Titrations

OXIDATION-REDUCTION TITRATIONS

REDOX TITRATION
• involves oxidizing agents and reducing agents titrants and analytes • oxidizing agents used as standard solutions:  potassium permanganate, KMnO4  potassium dichromate, K2Cr2O7  iodine, I2  ceric sulfate, Ce(SO4) 2  potassium iodate, KIO3

REDOX TITRATION
• reducing agents used as standard solutions:  ferrous sulfate, FeSO4  oxalic acid, H2C2O4  sodium oxalate, Na2C2O4  sodium thiosulfate, Na2S2O3  titanous chloride, TiCl2

REDOX TITRATION
• important redox titration combinations:  potassium permanganate and ferrous salts  potassium dichromate and ferrous salts  potassium permanganate and sodium oxalate  iodine and sodium thiosulfate

REDOX TITRATION CURVES
• resembles the curve for acid-base titration  x-axis: volume of the titrant  y-axis: half-cell potential (instead of pH) • exhibits a sigmoidal shape, as in acid-base titration

REDOX TITRATION CURVES
Reaction: Fe2+ + Ce4+ → Fe3+ + Ce3+
(analyte) (titrant)
Excess

Ce
4+


Equivalence
point




Fe Excess

2+


REDOX TITRATION CURVES
Reaction: Fe2+ + Ce4+ → Fe3+ + Ce3+
(analyte) (titrant)

Before the equivalence point:  Ce(IV) titrant: converted fully to Ce(III)  Fe(II) analyte: some converted to Fe (III); some unreacted  Potential is due to Fe(II) / Fe(III) couple E = 0.78 RT ln nF [Fe2+] [Fe3+]

REDOX TITRATION CURVES
Reaction: Fe2+ + Ce4+ → Fe3+ + Ce3+
(analyte) (titrant)

At the equivalence point:  Ce(IV) titrant: converted fully to Ce(III)  Fe(II) analyte: converted fully to Fe(III)  Fe(II) and Ce(IV): very small amounts are left o [Fe(II)] = [Ce(IV)]; [Fe(III)] = [Ce(III)]  Potential is due to all the species in the titration reaction: E = n1EFeo + n2ECeo n1+ n2

REDOX TITRATION CURVES
Reaction: Fe2+ + Ce4+ → Fe3+ + Ce3+
(analyte) (titrant)

After the equivalence point:  Ce(IV) titrant: additional amount remains unconverted  Fe(II) analyte: all converted to Fe (III)  Potential is due to Ce(III) / Ce(IV) couple E = 1.44 [Ce3+] RT l ln nF [Ce4+]

MnO4- + 5 Fe2+ + 8 H+ → Mn2+ + 5 Fe3+ + 4 H2O
("trant)
 (analyte)


(2)

(1)

TITRATIONS WITH KMnO4
• Potassium permanganate: a strong oxidizing agent • reduced to colorless Mn2+ in acidic medium: MnO4- + 8 H+ + 5 e → Mn2+ + 4 H2O • reduced to brown MnO2 in neutral or alkaline solution: MnO4- + 4 H+ + 3 e → MnO2 + 2 H2O • reduced to green MnO42- in strongly alkaline solution: MnO4- + e → MnO42-

TITRATIONS WITH KMnO4
• standardized commonly using Na2C2O4 2 MnO4- + 5 H2C2O4 + 6 H+ → 2 Mn2+ + 10 CO2 + 8 H2O • requires no indicator solution; self-indicating • decomposes slowly during storage; catalyzed by light

EXERCISE
A 0.1278 g of primary standard Na2C2O4 (M = 134) required exactly 31.31 mL of KMnO4 to reach the equivalence point. What is the molarity of the KMnO4 solution?

0.01218M


EXERCISE
A sample of commercial hydrogen peroxide weighing 1.25 g was dissolved in 25.0 mL of pure water and titrated with 0.0115 M KMnO4 solution. The titration consumed 38.5 mL of the KMnO4 solution. Calculate the percentage of H2O2 in the sample. 2 MnO4- + 5 H2O2 + 6 H+ → 2 Mn2+ + 5 O2 + 8 H2O

3.01%


TITRATIONS WITH K2Cr2O7
• Potassium dichromate: a powerful oxidizing agent, but not as strong as KMnO4 • reduced to green Cr3+ in acidic medium: Cr2O72- + 14 H+ + 6 e → 2 Cr3+ + 7 H2O • requires indicator for equivalence point: diphenylamine sulfonate • employed chiefly for the determination of Fe(II): Cr2O72- + 6 Fe2+ + 14 H+ → 2 Cr3+ + 6 Fe3+ + 7 H2O

EXERCISE
Treatment of hydroxylamine (NH2OH) with an excess of Fe(III) results in the formation of N2O and Fe(II). 2 NH2OH + Fe3+ → N2O + Fe2+ + 4 H+ + H2O The Fe(II) produced by the reaction of a 50.00 mL sample of NH2OH required 23.61 mL of 0.02170 M K2Cr2O7 solution for titration to the equivalence point using diphenylamine sulfonate as indicator. Calculate the molar concentration of the NH2OH solution.

0.1230
M


TITRATIONS INVOLVING I2
• Iodine: a weak oxidizing agent used mainly for the determination of strong reducing agents I2 + 2 e → 2 I• low solubility in water; increased through addition of KI and formation of the tri-iodide ion I2 + I - → I3• titrated with sodium thiosulfate with starch as indicator I3‐
+
2
S2O32‐
→ 3 I
‐ + S4O62
‐




EXERCISE
Calculate the percentage of MnO2 in a mineral specimen if the I2 liberated by a 0.1344g sample in the net reaction MnO2(s) + 4H+ + 2I-  Mn2+ + I2 + 2H2O required 32.30mL of 0.07220M Na2S2O3.

75.42%