chapter 11: phenomena · chapter 11: electrochemistry redox reaction review assigning oxidation...

Chapter 11: Electrochemistry

Chapter 11: Phenomena

Phenomena: Two electrochemical cells were constructed. Experiments were

done by changing mass, concentration, and/or temperature. Examine the data

to determine patterns in how these variables affect the voltage of the cell.

Half Cell 1:

Zn(s) & Zn2+(aq)

Half Cell 2:

Cu(s) & Cu2+(aq)

Cell Reaction: Zn(s)+Cu2+(aq)

Zn2+(aq)+Cu(s)

Exp.Mass

Zn(s)[Zn2+]

Volume

Zn2+

Mass

Cu(s)[Cu2+]

Volume

Cu2+ Temp Voltage

1 1.0 g 1.0 M 0.50 L 1.0 g 1.0 M 0.50 L 298 K 1.10 V

2 1.0 g 1.0 M 0.50 L 1.0 g 5.0 M 0.50 L 298 K 1.12 V

3 2.0 g 1.0 M 0.50 L 1.0 g 1.0 M 0.50 L 298 K 1.10 V

4 2.0 g 1.0 M 0.50 L 3.4 g 1.0 M 0.50 L 298 K 1.10 V

5 5.0 g 1.0 M 0.50 L 2.5 g 1.0 M 0.50 L 415 K 1.07 V

6 1.0 g 0.2 M 0.50 L 1.0 g 1.0 M 0.50 L 298 K 1.12 V

7 3.4 g 5.0 M 0.50 L 1.0 g 1.0 M 0.50 L 415 K 1.05 V

Half Cell 1:

Ca(s) & Ca2+(aq)

Half Cell 2:

Ag(s) & Ag+(aq)

Cell Reaction: Ca(s)+2Ag+(aq)

Ca2+(aq)+2Ag(s)

Exp.Mass

Ca(s)[Ca2+]

Volume

Ca2+

Mass

Ag(s)[Ag+]

Volume

Ag+ Temp Voltage

1 1.0 g 1.0 M 0.50 L 1.0 g 1.0 M 0.50 L 298 K 3.67 V

2 1.0 g 1.0 M 0.50 L 1.0 g 5.0 M 0.50 L 298 K 3.71 V

3 1.0 g 1.0 M 1.00 L 1.0 g 1.0 M 0.50 L 298 K 3.67 V

4 1.0 g 1.0 M 1.00 L 1.0 g 1.0 M 1.70 L 298 K 3.67 V

5 1.0 g 1.0 M 2.50 L 1.0 g 1.0 M 2.50 L 415 K 3.57 V

6 1.0 g 0.2 M 0.50 L 1.0 g 1.0 M 0.50 L 298 K 3.69 V

7 1.0 g 5.0 M 1.70 L 1.0 g 1.0 M 0.50 L 415 K 3.55 V

Chapter 11 Electrochemistry

o Redox Reaction Review

o Galvanic Cells

o Thermo of

Electrochemistry

o Nernst Equation

o Batteries / Fuel Cells

o Electrolytic Cells

2

Big Idea: Electron transfer in a chemical reaction is

both material and

concentration

specific. If the process

is spontaneous the

transfer of electrons

can be used to

produce a current and drive electrical devices.


Redox Reaction Review

Assigning Oxidation Numbers

The oxidation number (ON) of an element uncombined with

another element is zero: Na(s) ON = 0, and H2(g) ON = 0.

For monatomic ions, the charge is the ON: Na+ ON = +1.

The ONs of elements in group 1 equal 1 (ex. Lithium ON = +1)

ONs of elements in group 2 equal 2 (ex. Magnesium ON = +2),

when the atoms are in a compound.

The ON of fluorine is always -1 in compounds.

The ON of the other elements in group 7 usually equal -1

when the atoms are in a compound.

The ON of oxygen is usually -2 in compounds. Exceptions are

fluorine compounds and peroxide (a compound that

contains an O-O single bond).

Hydrogen's ON is +1 when combined with non metals and -1

when combined with metals.

The sum of the ON’s of all the atoms in a species is equal to its

total charge.

3



Assigning Oxidation Numbers

4

NaCl

Element Oxidation Number Reason

Na

Cl

Fe2(SO4)3


SO4

O

S

Fe

Br2


Br



Determining Which Element is Oxidized

and Which is Reduced

Step 1: Assign oxidation numbers.

Step 2: Use oxidation numbers and ‘OIL RIG’ to

identify which element is oxidized and which is

reduced.

5

Note: If the question asks which “substance is oxidized;” instead of giving the

element that is oxidized give the entire compound.



Oxidizing Agent: A species that removes

electrons from a species being oxidized in

a redox reaction.

Reducing Agent: The species that supplies

electrons to a substance being reduced

in a redox reaction.

6

Note: Oxidizing agent is the species being reduced.

Note: Reducing agent is the species being oxidized.



Balancing Redox Reactions in Acidic Conditions

Step 1: Write unbalanced half reactions.

Step 2: Balance half reactions except for O and

H.

Step 3: Balance O by using H2O.

Step 4: Balance H by using H+.

Step 5: Balance electrons in each half reaction.

Step 6: Multiply half reactions by an integer so

that number of electrons match, then add the

reactions together.

7



Balancing Redox Reactions in Basic Conditions

Step 1: Balance the reaction as if it were in

acidic conditions.

Step 2: Determine the number of H+ in the

balanced equation.

Step 3: Add the same number OH- as there are

H+ to BOTH sided of the equation.

