1601 redox titration - san diego miramar...

1 Aug ‘17Redox Titration

1601 Redox TitrationRedox Titration in Analytical Chemistry

Dr. Fred Omega GarcesAnalytical Chemistry 251

Miramar College, SDCCD

Recommended:


IntroductionOxidation and reduction titrations may be used to measure many species especially metals in high or low valency states, iodine and iodides and easily oxidized organic compounds.

In redox titration the concentration of oxidizing agent and reducing agent will change simultaneously. Consider the following redox reaction: SCE || a (Ox), a (Red) | Pt The cell emf is: E = E° + (RT/F) ln [a (Ox) / a (Red)] - ESCE

As the reaction proceeds during titration, the ratio of the concentration will change as well as the emf.


Titration PreparationFor standard solutions, sodium oxalate, iron(II) ammonium sulfateor potassium iodate can be used. Potassium permanganate and iodine solution must be standardized before using because of decomposition.

An indicator for redox reactions are reagents whose oxidized and reduced form differ in color: ln (Ox) [color 1] + ne- = ln (Red) [color 2]

Consider the example of 1,10-phenanthroline iron(II): [Fe(C12H8N2)3]3+

(pale blue) + ne- = [Fe(C12H8N2)3]2+(deep red)

In some reactions the H3O+ results in a change in pH.


Redox IndicatorsBelow are common redox indictors

Redox Oxidizing Reduced Eln (V) Solution 1,10-phen iron(III) Pale blue Red 1.11 1 M H2SO4

Diphenylamine Violet Colorless 0.76 Dilute acidMethylene blue Blue Colorless 0.53 1M acidPhenosafranine Red Colorless 0.28 1M acid

Another indicator involving iodine is starch solution (or synthetically similar material) . The starch forms a blue-black complex with iodine which but is colorless when there no iodine is present, i.e., iodine is converted to triiodide (I3

-).

For a redox indicator at 25°C, the color change takes place at the potential range of E = Eln + 0.059


Starch as an IndicatorsStarch is often used in chemistry as an indicator for redox titration where triiodide is present. Starch forms a very dark blue-black complex with triiodide which can be made by mixing iodine with iodide (often from potassium iodide). However, the complex is not formed if only iodine (I2) or only iodide (I-) is present. The color of the starch complex is so deep, that it can be detected visually when the concentration of the iodine is as low as 0.00002 M at 20 °C.

During iodine titrations, concentrated iodine solutions must be reacted with some titrant, often thiosulfate, in order to remove most of the iodine before the starch is added. This is due to the insolubility of the starch-iodine complex which may prevent some of the iodine reacting with the titrant. Close to the end-point, the starch is added, and the titration process is resumed taking into account the amount of thiosulfate added before adding the starch

http://www.elmhurst.edu/~chm/vchembook/548starchiodine.html


Experiment 4Tri iodide is formed by:

KIO3 + KI g I3-

http://www.elmhurst.edu/~chm/vchembook/548starchiodine.html

C6H

8O

6 + I

3− C

6H

6O

6 + I−

VitaminC dehydroascorbic acid

I2 (aq)

+ I − I3−


Redox TitrationBalance redox chem eqn: Solve problem using stoichiometric strategy.Q: 1.225 g Fe ore requires 45.30 ml of 0.0180 M KMnO4. How pure is the ore sample?

When iron ore is titrated with KMnO4 . The equivalent point results when:KMnO4 (purple) g Mn2+ (pink)

Mn (+7) Mn(+2)


Redox TitrationBalance redox chem eqn: Solve problem using stoichiometric strategy.Q: 1.225 g Fe (Fe+2) ore requires 45.30 ml of 0.0180 M KMnO4.

How pure is the ore sample?


Redox TitrationBalance redox chem eqn: Solve problem using stoichiometric strategy.Q: 1.225 g Fe (Fe+2) ore requires 45.30 ml of 0.0180 M KMnO4.

How pure is the ore sample?

When iron ore is titrated with KMnO4 . The equivalent point results when:KMnO4 (purple) g Mn2+ (pink)

Mn (+7) Mn(+2)

Rxn: Fe+2 + MnO4- g Fe+3 + Mn2+

Bal. rxn: 5 Fe2+ + MnO4- + 8 H+ g 5 Fe3+ + Mn2+ + 4 H2O

Note Fe2+ g 5 Fe3+ : Oxidized Lose e- : Reducing Agent

Mol of MnO4- = 45.30 ml • 0.180(mol/L) = 0.8154 mmol MnO4-

Amt of Fe: = 0.8154 mmol • 5 mol Fe+2 • 55.8 g = 0.2275 g1 mol MnO4- 1 mol Fe2+

% Fe = (0.2275 g / 1.225 g) • 100 = 18.6 %


Redox Titration: Example20.28. Titration of 0.1809 g of pure iron wire was dissolve in acid, reduced to +2 state and titrated with 31.33 mL of cerium (IV). Calculate the molar concentration of Ce4+ solution.

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Reaction :Ce4 + + Fe2+ →Ce3+ + Fe3+


Redox Titration: Example20.28. Titration of 0.1809 g of pure iron wire was dissolve in acid, reduced to +2 state and titrated with 31.33 mL of cerium (IV). Calculate the molar concentration of Ce4+ solution.

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[Ce4 +] =0.1809g Fe

31.33mL Ce4 + ×1000 mmolFe

55.847 g×

1mmolFe2+

mmolFe×

1mmol Ce4 +

mmolFe2+

[Ce4 +] = 0.1034 M Ce4 +

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Reaction :Ce4 + + Fe2+ →Ce3+ + Fe3+


Redox Titration: ExamplePetrucci 7th Ed. p 1521. A piece of iron wire weighting 0.1568 g is converted to Fe2+ (aq) and requires 26.24 mL of a KMnO4 (aq) solution for its titration. What is the molarity of the KMNO4 (aq) ?

2. Another substance that may be used to standardized KMNO4 (aq) is sodium oxalate, Na2C2O4. If 0.2482 g of Na2C2O4 is dissolved in water and titrated with 23.68 mL KMnO4, what is the molarity of the KMnO4 (aq) ?

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5 Fe+2 + MnO4

- + 8H+ → Mn+2 + 5Fe+3 + 4H2O

MKMnO4 = 0.1568 g ∗ 1 mol Fe

55.8 g ∗ 1 mol MnO4

-

5 mol Fe+2 ∗ 1

0.02624 L =

MKMnO4 = 0.0212 M

5 C2O

4-2 → 10 CO

2 + 10 e−

______________________________________________ 2 MnO

4- + 16H+ + 10e− → 2Mn+2 + 8 H

2O n=10

0.2482 g Na2C

2O

4 ∗

mol Na2C

2O

4

134.00 g∗

2 mol KMnO4

5 mol Na2C

2O

4 ∗

10.02368 L KMnO

4 = 0.0312 M

1601 redox titration - san diego miramar...

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