CHEM1612 - PharmacyWeek 9: Nernst EquationDr. Siegbert Schmid

School of Chemistry, Rm 223

Phone: 9351 4196

E-mail: siegbert.schmid@sydney.edu.au

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

Blackman, Bottle, Schmid, Mocerino and Wille,     Chemistry, John Wiley & Sons Australia, Ltd. 2008

     ISBN: 9 78047081 0866

Lecture 25-3

Electrochemistry Blackman, Bottle, Schmid, Mocerino & Wille:

Chapter 12, Sections 4.8 and 4.9

Key chemical concepts: Redox and half reactions Cell potential Voltaic and electrolytic cells Concentration cells

Key Calculations: Calculating cell potential Calculating amount of product for given current Using the Nernst equation for concentration cells

Lecture 26 - 4

The measured voltage across the cell under standard conditions is the standard cell potential E0

cell (also called emf).

Recap: Standard cell potential

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

(aq) + Cu(s)

E0cell = E0

cathode – E0anode = 0.34 - ( - 0.76)= 0.34 +0.76 = 1.10 V

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

Lecture 26 - 5

Tricks to memorise anode/cathode 1. Anode and Oxidation begin with a vowels,

Cathode and Reduction with consonants.

2. Alphabetically, the A in anode comes before the C in cathode, as the O in oxidation comes before the R in reduction.

3. Think of this picture:

AN OX and a RED CAT

(anode oxidation reduction cathode)

Lecture 26 - 6

Standard cell potential and free energy For a spontaneous reaction, E0

cell > 0 and also ΔG0 < 0

For a non-spontaneous reaction, E0cell <0 and also ΔG0 > 0

So there is a proportionality between E0 and -ΔG0.

You also know that the maximum work done on the surroundings is

-wmax = ΔG

Electrical work done by the cell is w = Ecell × charge

Lecture 26 - 7

Standard cell potential and free energy The emf E0

cell is related to the change in free energy of a reaction:

∆G0 = –nFE0cell

∆G0 = Standard change in free energy

n = number of electrons exchanged

F = 96485 C/mol e- (Faraday constant)

∆G = –nFEcell

Also, away from standard conditions:

But what is Ecell?

Lecture 26 - 8

ExampleCalculate ∆G0 for a cell reaction:

Cu2+(aq) + Fe (s) Cu(s) + Fe2+ (aq)

Is this a spontaneous reaction?

This process is spontaneous as indicated by the negative sign of G0 and the positive sign for E0

cell.

Cu2+ + 2e– Cu E0 = 0.34 V

Fe2+ + 2e– Fe E0 = –0.44 V

E0cell = 0.34 - (-0.44) = 0.78 V

∆G0 = –nFE0cell

∆G0 = – 2 · 96485 · 0.78 = – 1.5 · 105 J

Lecture 26 - 9

Al is very easily oxidised,

Al3+ + 3e− Al Eo = -1.66V.

The filling is an inactive

cathode for the

reduction of oxygen,

O2 + 4H+ + 4e− 2H2O

and saliva is an electrolyte.

Put the three together (biting on a piece of foil) results in generation of a current and possible pain.

Al(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg

Example: a Dental Galvanic Cell

Lecture 26 - 10

Example: a Dental Galvanic CellAl(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg

O2 + 4H+ + 4e– 2H2O E0 = 1.23 V

Al3+ + 3e– AlE0 = –1.66 V

12H+ + 3O2 + 4Al 6H2O + 4Al3+

E0cell = 2.89 V

∆G0 = –nFE0cell=

–12 · 96485 · 2.89 = –3346 kJmol–1

Lecture 26 - 11

Recall ΔG = ΔG0 + RT ln(Q) (1)

Since ΔG0 = -nFE0cell and ΔG = -nFE

Equation (1) becomes

-nFE = -nFE0cell + RTln(Q) Q = [products] / [reactants]

dividing both sides by –nF gives:

E = E0 – RT ln(Q) nF

Nernst Equation

Walther Nernst

Nobel Prize 1920

But what is Ecell?

