added = 0.1 M x 36 ml = 3.6 mmol

[Fe2+] formed = 3.6 mmol/136 ml = 0.0265 M

[Fe3+] remain = 1.4 mmol/136 ml = 0.0103 M

Ecell = Eo - 0.0592 log([Fe2+]/[Fe3+])

         = 0.767 - 0.0592 log(0.0265/0.0103)

         = 0.743 V

(c) 50 ml Ce4+ added

equivalence point

Ecell = (1.70 + 0.767)/2 = 1.233 V

(d) 51 ml of 0.1 M Ce4+ added

initial moles Fe2+ = 0.05 m x 100 ml = 5 mmol

moles of Ce4+ added = 0.1 M x 51 ml = 5.1 mmol

[Ce3+] formed = 5 mmol/151 ml = 0.033 M

[Ce4+] remain = 0.1 mmol/151 ml = 0.00066 M

Ecell = Eo - 0.0592 log([Ce3+]/[Ce4+])

         = 1.70 - 0.0592 log(0.033/0.00066)

         = 1.60 V

(e) 63 ml of 0.1 M Ce4+ added

initial moles Fe2+ = 0.05 m x 100 ml = 5 mmol

moles of Ce4+ added = 0.1 M x 63 ml = 6.3 mmol

[Ce3+] formed = 5 mmol/163 ml = 0.0307 M

[Ce4+] remain = 1.3 mmol/163 ml = 0.008 M

Ecell = Eo - 0.0592 log([Ce3+]/[Ce4+])

         = 1.70 - 0.0592 log(0.0307/0.008)

         = 1.665 V