You are titrating 100.0 mL of 0.0400 M Fe^2+ in 1 M HClO_4 with 0.100 M Ce^4+ to

Question

You are titrating 100.0 mL of 0.0400 M Fe^2+ in 1 M HClO_4 with 0.100 M Ce^4+ to give Fe^3+ and Ce^3+ using Pt and calomel electrodes to find the endpoint (a) Write the balanced titration reaction. (b) Complete the two half reactions for the Pt electrode. (c) From the list in the column at the right, select the two correct Nemst equations for the cell voltage. (Each applying at different points in the titration.) E degree of the calomel electrode is 0.241 V. (A) (B) (C) (D) (E) (F) (G) (H) (d) Calculate the values of E for the cell when the following volumes of the Ce^4+ titrant have been added (Activity coefficients may be ignored as they tend to cancel when calculating concentration ratios.) Volume E 2 50 mL V 20.0 mL V 36.0 mL V 40.0 mL V 43.0 mL V (A) E = 0.767 - 0.05916log([Fe^2+]/[Fe^3+]) - 0.241 (B) E = 0.767 - 0.05916log([Fe^3+]/[Fe^2+]) - 0.241 (C) E = 1.70 - 0.05916log([Fe^2+]/[Fe^3+]) - 0.241 (D) E = 1.70 - 0.05916log([Fe^3+]/[Fe^2+]) - 0.241 (E) E = 0.767 - 0.05916log([Ce^3+]/[Ce^4+]) - 0.241 (F) E = 0.767 - 0.05916log([Ce^4+]/[Ce^3+]) - 0.241 (G) E = 1.70 - 0.05916log([Ce^3+]/[Ce^4+]) - 0.241 (H) E = 1.70 - 0.05916log([Ce^4+]/[Ce^3+]) - 0.241

Explanation / Answer

For the given electrochemical cell,

(a) balanced equation,

Fe2+ + Ce4+ <==> Fe3+ + Ce3+

(b) half reactions,

Ce4+ + e- <==> Ce3+

Fe3+ + e- <==> Fe2+

(c) correct Nernst equations for the cell would be,

(A)

and,

(G)

(d) Calculate E value for the cell

2.50 ml of 0.1 M Ce4+ added

initial moles Fe2+ = 0.04 M x 100 ml = 4 mmol

moles Ce4+ added = 0.1 M x 2.50 ml = 0.250 mmol

[Fe3+] formed = 0.250 mmol/102.50 ml = 0.00244 M

[Fe2+] remained = (4 - 0.250) mmol/102.50 ml = 0.0366 M

using,

E = 0.767 - 0.05916 log([Fe2+]/[Fe3+]) - 0.241

   = 0.767 - 0.05916 log(0.0366/0.00244) - 0.241

= 0.456 V

20 ml of 0.1 M Ce4+ added

initial moles Fe2+ = 0.04 M x 100 ml = 4 mmol

moles Ce4+ added = 0.1 M x 20 ml = 2.0 mmol

[Fe3+] formed = 2.0 mmol/120.0 ml = 0.0167 M

[Fe2+] remained = (4 - 2.0) mmol/120.0 ml = 0.0167 M

using,

E = 0.767 - 0.05916 log([Fe2+]/[Fe3+]) - 0.241

   = 0.767 - 0.05916 log(0.0167/0.0167) - 0.241

= 0.526 V

36 ml of 0.1 M Ce4+ added

initial moles Fe2+ = 0.04 M x 100 ml = 4 mmol

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

[Fe3+] formed = 3.6 mmol/136.0 ml = 0.0265 M

[Fe2+] remained = (4 - 3.6) mmol/136.0 ml = 0.00294 M

using,

E = 0.767 - 0.05916 log([Fe2+]/[Fe3+]) - 0.241

   = 0.767 - 0.05916 log(0.00294/0.0265) - 0.241

= 0.582 V

40 ml of 0.1 M Ce4+ added

initial moles Fe2+ = 0.04 M x 100 ml = 4 mmol

moles Ce4+ added = 0.1 M x 40 ml = 4.0 mmol

Equivalence point

[Ce3+] = [Fe3+]

[Ce4+] = [Fe2+]

So,

E = [(0.767 + 1.70)/2] - 0.241

= 0.992 V

43 ml of 0.1 M Ce4+ added

initial moles Fe2+ = 0.04 M x 100 ml = 4 mmol

moles Ce4+ added = 0.1 M x 43 ml = 4.3 mmol

[Ce3+] formed = 4.0 mmol/143.0 ml = 0.028 M

[Ce4+] remained = (4.3 - 4.0) mmol/143.0 ml = 0.0021 M

using,

E = 1.70 - 0.05916 log([Ce3+]/[Ce4+]) - 0.241

   = 0.767 - 0.05916 log(0.028/0.0021) - 0.241

= 1.392 V

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