Understanding Redox Titration: Role of Agents & Deriving Curves, Study notes of Electrochemistry

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An Introduction to

Electrochemistry

Redox Reaction

ox 1 + red 2 <=> red 1 + ox 2

Outline

1. Review

 Concepts: Redox Reaction, Cell, Potential  Balance an Equation  Nernst Equation

2. Redox Titration

3. Potentiometry

4. Voltammetry and Electrochemical Sensor

• Mass Balance : the amount of each element present at the beginning of the reaction must be present at the end.
• Charge Balance : electrons are not lost in a chemical reaction. Half Reactions
• Half-reactions are a convenient way of separating oxidation and reduction reactions.

12.1 Balancing Oxidation- 12.1 Balancing Oxidation-

Reduction ReactionsReduction Reactions

How?

1. Write down the two half reactions.
2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H+^. d. Finish by balancing charge by adding electrons.
3. Multiply each half reaction to make the number of electrons equal.
4. Add the reactions and simplify.
5. Check!

Balancing Equations by the Method of Half Reactions

16H +^ ( aq ) + 2MnO 4 -^ ( aq ) + 5C 2 O 4 2-( aq )
2Mn2+( aq ) + 8H 2 O( l ) + 10CO 2 ( g )

Exercise:

Try to Balance:

Sn+2^ + Fe+3^ <=> Sn+4^ + Fe+

Fe +2^ + MnO 4 -^ <=> Fe+3^ + Mn+

Sn +2^ + Fe+3^ <=> Sn+4^ + Fe+

Important Redox Titrants and the Reactions

Oxidizing Reagents (Oxidants) (2) Potassium Dichromate

Cr 2 O 72
^ + 3 U^4 +^ + 2 H +^  2 Cr^3 +^ + 3 UO 22 +^ + H 2 O

Cr 2 O 72
^ + 14 H +^ + 6 e
^  2 Cr^3 +^ + 7 H 2 O

Important Redox Titrants and the Reactions

Oxidizing Reagents (Oxidants)

(2) Potassium Iodate

IO 3
^ + 6 H +^ + 5 e
^  1 2

I 2 + 3 H 2 O

Important Redox Titrants and the

Reactions

Reducing Reagent ( Reductants ) (2) Potassium Iodide

I

I 2 + e

Important Redox Titrants and the

Reactions

Reducing Reagent ( Reductants )

(2) Sodium Thiosulfate

2S 2 O 3 2-^ S 4 O 6 2-^ +2e

Fig. 12.1. Voltaic cell.

The salt bridge allows charge transfer through the solution and prevents mixing.

Galvanic Cells

spontaneous redox reaction

anode oxidation

cathode reduction

Standard Electrode Potentials

Standard reduction potential (E^0 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm.

E^0 = 0 V

Standard hydrogen electrode (SHE)

2e -^ + 2H +^ (1 M ) H 2 (1 atm )

Reduction Reaction

Standard Reduction (Half-Cell) Potentials

• The SHE is the cathode. It consists of a Pt electrode in a tube placed in 1 M H +^ solution. H 2 is bubbled through the tube.
• For the SHE, we assign 2H+^ ( aq , 1 M ) + 2e -^
  H 2 ( g , 1 atm)
• E °red of zero.
• The potential of a cell can be calculated from standard reduction potentials: E ° cell = E ° red ( cathode )
  E ° red ( anode )

Standard Reduction (Half-Cell) Potentials

• Consider Zn( s )
  Zn 2+( aq ) + 2e -^. We measure E cell relative to the SHE (cathode): E °cell = E °red (cathode) - E °Ox(anode) 0.76 V = 0 V - E °red(anode).
• Therefore, E °red (anode) = -0.76 V.
• Standard reduction potentials must be written as reduction reactions: Zn 2+( aq ) + 2e-^
  Zn( s ), E °red = -0.76 V.

The more positive the E o, the better oxidizing agent is the oxidized form (e.g., MnO 4 -^ ). The more negative the E o, the better reducing agent is the reduced form (e.g., Zn).

©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

12.4 Redox Titration Curve

12.4 Redox Titration Curve

Derive the titration curve for 50.00 mL of 0.0500 M Fe+2^ with 0.1000 M Ce +4^ in a medium that is 1.0 M in H 2 SO 4. Fe+2^ + Ce+4^ <=> Ce+3^ + Fe + At 15.00 mL of Ce+4^ added, V (^) FeM (^) Fe > V (^) Ce M (^) Ce

= 2.308 x 10 -2^ M

VCe M Ce (15.00 mL)(0.1000 M)

[Fe+3] = --------------- = ----------------------------

V Fe + V Ce (50.00 + 15.00)mL

EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M

Fe+2^ with 0.1000 M Ce +4^ in a medium

that is 1.0 M in H 2 SO 4.

Fe +2^ + Ce +4^ <=> Ce +3^ + Fe+

At 15.00 mL of Ce +4^ added, VFeMFe > VCe M Ce

[Fe+3] = 2.308 x 10-2^ M

V Fe M Fe - V Ce MCe

[Fe+2] = ------------------------

V Fe + V Ce

EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+

with 0.1000 M Ce+4^ in a medium that

is 1.0 M in H 2 SO 4.

Fe +2^ + Ce +4^ <=> Ce +3^ + Fe+

At 15.00 mL of Ce +4^ added, VFeMFe > VCe M Ce

[Fe+3] = 2.308 x 10-2^ M

(50.00 mL)(0.0500 M) - (15.00 mL)(0.1000 M) [Fe +2] = -------------------------------------------------------- (50.00 + 15.00)mL = 1.54 x 10-2^ M

EXAMPLE: Derive the titration

curve for 50.00 mL of 0.0500 M Fe+

with 0.1000 M Ce+4^ in a medium that

is 1.0 M in H 2 SO 4.

Fe +2^ + Ce +4^ <=> Ce +3^ + Fe+

At 15.00 mL of Ce +4^ added, VFeMFe > VCe M Ce

[Fe+3] = 2.308 x 10-2^ M [Fe+2] = 1.54 x 10-2^ M

for Fe +2^ -> Fe+3^ E o^ = 0.69 v

0.0592 [Fe +2]

E cell = E cello^ - ----------- log -----------

n [Fe +3]