CHEM1612 - Pharmacy Week 9: Concentration Cells

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CHEM1612 - Pharmacy Week 9: Concentration Cells Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196 E-mail: [email protected]

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CHEM1612 - Pharmacy Week 9: Concentration Cells. Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196 E-mail: [email protected]. Unless otherwise stated, all images in this file have been reproduced from: - PowerPoint PPT Presentation

Transcript of CHEM1612 - Pharmacy Week 9: Concentration Cells

Page 1: CHEM1612 - Pharmacy Week 9: Concentration Cells

CHEM1612 - PharmacyWeek 9: Concentration CellsDr. Siegbert Schmid

School of Chemistry, Rm 223Phone: 9351 4196E-mail: [email protected]

Page 2: CHEM1612 - Pharmacy Week 9: Concentration Cells

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

Page 3: CHEM1612 - Pharmacy Week 9: Concentration Cells

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

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Lecture 26 - 4

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–

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Lecture 26 - 5

Pop Quiz 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 respect to the SHE?

Answer.The reaction that occurs 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 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+

?

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Lecture 26 - 6

Measurement of pHWe 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) cathode2H+ (1M) 2H+ (unknown) Ecell = ?

unknown

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Lecture 26 - 7

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 ||.

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Lecture 27 - 8

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.

Page 9: CHEM1612 - Pharmacy Week 9: Concentration Cells

Lecture 27 - 9

Potential of an electrochemical cell

]C[]Z[

2

2

unQ

0 V

~1037

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

(aq) + Cu(s)

Ece

ll (Vo

lts)

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

.)log(0592.00 Kn

E

K

)Qlog(n.

EEcell059200= -

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Lecture 27 - 10

Recap: Examining Q To K RatiosIf 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

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Lecture 27 - 11

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

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Lecture 27 - 12

Redox reactions are special

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

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Lecture 27 - 13

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.

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Lecture 27 - 14

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 Ni

2+/Ni = -0.25 V; E 0 Co

2+/Co = -0.28 V

Ni2+ + 2e- → Ni E 0 = -0.25V

Co → Co2+ + 2e- E 0 = +0.28V

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Lecture 27 - 15

Example question 2a

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

a) What is the initial Ecell?

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Lecture 27 - 16

Example question 2b

)log(0592.00 Qn

EEcell )log(0296.003.0025.0 Q

169.00296.0005.0)log( Q

Q = 1.47Co(s) + Ni2+(aq) Co2+(aq) + Ni(s)

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

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Lecture 27 - 17

Example question 2c

)log(0592.00 0 Kn

E )log(0296.003.0 K

014.10296.003.0)log( K

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

0.80-x 0.20+x

c) What are the equilibrium concentrations of the ions?

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Lecture 27 - 18

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

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Lecture 27 - 19

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.

Page 20: CHEM1612 - Pharmacy Week 9: Concentration Cells

Lecture 27 - 20

Nerve cells Nernst Equation gives the membrane potential generated by the

differing extracellular versus intracellular concentrations of each ion:

inside

outsideion

inside

outsideion

ionionE

ionion

nFRTE

][][log5.61

][][log303.2

Substituting n = 1, T = 37oC:

( in mV)

1 mV = 10-3 V

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Lecture 27 - 21

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

HHHH

OPO

O-

O

POP-O

O- O-

OO

adenosinetriphosphate

N

NN

N

NH2

O

OHOH

HHHH

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.)

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Lecture 27 - 22

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.

Figure from S

ilberberg, “Chem

istry”,

McG

raw H

ill, 2006.

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Lecture 27 - 23

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

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Lecture 27 - 24

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)

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Lecture 27 - 25

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.

Page 26: CHEM1612 - Pharmacy Week 9: Concentration Cells

Lecture 27 - 26

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