Slide 34-1

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 34-2

Chem 1101

A/Prof Sébastien PerrierRoom: 351

Phone: 9351-3366

Email: s.perrier@chem.usyd.edu.au

Prof Scott KableRoom: 311

Phone: 9351-2756

Email: s.kable@chem.usyd.edu.au

A/Prof Adam BridgemanRoom: 222

Phone: 9351-2731

Email: a.bridgeman@chem.usyd.edu.au

Slide 34-3

Highlights of last lecture

Voltaic cells...

CONCEPTS� Half-reaction� Table of standard reduction potential� Effect of concentration� Link between E , Q and K

� CALCULATIONS� Work out cell potential from reduction potentials;� Work out cell potential for any concentration (Nernst equation)

)log(303.20 QnFRT

EEcell ×−=“Nernst equation”

Slide 34-4

Concentration cells

� If the electrochemical potential is a function of concentration (as in the Nernst eq’n), then can we build a voltaic cell using the same half-reaction, but with different concentrations in each half-cell?

How does this work?

Afterall:

Cu → Cu2+ + 2e- E 0=-0.34

Cu2+ + 2e- → Cu E 0=+0.34

???? E = 0.0V

0.75V

Slide 34-5

Concentration cells

� It works because the standard half-cell potential is based on 1.0 M concentrations. Even though in this experiment E 0 = 0, the potential for Q ≠ 1 is non-zero.

� The measured cell potential in our experiment was 0.75V. Let’s work out what the [Cu2+] was in the beaker:

Cu → Cu2+ (y M) + 2e-

Cu2+ (0.1M) + 2e- → Cu

Cu2+ (0.1M) → Cu2+ (y M) Ecell = ____V

)log(0592.00

Qn

EEcell −=

−=

1.0log0296.00.0

y= ____ V

Solve for y: _______________________

0.75

0.75

1E-26 M

Slide 34-6

×−=−=

+

0.1

][log0296.0)log(

0592.0 20 H

Qn

EEcell

]log[0592.0 +−= H

E cell = 0.0592 x pH

i.e. measurement of the cell potential provides pH directly!

pH ≡ -log[H+]

Applications of concentration cells

Consider the following concentration cell:

H2(g) → 2H+(aq, unknown) + 2e-

2H+ (aq, 1M) + 2e- → H2(g)

2H+ (1M) → 2H+ (unknown) Ecell = ?

pH meter

In case you missed it!

Slide 34-7

Applications of concentration cells

Concentration cells are used all around us, e.g.� nerve signalling

� concentration gradients produce electrical current

� ion pumps across cell membranes� Na+ / K+ pump, Ca2+ pump� energy production and storage in cells

� ATP

� pH meters� these usually use a silver/silver chloride reference rather than SHE

� ion selective electrodes� different choice of reference can then be specific to specific ions.

Slide 34-8

Other applications of voltaic cells

� Batteries (later lecture)Refs: Most General Chemistry texts have good sections on batteries, e.g.

� Silberberg, pp.923-5� McMurray and Fay, pp.781-6� Petrucci, Harwood and Herring, pp.844-848� However, Housecroft and Constable does not(!)

� Corrosion (later lecture)Ref: Most General Chemistry texts also have a section on corrosion:

� Silberberg, pp.926-8� McMurray and Fay, pp.786-9� Petrucci, Harwood and Herring, pp.849-50� Again, Housecroft and Constable does not(!)

Slide 34-9

Electrolysis

So far we have considered electrochemical cells where: � the reaction is spontaneous, or� the cell potential is positive, or� the cell produces electricity

Many of these reactions are equilibrium reactions. Can we choose conditions to force the reverse reaction?

YES! Just force electrons the other way around the circuit…

�This is called “electrolysis”.

�A cell that performs this is called an “electrolytic cell”.

�This process is responsible for several of the industrial processes that we have glossed over, including #9 and #10 on “top ten” list (NaOH and Cl2), refining of metals, and production of aluminium.

