highlights of last lecture 2 cases from last 2...

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Unless otherwise stated, all images in this file have been reproduced from:

Blackman, Bottle, Schmid, Mocerino and Wille,Chemistry, 2007 (John Wiley)

ISBN: 9 78047081 0866

Slide 25-2

Chem 1101

A/Prof Sébastien PerrierRoom: 351

Phone: 9351-3366

Email: [email protected]

Prof Scott KableRoom: 311

Phone: 9351-2756


A/Prof Adam BridgemanRoom: 222

Phone: 9351-2731


Slide 25-3

Highlights of last lecture

Equilibrium and thermochemistry in production of the world’s Top 10 chemicals

CONCEPTS� Control of equilibrium in industrial processes.� Energy recycling� Endo- and exothermic processes in industry� Methods of manufacture of H2SO4 and NH3

� Uses of these chemicals

CALCULATIONS� Industrial-type equilibrium and thermochemistry problems

Slide 25-4

2 cases from last 2 lectures…

� Aluminium metal cannot be refined using smelting;

� NaOH and Cl2 production;

An understanding of both these processes requires knowledge of electrochemistry.

Slide 25-5

Electrochemistry

Themes: generating electricity from chemical reactions

driving unfavorable chemical reactions using electricity

References: Any general chemistry text will have a satisfactory chapter on redox and electrochemistry, e.g.

Blackman, et al.: Chap 12

Slide 25-6

Electrochemistry

Key Calculations:

• Calculating cell potential;

• Calculating amount of product for given current;

• Using the Nernst equation for concentration cells.

Key chemical concepts:

• Redox and half reactions;

• Cell potential;

• Voltaic and electrolytic cells;

• Concentration cells.

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� OXIDATION� Loss of Electrons� Reducing agent is oxidised

� Oxidation number increases

An example:Zn(s) → Zn2+(aq) + 2e−

� REDUCTION� Gain of Electrons

� Oxidizing agent is reduced

� Oxidation number decreases

An example:

2H+(aq) + 2e−→ H2(g)

Half-Reactions

LEO, the lion… … goes GER

Loss of Electrons = Oxidation Gain of Electrons = Reduction

Slide 25-8

Activity series of metals

Consider the same reaction for 5 metals:

Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)

Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g)

Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)

Observations

Slide 25-9


Consider the same reaction for 5 metals:

Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)

Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g)

Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)

Activity: Mg > Zn > Fe > Sn > Cu as reducing agents

Observations

Movie 32_1

Slide 25-10


Activity of metals, compared to H+:

Mg > Zn > Fe > Sn > Cu

Consider:

(1) Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

(2) Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)

(1)-(2) Zn(s) + Cu2+(aq) � Zn2+(aq) + Cu(s)

� Which direction? �

Zn is a stronger reducing agent than Cu, so reaction will proceed to right.

Movie 32_1 cont’d

Slide 25-11


Can these electrons be harnessed?

The molecular interpretation…

Figure 12.1 and 12.2 Blackman Slide 25-12

Half reactions

Consider the two half-reactions:

Zn →→→→ Zn2+ + 2e-

Cu2+ + 2e- →→→→ Cu

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

If we can separate the half-reactions then we can harness the electrons.

In a redox reaction we consider the electrons to be transferred from one element to another (obvious in this case).

Note: state labels are

left off half-reactions

Movie 32_2

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Galvanic cells

~1.1 V

Slide 25-14

Galvanic cells

Movie 32_3

Slide 25-15

Galvanic cells

Anode:

Oxidation is occuring

Negative terminal

Cathode:

Reduction is occuring

Positive terminal

Slide 25-16

Short-hand nomenclature

� Rather than writing out the full chemical equation, we can use a short-hand notation:

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Cathode half-cellAnode half-cell

Salt bridge

Phase boundary Phase boundaryElectrons flow this way

Slide 25-17

Cell potentials

Exp’t 1: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)Exp’t 2a:

Potential(E0)

Exp’t 3: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Conclusion 1: Reverse the reaction – reverse the sign of the potential

Conclusion 2: You can add cell potentials when you add chemical reactions.

all reagents at standard concentration of 1.0 M

Cu2+(aq) + H2(g) → Cu(s) + 2H+(aq)Exp’t 2b:

= Exp’t 1 + Exp’t 2b

+0.76V-0.34V

+0.34V

+1.10V

Slide 25-18

Galvanic cells

� Every different galvanic cell has a different cell potential.

� We have seen that you can add cell potentials, so…

� We could tabulate the cell potential against H2 and then calculate any other combination…

� or, we could assign a “half-cell” potential to each “half-reaction”, and then add up two half-reactions to get the full reaction and full potential.