Ima

ge

from

no

be

lprize

.org

Lecture 26 - 12

Ecell = E0 – RT ln(Q)nF

E = cell potentialE0 = Standard cell potentialR = Real Gas Constant= 8.314 JK-1mol-1

T =Temperature (K)n = no. of e- transferredF = Faraday constant = 96485 C mol-1

Q = Reaction quotient(Q = K at equilibrium)

Nernst Equation

Since ln (x) = 2.303 log (x)

E = E0 – 2.303 · RT log(Q) nF

At 25°C, (2.303·R·298)/96485 = 0.0592

)log(0592.00 Qn

EEcell Nernst equation more commonly

written like this (note: only at 25°C)

Lecture 26 - 13

Calculate the expected potential for the following cell:

i) [Cu2+] = 1.0 M; [Zn2+] = 10-5M

ii) [Cu2+] = 10-5M; [Zn2+] = 1.0 M

Example calculation 1

)log(0592.00 Qn

EEcell

Firstly, work out the value of n :

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

Cu2+ + 2e- Cu

Zn Zn2+ + 2e-

2 mol e- transferred per mole of reaction: n = 2

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) E0 = 1.1 V

Lecture 26 - 14

[Zn2+] (M) [Cu2+] (M) Q log(Q) Ecell (V)

1.0 1.0

10-5 1.0

1.0 10-5

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

)log(0592.00 Qn

EEcell

Example calculation

(n=2)

1.0

10-5

105

0.0

-5.0

5.0

1.10

1.25

0.95

][

][2

2

Cu

ZnQ

Lecture 26 - 15

Demo: The effect of concentration

What happens when the concentration of one cell is changed?

Cu|Cu2+||Cu2+|Cu

Both compartments of the voltaic cell are identical.

E0cell = E0

copper – E0copper = 0 (in standard conditions, 1M concentrations)

0.00 V

Lecture 26 - 16

Demo: Cu Concentration Cell

Can we explain this?…

Cu2+ + 2e- Cu High [Cu2+]Low [Cu2+]

Cu Cu2++ 2e-

Cu |Cu2+||Cu2+|Cu

E0 same for both half-reactions, so E0cell= 0.

However, we have reduced the concentration of Cu2+ in one cell = non-standard conditions.

Electrical energy is generated until the concentrations in each half-cell become equal (equilibrium is attained).

Add Na2S –precipitate forms.

Lecture 26 - 17

Cu Concentration Cell

Cu|Cu2+||Cu2+|Cu

Low [Cu2+] High [Cu2+]

Cu Cu2++ 2e- Cu2+ + 2e- Cu

E0cell is same on both sides, but the Cu concentrations are different.

More charge carriers in one half-cell. If we poured the two solutions together, we would expect

spontaneous mixing of two solutions of different concentrations to give one of equal concentration.

The electrical connection allows electrons to pour from one half-cell to the other.

Lecture 26 - 18

Concentration cells The measured cell potential in our experiment was “Voltage”.

Let’s work out what the voltage should be:

Cu Cu2+ (0.01 M) + 2e-

Cu2+ (0.1M) + 2e- Cu

Cu2+ (0.1M) Cu2+ (0.01 M) Ecell =

)log(0592.00 Qn

EEcell

1.0log0296.00.0

y= “Voltage”

Solve for “Voltage”:

“Voltage”

“Voltage” = 0.0296 V

“Voltage”

0.01

Lecture 26 - 19

Concentration cells

The cell potential depends on the concentration of reactants.

Implication: We need to specify concentration when referring to the cell potential.

The overall potential for the Cu/Cu2+ concentration cell is:

E = E0cell – 0.0592/2 · log [Cu2+]dil / [Cu2+]conc

Corollary: It is useful to define a standard concentration, which is 1 M.