Slide 34-10

Comparison – voltaic vs electrolytic

Electrolytic CellVoltaic Cell

Anode = oxidation = negative

Cathode = reduction = positive

Ecell > 0

Anode = oxidation = positive

Cathode = reduction = negative

Ecell < 0

++- -

Zn→→→→Zn2++2e- Cu2++2e-→→→→Cu Zn2++2e-→→→→Zn Cu→→→→Cu2++2e-

Slide 34-11

Why does the sign change?

� Assigning the term “anode” and “cathode” is not based on sign, but on whether the chemical species is oxidised or reduced.

� Oxidation ALWAYS occurs at the anode.

� In a voltaic cell, the anode liberates the electrons from solution and the electrode becomes negatively charge.

� In an electrolytic cell, the electrons are withdrawn from the electrode, which therefore becomes positive.

� Reduction ALWAYS occurs at the cathode.

� In a voltaic cell, the cathode loses electrons to the solution and the electrode becomes positive.

� In an electrolytic cell, the cathode has electrons pumped into it from the battery, and the electrode becomes negative.

Slide 34-12

Electrolysis of water

Oxidation: 2H2O → O2 + 4H+ + 4e- E 0 = -1.23 V

Reduction: 4H2O + 4e- → 2H2 + 4OH

- E 0 = -0.83 V

Overall: 6H2O(l) → 2H2(g) + O2(g) + 4H+(aq) + 4OH-(aq)

E = -1.23 V @ pH 7

At standard state (all @ 1 mol/L):

Oxidation: 2H2O → O2 + 4H+ + 4e- E = -0.82 V

Reduction: 4H2O + 4e- → 2H2 + 4OH

- E = -0.41 V

But at [H+] and [OH-] = 10-7 M (use Nernst eq):

Overall: 6H2O(l) → 2H2(g) + O2(g) + 4H+(aq) + 4OH-(aq)

E 0 = -2.06 V @ [H+] = [OH-] = 1 M

Slide 34-13

Electrolysis of water

Vmin = 1.23-2.06 V

but Vexp ≥≥≥≥ 3 V??

At standard state (all @ 1 mol/L): E 0 = -2.06 VBut at [H+] and [OH-] = 10-7 M : E = -1.23 V

Oxidation: 2H2O → O2 + 4H+ + 4e-

Reduction: 4H2O + 4e- → 2H2 + 4OH

-

So it should take between 1.23 and 2.06 V to drive the electrolysis of water

Turns redTurns blue

Slide 34-14

Over-potential

It is frequently observed that the required voltage for electrolysis is higher than predicted using the reduction potential tables. Several reasons:

� Resistance in the electrical circuit;

� Rate of electron transfer limited by electron transfer at the electrode interface;

� If half-reaction has a high barrier for electron transfer then the rate is slowed.

Empirically:

� Deposition and dissolution of metals usually involves only a very small over-potential;

� Production of gases at the electrode can require quite a high over-potential. For example, H2 and O2 require an over-potential of typically 0.4-0.6 V

Present theory is still unable to predict the size of the over-potential

Slide 34-15

Predicting electrolysis reactions

� Given any combination of anode and cathode materials and electrolyte, can you predict what reaction will result?

Things to think about:

- Is the battery providing enough voltage?

- Which species are oxidised / reduced most easily?

- Is an over-voltage needed?

Slide 34-16

What can be electrolysed?

� Cations in solution that have a more positive reduction potential than water (E 0 > -0.41 to -0.82 V),can be reduced.

� Anions in solution that have a more positive oxidationpotential than water (E 0 > -0.82 to -1.23 V), can be oxidised.

� If the redox potential is in the “water range” then probably both will happen.

