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Chapter 24: Q 24.12 (page 651)

Construct a coulometric titration curve of 100.0 mL of a 1 M H2SO4 solution containing Fe(II) titrated
with Ce(IV) generated from 0.075 M Ce(III). The titration is monitored by potentiometry. The initial
amount of Fe(II) present is 0.05182 mmol. A constant current of 20.0 mA is used. Find the timecorresponding to the equivalence point. Then, for about ten values of time before the equivalence point,use the stoichiometry of the reaction to calculate the amount of Fe31 produced and the amount of Fe21 remaining. Use the Nernst equation to find the system potential. Find the equivalence point potential
in the usual manner for a redox titration. For about ten times after the equivalence point, calculate theamount of Ce41 produced from the electrolysis and the amount of Ce31 remaining. Plot the curve of system potential versus electrolysis time.

Short Answer

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Most popular questions from this chapter

A 6.27-g sample of an ant-control preparation was decomposed by wet ashing with H₂SO₄ and HNO₃. The arsenic in the residue was reduced to the trivalent state with hydrazine. After the excess reducing agent had been removed, the arsenic(III) was oxidized with electrolytically generated I₂ in a faintly alkaline medium:

${HAsO}_{3}^{2 -} + I_{2} + 2 {HCO}_{3}^{-} \to {HAsO}_{4}^{2 -} + 2 I^{-} + 2 CO + H_{2} O$

The titration was complete after a constant current of 150.2 mA had been passed for 10 min and 34 s. Express the results of this analysis in terms of the percentage As₂O₃ in the original sample.

A 0.0712-g sample of a purified organic acid was dissolved in an alcohol-water mixture and titrated with coulometrically generated hydroxide ions. With a current of 0.0392 A, 241 s was required to reach a phenolphthalein end point. Calculate the equivalent mass of the acid.

What cathode potential (versus SCE) would be required to lower the total Hg(II) concentration of the following solutions to 1.00 x 10^-6 M (assume reaction product in each case is elemental Hg):

(a) an aqueous solution of Hg²⁺?

(b) a solution with an equilibrium SCN^- concentration of 0.100 M?

${Hg}^{2 +} + 2 {SCN}^{-} ⇌ Hg {(SCN)}_{2} (aq) K_{f} = 1.8 \times 10^{7}$

(c) a solution with an equilibrium Br2 concentration of 0.100 M?

${HgBr}_{4}^{2 -} + 2 e ⇌ Hg (l) + 4 {Br}^{-} E^{0} = 0.223 V$

At a potential of 21.0 V (versus SCE), carbon tetrachloride in methanol is reduced to chloroform at a Hg cathode:

$2 {CCl}_{4} + 2 H^{+} + 2 e^{-} + 2 Hg (l) \to 2 {CHCl}_{3} + {Hg}_{2} {Cl}_{2} (s)$

At 21.80 V, the chloroform further reacts to give methane:

$2 {CHCl}_{3} + 6 H^{+} + 6 e^{-} + 6 Hg (l) \to 2 {CH}_{4} + 3 {Hg}_{2} {Cl}_{2} (s)$

Several 0.750-g samples containing CCl₄, CHCl₃, and inert organic species were dissolved in methanol and electrolyzed at 21.0 V until the current approached zero. A coulometer indicated the charge required to complete the reaction, as given in the second column of the following table. The potential of the cathode was then adjusted to 21.80 V. The additional charge required to complete the reaction at this potential is given in the third column of the table. Calculate the percent CCl₄ and CHCl₃ in each mixture.

Lead is to be deposited at a cathode from a solution that is 0.150 M in Pb²⁺ and 0.215 M in HClO₄. Oxygen is evolved at a pressure of 0.850 atm at a 30-cm²platinum anode. The cell has a resistance of 0.900 V.

(a) Calculate the thermodynamic potential of the cell.

(b) Calculate the IR drop if a current of 0.220 A is to be used.

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