Lecture 26 - 20

1. Standard Hydrogen Electrode (SHE)

2. Metal-Insoluble Salt Electrode: Standard Calomel Electrode (SCE) and Silver Electrode

3. Ion-Specific: pH electrode

Reference Electrodes

Lecture 26 - 21

Standard Hydrogen Electrode (SHE)

Platinum – gas electrode

H2 electrode

H2 2H+ + 2 e- Eo = 0.00 V

Metal – Metal ion Electrode

Cu2+ + 2e- Cu Eo = 0.34 V

Pt|H2(g)|H+(aq)||Cu2+(aq)|Cu(s) Eocell = 0.34 – 0 = 0.34 V

anode cathode

Finely divided surface Pt electrode.

HCl solution with [H+] =1,

H2 p= 1 atm bubbling over the electrode.

H2 absorbs on the Pt, forming the equivalent of a 'solid hydrogen‘ electrode in equilibrium with H+.

Lecture 26 - 22

The Standard Hydrogen Electrode (SHE) isn't convenient to use in practice (can be contaminated easily by O2 or organic substances).

There are more practical choices, like metal - insoluble salt electrodes. The potential of these electrodes depends on the concentration of the

anion X- in solution.

In practice 2 interfaces:

1. M / MX insoluble salt: MX (s) + e- M (s) + X-(s)

2. coating/solution: X- (s) X- (aq)

Overall: MX (s) + e-(metal) M (s) + X- (aq)

Metal-insoluble salt electrodes

M

X-

X-

MXC+

C+

C+

C+

X-

X-

Lecture 26 - 23

The concentration of anions in solution is controlled

by the salt's solubility:

[Ag+] [X-] = Ksp

Normal calomel electrode, Pt | Hg | Hg2Cl2 | KCl (1M) E 0 = 0.28 V

Saturated calomel electrode Pt | Hg | Hg2Cl2 | KCl(sat.) E0 = 0.24 V

Silver/Silver chloride, Ag | AgCl | Cl- (1M) E 0 = 0.22 V

(used in pH meters)

Metal-insoluble salt electrodes

M

X-

X-

MXC+

C+

C+

C+

X-

X-

Lecture 26 - 24

2

The ‘saturated calomel electrode’ (SCE) features the reduction half-reaction:

Hg+ + e– Hg

Hg2Cl2 2Hg+ + 2Cl– Overall:

Hg2Cl2 (s) + 2e– 2 Hg (s) + 2Cl– (sat)

Pt | Hg | Hg2Cl2 | KCl ||

Standard cell potential of

E0 = 0.24 V.

Saturated Calomel Electrode

5 M

Lecture 26 - 25

Calomel: Hg2Cl2(s) + 2e- 2Hg(l) + 2Cl-(aq) E0 = 0.24V Zn Zn2+ + 2e-E0 = + 0.76V

(reversed because it is written as an oxidation)

Q: The standard reduction potential of Zn2+/Zn is - 0.76 V. What would be the observed cell potential for the Zn/Zn2+ couple when measured using the SCE as a reference?

Ans: The Zn will be oxidised (lower reduction potential), soE (cell) = 0.76 + 0.24 = 1.00 V respect to the SCE

So to get the oxidation half-reaction E0 using the SCE as cathode, subtract 0.24 V from the volt meter reading.

Cell Potentials 1

0.24 Hg+/Hg

0.0V H2/H+

-0.76 Zn2+/Zn

Lecture 26 - 26

Summary

CONCEPTS Concentration cells Nernst equation

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

Lecture 26 - 27

Ag+ + e– AgAgCl Ag+ + Cl–

Overall: AgCl + e– Ag + Cl– E0 = 0.22 V

AgCl (s) + e– Ag (s) + Cl– (sat)

A thin coating of AgCl is deposited

on the pure metal surface.

