galvanic nernst

8/3/2019 Galvanic Nernst

1/19

ElectrochemistryThe study of the interchange

of chemical and electrical energy.

Sample electrochemical processes:

1) Corrosion

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

2) Biological processes

C6H12O6 + 6 O2 6 CO2 + 6 H2O

3) Batteries (Galvanic or Voltaic cells) Electrochemical cells that produce a current (flow of electrons)

as a result of a redox reaction

4) Electrolytic cells Electrical energy is used to produce chemical change

Used to prepare or purify metals (such as sodium, aluminum,


2/19

Chemical Change Electron Flow Copper: Cu(s) , Cu

2+(aq)

Cu(s) Cu2+

(aq) + 2e-

Grxn = Gf(Cu2+ ) = 65.6 kJ

Silver: Ag(s) , Ag+

(aq)

Ag(s) Ag+

(aq) + e-

Grxn = Gf(Ag+) = 77.2 kJ

Cu(s) Cu2+ (aq) + 2e- G = +65.6 kJ

Ag+(aq)

+ e- Ag(s)

G = -77.2 kJ

Cu(s) + 2 Ag+

(aq) Cu2+ (aq) + 2 Ag(s) G = -88.8 kJSpontaneous

wmax = -88.8 kJ

Cu2+ in

solution

2( ) 2( )

Ag(s)


3/19

Harnessing the Energy

Separate the half-reactions

Creates a galvanic orvoltaic cell

Cu Ag

1 M CuSO4

Cu(s) Cu2+

(aq) + 2e-

1 M AgNO3

Ag+

(aq) + e-

Ag(s)

Luigi Galvani AlessandroVolta

Cu2+

SO42-

Ag+

NO3-

KNO3(aq)

K+NO3-

e-

salt bridge

Oxidation Reduction

Anode Cathode

cathode and

reduction

begin with

consonants

anode and

oxidation

begin with

vowels

(produces electrons)

+(attracts electrons)

Red Cat


4/19

Line Notation for Galvanic Cells

Cu Ag

1 M CuSO4

Cu(s) Cu2+

(aq) + 2e-

1 M AgNO3

Ag+(aq) + e- Ag(s)

Cu2+

SO42-

Ag+

NO3-

K+NO3-

e-

Oxidation Reduction

Anode() Cathode(+)

Anode always on the left

Cu(s) Cu2+ (1 M)Ag+ (1 M)Ag(s)Cathode always on the right


5/19

Chemical Change Electrical Work Chemical change produces electrical energy

Electrical energy can be used to do work!

G = wmax

Electrical work: w = -nFn = # of moles e- transferred

F = charge on a mole of e-

= electrical potential

(electromotive force)

Cell Potential () or Electromotive Force (emf): The drivingforce pushing the electrons from the anode to the cathode.

Units = Volts 1 Volt = 1 joule/coulomb


6/19

Standard Reduction Potentials

The cell potential cell can be determined from

the standardreduction potentials (red) for

the half-reactions:

Reduction potential = tendency for reduction tohappen

Positive redspontaneous reduction reaction

Negative red non-spontaneous reduction orspontaneous oxidation (reverse reaction)

Standard (o) = standard conditions (1 M solutions,

1 atm gases)


7/19


Half-Reaction (V)

F2 + 2e- 2F- 2.87

Au3+ + 3 e- Au 1.50

Ag+ + e- Ag 0.80

Cu2+

+ 2e-

Cu 0.342H+ + 2e- H2 0.00

Ni2+ + 2e- Ni -0.23

Zn2+ + 2e- Zn -0.76

Al3+ + 3e- Al -1.66

Li+ + e- Li -3.05

= 0 (SHE)Standard Hydrogen Electrode

> 0Spontaneous reduction

< 0Non-Spontaneous reduction

Reduction potential = tendency for reduction to happen

Standard = standard conditions (1 M solutions, 1 atm gases)

Spontaneous oxidation

(reverse rxn)


8/19


Half-Reaction (V)

F2 + 2e- 2F- 2.87

Au3+ + 3 e- Au 1.50

Ag+ + e- Ag 0.80

Cu2+

+ 2e-

Cu 0.342H+ + 2e- H2 0.00

Ni Ni2+ + 2e- +0.23

Zn Zn2+ + 2e- +0.76

Al Al3+ + 3e- +1.66

Li Li+ + e- +3.05

> 0Spontaneous reduction

But remember, an oxidation CANNOT happen without a

reduction

Spontaneous oxidation


9/19


Half-Reaction (V)

F2 + 2e- 2F- 2.87

Au3+ + 3 e- Au 1.50

Ag+ + e- Ag 0.80

Cu2+

+ 2e-

Cu 0.342H+ + 2e- H2 0.00

Ni2+ + 2e- Ni -0.23

Zn2+ + 2e- Zn -0.76

Al3+ + 3e- Al -1.66

Li+ + e- Li -3.05

Strongest Oxidizing Agent(most easily reduced)

Reduction potential = tendency for reduction to happen Standard = standard conditions 1 M solutions 1 atm ases

Strongest Reducing Agent(most easily oxidized)