Let’s see how this is done…

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Half-cell potentials

Consider the half-reactions for each exp’t:

Zn → Zn2+ + 2e-

2H+ + 2e- → H2

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

ε0 (V)

Calculated:

Definition:

Observed: +0.76V

Cu2+ + 2e- → CuH2 → 2H+ + 2e-

Cu2+(aq) + H2(g) → Cu(s) + 2H+(aq)

Calculated:

Definition:

Observed: +0.34V

Zn → Zn2+ + 2e-

Cu2+ + 2e- → Cu

Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)

Calculated:

Calculated:

Calculated:

+0.00V

+0.76V

+0.34V

+0.00V

+0.76V

+0.34V

+1.10V

Slide 25-20

Half-cell potentials

Slide 25-21

Reduction potential table

Half-reaction Half-cell potential (V)

Au3+(aq) + 3e− � Au(s) +1.50

Cl2(g) + 2e− � 2Cl− (aq) +1.36

O2(g) + 4H+(aq) + 4e− � 2H2O(l) +1.23

Ag+(aq) + e− � Ag(s) +0.80

Cu2+(aq) + 2e− � Cu(s) +0.34

2H+(aq) + 2e− � H2(g) 0.00*

Sn2+(aq) + 2e− � Sn(s) -0.14

Fe2+(aq) + 2e− � Fe(s) -0.44

Zn2+(aq) + 2e− � Zn(s) -0.76

2H2O(l) + 2e− � H2(g) + 2OH−(aq) -0.83

Mg2+(aq) + 2e− � Mg(s) -2.37

* by definition

Strong oxidising agent

Weak oxidising agent

Weak reducing agent

Strong reducing agent Slide 25-22

Using the reduction tables

� Write down the two half-reactions� Work out which is the oxidation and which is the reduction half-reaction.

� Balance the electrons� Add up the half-reactions to get full reaction

� Add up half-cell potentials to get E 0

� A spontaneous (working) voltaic cell, ALWAYS has a positive cell potential, E 0

Slide 25-23

Example calculation (1)What will the cell potential be for a galvanic cell with Mg2+(aq) | Mg(s)

in one half-cell and Sn2+(aq) | Sn(s) in the other?

Which half reaction is turned around?We saw that Mg is a stronger reducing agent than Sn, i.e. it likes to be oxidised. Turn the Mg reaction around to an ox. reaction.

In general, you turn around the reaction lowest in the table of reduction potentials

Sn2+ + 2e- → Sn E 0 = -0.14V

Mg2+ + 2e- → Mg E 0 = -2.37V

Sn2+ + 2e- → Sn E 0 = -0.14V

Mg → Mg2+ + 2e- E 0 = +2.37V

Mg(s) + Sn2+(aq) → Mg2+(aq) + Sn(s) E 0 = +2. 23V

Slide 25-24

Reduction potential table

Half-reaction Half-cell potential (V)

Au3+(aq) + 3e− � Au(s) +1.50

Cl2(g) + 2e− � 2Cl− (aq) +1.36

O2(g) + 4H+(aq) + 4e− � 2H2O(l) +1.23

Ag+(aq) + e− � Ag(s) +0.80

Cu2+(aq) + 2e− � Cu(s) +0.34

2H+(aq) + 2e− � H2(g) 0.00*

Sn2+(aq) + 2e− � Sn(s) -0.14

Fe2+(aq) + 2e− � Fe(s) -0.44

Zn2+(aq) + 2e− � Zn(s) -0.76

2H2O(l) + 2e− � H2(g) + 2OH−(aq) -0.83

Mg2+(aq) + 2e− � Mg(s) -2.37

* by definition

Strong oxidising agent

Weak oxidising agent

Weak reducing agent

Strong reducing agent

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Example calculation (2)What will the cell potential be for a galvanic cell with Ag+(aq) | Ag(s) in one half-cell and Cr3+(aq) | Cr(s) in the other?

Data: Ag+(aq)|Ag(s) = +0.80V; Cr3+(aq)|Cr(s) = -0.74V

Cr half reaction is lowest (most negative), turn it around…

Cr(s) + 3Ag+(aq) → Cr3+(aq) + 3Ag(s) E 0 = +1.54V

Balance the electrons… Note!

Note: E 0 does not depend on stoichiometry!

Ag+ + e- → Ag E 0 = +0.80V

Cr → Cr3+ + 3e- E 0 = +0.74V

3Ag+ + 3e- → 3Ag E 0 = +0.80V

Cr → Cr3+ + 3e- E 0 = +0.74V

Slide 25-26

Summary

CONCEPTS� Oxidation and reduction

� Half-reactions

� Galvanic cell

� Cell notation

CALCULATIONS� Work out cell potential from reduction potentials;

highlights of last lecture 2 cases from last 2...

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