Silver electrode

Ag | AgCl | Cl–

Lecture 26 - 28

Cell Potentials 2 A Fe3+/Fe2+ cell with [Fe3+]=[Fe2+] =1 M has a potential of

0.55V respect to the Ag/AgCl electrode (E0= 0.22 V). What is the potential of this electrode with respect to the SHE?

Answer.

The reactions that occur in the two half-cells are:

Fe3+ + e- → Fe2+ at the cathode E = 0.55 V; E0 = ?

Ag + Cl- +e-→ AgCl at the anode E0= 0.22 V

The potential of this electrode with respect to the SHE is the difference of the two electrode potentials:

E = E0 - 0.22 E0Fe3+/Fe2+ = 0.55 + 0.22 = 0.77 V

0.22 Ag+/Ag

0.0V H2/H+

?

Lecture 26 - 29

Measurement of pH

We could construct a concentration cell, using the standard hydrogen electrode (SHE) and a hydrogen electrode:

H2(g, 1 atm) 2H+(aq, unknown) + 2e- anode

2H+ (aq, 1M) + 2e- H2(g, 1 atm) cathode

2H+ (1M) 2H+ (unknown) Ecell = ?

Using Nernst equation:

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

pH 0.0592 ][H ln 22F

RT-

][H

][H ln

nF

RT E E un2

ref

2un0

cell

at 25°C,

unknown

Lecture 26 - 30

pH electrodeThe pH electrode potential is typically measured versus a fixed reference calomel electrode.

E’ is the sum of the constant offset potentials of the inner glass surface/solution, the Ag/AgCl electrode, and the calomel electrode.

Eglass electrode= E’ + RT/2.303F log [H+]

• Based on a thin glass membrane: a modified glass enriched in H+ and resulting in a hydration layer a few micrometers thick.

• Inside the membrane is a 'reference solution' of known [H+] (1M HCl).

• The potential difference relevant to pH measurement builds up across the outside glass/solution interface marked ||.

Lecture 27 - 31

Nernst Equation

]reactants[

]products[

)log(0592.00

Q

Qn

EEcell

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

(aq) + Cu(s)

The Nernst equation describes the effect of concentration on cell potential.

When Q < 1, [reactants] >[products] and the cell can do more work.

When Q = 1, Ecell = E0cell (standard conditions [x] = 1 M).

When Q > 1 , [products] > [reactants] and Ecell is lower.

Lecture 27 - 32

Difference between Q and K

Breakdown of N2O4 to NO2:

N2O4 (g, colourless) → 2 NO2 (g, brown)

][][

42

22

ONNOQ

Q =K

Figure from Silberberg, “Chemistry”,

McGraw Hill, 2006.

Lecture 27 - 33

Potential of an electrochemical cell

]C[

]Z[2

2

u

nQ

0 V

~1037

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

(aq) + Cu(s)

Ece

ll (V

olts

)

Ecell decreases as the reaction proceeds, until at equilibrium Ecell =0 and

.)log(0592.00 Kn

E

K

)Qlog(n

.EEcell

059200= -

Lecture 27 - 34

Recap: Examining Q To K Ratios

If Q/K < 1 Ecell is positive for the reaction as written. The smaller the Q/K ratio,

the greater the value of Ecell and the more electrical work the cell can do.

If Q/K = 1, Ecell = 0. The cell is at equilibrium and can no longer do work.

If Q/K > 1, Ecell is negative for the reaction as written. The cell will operate in

reverse – the reverse reaction will take place and do work until Q/K = 1 at equilibrium.

)log(0592.00 Qn

EEcell

Lecture 27 - 35

Large K products favoured large standard cell potential, E0

Link between E 0 and KQ: What happens if the reaction proceeds until equilibrium is reached?

A: The reaction stops, therefore the voltage, or electrical potential, is zero (the battery is flat). In mathematical terms:

0)log(0592.00 Qn

EEcell

So the equilibrium constant determines the cell potential.