10/19

Cell Potential

cell = reduction + oxidation

Ag+(aq) + e-Ag(s) = 0.80 V

Cu2+ (aq) + 2 e-Cu(s) = 0.34 V

Reduction reaction: 2(Ag+(aq) + e-Ag(s)) = +0.80 V

Oxidation reaction: Cu(s)Cu2+

(aq) + 2 e-

= - 0.34 VCu(s) + 2 Ag

+(aq) Cu2+ (aq) + 2 Ag(s) cell = +0.46 V

is intensive, unlike Go

Cu(s) Cu2+ (1 M)Ag

+ (1 M)Ag(s)

The cell MUST be + and thus spontaneous for Galvanic cells


11/19

Free Energy and Cell Potential

G= wmax = nF

n = number of moles of electrons transferred

F = Faradays constant

= 96,485 coulombs per mole of electrons (C/mol e-)

= standard cell potential (V or J/C)

Michael Faraday

Cu(s) Cu2+ (1 M)Ag+ (1 M)Ag(s)

cell = +0.46 V

G = -nFcell

G = -(2 mol e-)(96485 C/mole-)(0.46 V)

G = -88,800 J or-88.8 kJ
http://scienceworld.wolfram.com/biography/photo-credits.html


12/19

Practice Time

Given the following information, draw a galvanic

cell.

Fe(s)Fe2+ (1 M)Au3+ (1 M)Au(s)

Be sure to include the following:

Anode/Cathode reactions

Balanced overall reaction

Complete circuit (external wire with e- flow

direction, salt bridge)

Label all parts of the cell (solution, electrode,

etc.)


13/19

Fe(s)Fe2+ (1 M)Au3+ (1 M)Au(s)

Fe Au

1 M Fe2+

Fe(s) Fe2+ (aq) + 2e-

1 M Au3+

Au3+ (aq) + 3e- Au(s)

Fe2+ Au3+

anions

e-

Oxidation Reduction

Anode Cathode

3Fe(s) + 2Au3+ (aq) 3Fe2+ (aq) + 2Au(s)

cations

cell = +0.440V (Fe rxn) + 1.50 V (Au rxn) = 1.94 V


14/19

Reaction Quotient

The reaction quotient (Q) sets up a ratio of

products and reactants

For a reaction, A + 2B 3C + 4D

[C]3[D]4

[A]1[B]2

Only concentrations (aq) or pressures (g) are

used to solve for QSolids (s) and liquids (l) are not included in the

expression

Q =


15/19

Reaction Quotient practice

Write the Q expression for the following

reaction

CH4(g) + O2(g) CO2(g) + H2O(g)

Reaction must be balanced firstCH4(g) + 2O2(g) CO2(g) + 2H2O(g)

(CO2)(H2O)2

(CH4)(O2)2

Q =


16/19

Reaction Quotient practice

Write the Q expression for the following reaction

Cu(s) + 2Ag+(aq) Cu2+ (aq) + 2Ag(s)

(Cu2+ )(Ag)2

(Cu)(Ag+)2

Is this correct?

NO: Solids arent included in the equation!

(Cu2+ )(Ag+)2

Q =

Q =


17/19

Non-standard conditions:

The Nernst Equation

We can calculate the potential of a cell in which some orall of the components are not in their standard states

(not 1 M concentration or 1 atm pressure).

G = G + RT lnQ

G = -nF G = -nF

-nF = -nF + RT lnQ

Walther Nernst

= - lnQnF

RT R= 8.3415 J/mol KT = temperature

n = moles of e-

F = Faradays constant

96,485 C/mol e-
http://scienceworld.wolfram.com/biography/photo-credits.htmlhttp://scienceworld.wolfram.com/biography/photo-credits.html


18/19

Practice with the Nernst Equation What will be the cell potential of a Cu/Ag cell using

0.10 M Cu

2+

and 1.0 M Ag

+

solutions at 25C?

Cu(s) Cu2+ (aq) + 2e- Ag+(aq) + e- Ag(s)

Cu Ag

Cu2+

SO42-

Ag+

NO3-

Cu(s) + 2 Ag+

(aq)

Cu2+

(aq) + 2 Ag(s)

= 0.46 V (-0.03 V)

Cu(s) Cu2+ (0.10 M)Ag+ (1.0 M)Ag(s)

ln(0.10))(2)(96485K))(298(8.314V0.46

molC

Kmol

J

=

= 0.49 V

2

2

][Ag

][Cu

Q +

+

=

= - lnQnF

RT


19/19

Brain Warmup

Half-Reaction (V)

Ag+ + e- Ag 0.80Cu2+ + 2e- Cu 0.34

Zn2+ + 2e- Zn -0.76

Al3++ 3e- Al -1.66

What is for each of thefollowing reactions?

Which reaction(s) are

spontaneous?

3 Ag+(aq)

+ Al(s)

3 Ag(s)

+ Al3+(aq)

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

2 Al3+ (aq) + 3 Zn(s) 2 Al(s) + 3 Zn2+ (aq)

2.46 V

1.10 V

-0.90 V

Spontaneous?

Y

Y

N

Zn can reduce Cu2+ but not Al3+

galvanic nernst

Documents