Slide 34-17

Reduction potential tableHalf-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

O2(g) + 4H+(aq) + 4e−−−− ���� 2H2O(l) +0.82 @ pH7

Ag+(aq) + e− � Ag(s) +0.80

I2(s) + 2e−� 2I− (aq) +0.51

Cu2+(aq) + 2e− � Cu(s) +0.34

2H+(aq) + 2e− � H2(g) 0.00

Sn2+(aq) + 2e− � Sn(s) -0.14

2H2O(l) + 2e−−−− ���� H2(g) + 2OH

−−−−(aq) -0.41 @ pH7

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

Reduced by electrolysis

Oxidisedby e

lectrolysis

Slide 34-18

What can be electrolysed?

� Cations in solution that have a more positive reduction potential than water (E 0 > -0.41 to -0.82 V),can be reduced.

� Anions in solution that have a more positive oxidationpotential than water (E 0 > -0.82 to -1.23 V), can be oxidised.

� If the redox potential is in the water range then probably both will happen.

So many metal ions can be reduced to metal,

e.g. Cu, Pb, Sn, Ni, Co, Cd, …

But many cannot,

e.g. Na, Mg, Al, Ti, Mn, …

Slide 34-19

Electrolysis of aqueous solutions

� What products do you expect when the following aqueous solutions are electrolysed (all at 1 M)?a) KI; b) AgNO3 c) MgSO4

a) Two possible reduction reactions:

4H2O + 4e- → 2H2 + 4OH

- E 0 = -0.41 ⇒ -0.81 to -1.01 V with over-pot’l

K+ + e- → K E 0 = -2.92 V

2I− → I2(s) + 2e- E 0 = -0.51 V

2H2O → O2 + 4H+ + 4e- E 0 = -0.82

⇒ -1.2 to -1.4 V

Two possible oxidation reactions:

Therefore we expect H2 at the cathodeand I2 at the anode.

KI electrolysis

pure water (pH7)

�

�

Slide 34-20

b) 1 M AgNO3 ?

Electrolysis of aqueous solutions

What will be reduced?

Ag+(aq) or H2O(l) Check out their E 0 values…

What will be oxidised?

H2O(l) or NO3-.

c) 1 M MgSO4 ?

What will be reduced? Mg+(aq) or H2O(l)

What will be oxidised? SO42-(aq) or H2O(l)

NO3- cannot be oxidised

Slide 34-21

Faraday’s Laws of Electrolysis

� Faraday’s first law:� The mass of substance liberated at an electrode during electrolysis is proportional to the quantity of charge (in Coulombs) passing through the electrolyte.

� Faraday’s second law:� The number of Coulombs needed to liberate one mole of different products are in whole number ratios.

charge (Q) = current (I) x time (t)

Coulombs = Amperes x seconds

Faraday 1: mass ∝ charge

Faraday 2: Experimentally, the charge on 1 mol of electrons is 96,485 Coulombs. This value is now known as the “Faraday constant” and given the symbol “F”

Slide 34-22

Example question:

� In the electro-refining of Cu, what mass of Cu is deposited in 1.00 hr, by a current of 1.62 Amp?

Approach:

1. How many moles of electrons pass through the circuit?

2. How many moles of copper can be produced from this many electrons?

3. What is the mass of this many moles of copper?

Michael Faraday

(1791-1867)

Slide 34-23

Example question:

1. Q = I x t [1 hr = 3600 s]

= 1.62 x 3600

= 5830 C

96485 C per mol electrons, therefore

5830/96485 = 0.060 mol electrons

2. The copper half-reaction requires 2 electrons for each Cu:

Cu2+ + 2e- → Cu

Therefore 0.030 mol Cu produced

3. Atomic weight of Cu = 63.55 g/mol

therefore 0.030 x 63.55 = 1.92 g Cu.

Slide 34-24

Summary

CONCEPTS� Nernst equation� E 0 and K� Concentration cells� How pH meters work

CALCULATIONS� Work out cell potential for any concentration (Nernst equation)

� Work out K from E 0

� Work out pH from concentration cell