)log(0592.00 Kn

E

At equilibrium Q=K

Lecture 27 - 36

Redox reactions are special

For redox reactions there is a direct experimental method to measure K and ΔG°.

Figure from Silberberg, “Chemistry”,

McGraw Hill, 2006.

Lecture 27 - 37

Relation between E 0 and K

)log(0592.00 Kn

E

K is plotted on a logarithmic scale to give a straight line.

Figure from Silberberg, “Chemistry”,

McGraw Hill, 2006.

Lecture 27 - 38

Example question 2

Co(s) + Ni2+(aq) Co2+(aq) + Ni(s) E 0 = 0.03 V

Q: A voltaic cell consisting of a Ni/Ni2+ half-cell and Co/Co2+ half-cell is constructed with the following initial concentrations: [Ni2+] = 0.80 M; [Co2+]=0.2 M.

a) What is the initial Ecell?b) What is the [Ni2+] when the voltage reaches 0.025 V?c) What are the equilibrium concentrations of the ions?

Given: E 0 Ni2+

/Ni = -0.25 V; E 0 Co2+

/Co = -0.28 V

Ni2+ + 2e- → Ni E 0 = -0.25V

Co → Co2+ + 2e- E 0 = +0.28V

Lecture 27 - 39

Example question 2a

Co(s) + Ni2+(aq) Co2+(aq) + Ni(s) E 0 = 0.03 V

25.08.0

2.0

][

][2

2

Ni

CoQ

)log(0592.00 Qn

EEcell )25.0log(2

0592.003.0

)602.0(0296.003.0 V 048.0

a) What is the initial Ecell?

Lecture 27 - 40

Example question 2b

)log(0592.00 Qn

EEcell )log(0296.003.0025.0 Q

169.00296.0

005.0)log( Q

Q = 1.47

)8.0(

)2.0(

][

][47.1

2

2

x

x

Ni

CoQ

Co(s) + Ni2+(aq) Co2+(aq) + Ni(s)

0.80 - x 0.20+x

xx 2.047.1176.1

976.047.2 x x = 0.40

So when Ecell = 0.025 V

[Co2+] = 0.60 M

[Ni2+] = 0.40 M

b) What is the [Ni2+] when the voltage reaches 0.025V?

Lecture 27 - 41

Example question 2c

)log(0592.0

0 0 Kn

E )log(0296.003.0 K

014.10296.0

03.0)log( K

K = 10.24Co(s) + Ni2+(aq) Co2+(aq) + Ni(s)

0.80-x 0.20+x

)8.0(

)2.0(

][

][24.10

2

2

x

x

Ni

CoK

xx 2.024.10192.8

986.724.11 x x = 0.71

So at equilibrium,

[Co2+] = 0.91 M

[Ni2+] = 0.09 M

c) What are the equilibrium concentrations of the ions?

Lecture 27 - 42

Concentration Cells in Nature

Concentration cells are present 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

Lecture 27 - 43

Nerve cells

The membrane potential is more positive outside than inside the cell. On nerve stimulation, Na+ enters cell, the inside cell membrane

becomes > +ive, then K+ ions leave cell to re-equilibrate the outside. These rapid (ms) changes in charge across the membrane stimulate

the neighbouring region and the electrical impulse moves down the length of the cell.

Energy from ATP hydrolysis is used by ion pumps, so that across the nerve cell membrane concentration gradients are maintained:

Ion Concentration Gradient: Inside Outside

K+ High LowNa+ Low High

Figure from Silberberg, “Chemistry”,

McGraw Hill, 2006.

Lecture 27 - 44

Nerve cells Nernst Equation gives the membrane potential generated by the

differing extracellular versus intracellular concentrations of each ion:

inside

outsideion

inside

outsideion

ion

ionE

ion

ion

nF

RTE

][

][log5.61

][

][log

303.2

Substituting n = 1, T = 37oC:

( in mV)

1 mV = 10-3 V

Consider K+: [K+]outside = 3 mM, [K+]inside = 135 mM

102

65.15.61

135

3log5.61

][

][log5.61

K

inside

outsideK

E

K

KE

mV

mV

(Eo = 0)

Lecture 27 - 45

Cellular Electrochemistry Biological cells apply the principles

of electrochemical cells to generate energy in a complex multistep process.

Bond energy in food is used to generate an electrochemical potential.

The potential is used to create the bond energy of the high-energy molecule adenosine triphosphate (ATP) (energy currency for the cell).

N

NN

N

NH2

O

OHOH

HH

HH

OPO

O-

O

POP-O

O- O-

OO

adenosinetriphosphate

N

NN

N

NH2

O

OHOH

HH

HH

OPO

O-

O

P-O

O-

O

adenosinediphosphate

ATP

ADP

H2O

+ HPO42- +H+

ATP4- + H2O → ADP3- + HPO42- + H+

ΔG °’ = -30.5 kJ mol-1

ΔGo’ (solution at pH 7 and at human body temperature 37oC.)

Lecture 27 - 46

Bond Energy to Electrochemical Potential

Inside mitochondria, redox reactions are performed by a series of proteins that form the electron-transport chain (ETC) which contain redox couples, such as Fe2+/Fe3+.

Large potential differences provide enough energy to convert ADP into ATP.

Fig

ure

from

Silb

erb

erg

, “Ch

em

istry”,

McG

raw

Hill, 2

00

6.

Lecture 27 - 47

In nature, the most important reducing agent is a complex molecule named nicotinamide adenine dinucleotide, abbreviated NADH, which functions as a hydride donor (H-).

= NADH = biological reducing

agent

NAD+ = biological oxidising agent

Bond Energy to Electrochemical Potential

Lecture 27 - 48

ETC consists of three main steps, each of which has a high enough potential difference to produce one ATP molecule.

The reaction that ultimately powers ETC is the reduction of oxygen in the presence

of NADH:

2H+ + 2e- + ½ O2 → H2O Eo’ = +0.82

NADH + H+ → NAD+ + 2H+ + 2e- Eo’ = -0.32

NADH(aq) + H+(aq) + ½O2(aq) → NAD+

(aq) + H2O(l) Eo’overall = 1.14 V

Substantial energy release!

ΔGo’ = -nFEo’ = -2 · 96485 C mol-1· 1.14 V = - 220 kJ mol-1

Electron Transport Chain (ETC)

Lecture 27 - 49

ATP Synthesis

In short: e- are transported along the chain, while protons are forced into the intermembrane space.

This creates a H+ concentration cell across the membrane.

In this step, the cell acts as an electrolytic cell, i.e. uses a spontaneous process to drive a non-spontaneous process.

When [H+]intermembrane/[H+]matrix ~ 2.5 a trigger allows protons to flow back across membrane, and ATP is formed.

It’s not simple: Noble prize in 1997 to Boyer and Walker for elucidating this.

Figure from Silberberg, “Chemistry”,

McGraw Hill, 2006.

Lecture 27 - 50

SummaryCONCEPTS

Concentration cells Nernst equation E 0 and K Link between E , Q and K Applications of concentration cells

CALCULATIONS Work out cell potential from reduction potentials; Work out cell potential for any concentration (Nernst equation) Work out K from E 0

Work out pH from concentration cell

Lecture 26 - 51

Pop Quiz 1

Balance the following reaction in basic solution:

MnO4- + CN- MnO2 + CNO-

Answer: H2O + 2 MnO4- + 3 CN- --> 2 MnO2 + 3 CNO- + 2 OH-

Lecture 26 - 52

Pop Quiz 2

Balance the following reaction in basic solutions:

NO3- + Zn  Zn2+ + NH3

Answer: NO3- + 4 Zn + 6 H2O →  4 Zn2+ + NH3 + 9 OH-