Step 4: The H+ and OH- on one side of the

reaction will combine and form H2O.

Step 5: Simplify your reaction (combine waters) if

necessary.

8


Galvanic Cell

Electrochemical Cell: Device in which an electric current

is produced by either a spontaneous reaction or is used to

bring about a non spontaneous reaction.

Galvanic Cell: An electrochemical cell in which a spontaneous chemical reaction is used to generate an

electrical current.

9

Electrode: Metal contacts.

Electrolyte: Ionically

conducting medium.

Salt Bridge: Allows for the flow

of ions but prohibits reactions

from taking place.

Anode: Is where oxidation

occurs.

Cathode: Is where reduction

occurs.


Galvanic Cell

Short Hand Cell Notation

1) The anode is written on the left and cathode on the right.

2) Phase interfaces are separated with a ‘|’

3) Same phase species are separated by a ‘,’

4) For same phases, the species that is capable of being the

oxidizing agent is written 1st followed by the species that is

capable of being the reducing agent.

Example:

Fe3+(aq) + e- Fe2+(aq) Fe3+ is the oxidizing agent

(anode) Fe3+(aq), Fe2+(aq)||

(cathode) ||Fe3+(aq), Fe2+(aq)

5) The salt bridge is represented by ||

6) Never include water

7) Anode order of phase: solid | gas | liquid | aqueous

8) Cathode order of phase: aqueous | liquid | gas | solid

9) An inert electrode always goes on the outside (usually made

up of Pt(s) or C(gr))

10


Student Question

Galvanic Cell

Write the balanced reaction for the given cell Pt(s)|H2(g)|H+(aq)||Co3+(aq),Co2+(aq)|Pt(s)

a) H2(g)+Co3+(aq) 2H+(aq)+Co2+(aq)

b) Pt(s)+H2(g)+2Co3+(aq)2H+(aq)+2Co2+(aq)+Pt(s)

c) 2H+(aq)+Co2+(aq)H2(g)+Co3+(aq)

d) H+(aq)+Co2++e-(aq) H2(g)+Co3+(aq)

e) None of the above

11


Thermo of Electrochemistry

12

Standard Reaction Potentials at 298 K

Half -Reaction E°(V)

F2 + 2e- 2F- 2.87

Ce4+ + e- Ce3+ 1.70

MnO4- + 4H+ + 3e-

MnO2 + 2H2O 1.68

IO4- + 2H+ + 2e-

IO3- + H2O 1.60

MnO4- + 8H+ + 5e-

Mn2+ + 4H2O 1.51

Au3+ + 3e- Au 1.50

Cl2 + 2e- 2Cl- 1.36

Cr2O72- + 14H+ + 6e-

2Cr3+ + 7H2O 1.33

O2 + 4H+ + 4e- 2H2O 1.23

MnO2 + 4H+ + 2e- Mn2+ + 2H2O 1.21

IO3- + 6H+ + 5e-

½I2 +3H2O 1.20

Br2 + 2e-2Br- 1.09

VO2+ + 2H+ + e-

VO2+ + H2O 1.00

AuCl4- + 3e-

Au + 4Cl- 0.99

NO3- + 4H+ + 3e-

NO + 2H2O 0.96

ClO2 + e- ClO2

- 0.95

2Hg2+ + 2e- Hg2

2+ 0.91

Ag+ + e- Ag 0.80

Hg22+ + 2e-

2Hg 0.80

Fe3+ + e- Fe2+ 0.77

MnO4- + e-

MnO42- 0.56

I2 + 2e- 2I- 0.54

Cu+ + e- Cu 0.52

Cu2+ + 2e- Cu 0.34

Standard Reaction Potentials at 298 K

Half –Reaction E°(V)

Hg2Cl2 + 2e- 2Hg + 2Cl- 0.27

AgCl + e- Ag + Cl- 0.22

SO42- + 4H+ + 2e-

H2SO3 + H2O 0.20

Cu2+ +e- Cu+ 0.16

2H+ + 2e- H2 0.00

Fe3+ + 3e- Fe -0.04

Pb2+ + 2e- Pb -0.13

Sn2+ + 2e Sn -0.14

Ni2+ + 2e- Ni -0.23

PbSO4 + 2e- Pb + SO4

2- -0.35

Cd2+ +2e- Cd -0.40

Fe2+ + 2e- Fe -0.44

Cr3+ + e- Cr2+ -0.50

Cr3+ + 3e- Cr -0.73

Zn2+ + 2e- Zn -0.76

2H2O + 2e- H2 + 2OH- -0.83

Mn2+ + 2e- Mn -1.18

Al3+ + 3e- Al -1.66

H2 + 2e- 2H- -2.23

Mg2+ + 2e- Mg -2.37

La3+ + 3e- La -2.37

Na+ + e- Na -2.71

K+ +e- K -2.92

Li+ + e- Li -3.05



All reaction of referenced to:

H+(aq) + 2e- H2(g) E˚ = 0 V

Things to remember:

If you flip a reaction you change the sign of E˚

If you multiply a reaction by a constant DO NOT DO

ANYTHING TO E˚

13

Example:

F2 + 2e- 2F- E˚= 2.87

2F-F2 + 2e- E˚= -2.87

Example:

F2 + 2e- 2F- E˚= 2.87

2F2 + 4e- 4F- E˚= 2.87

Note: The book gives you the following equation:

𝐸𝑐𝑒𝑙𝑙° = 𝐸° 𝑐𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸° 𝑎𝑛𝑜𝑑𝑒

DO NOT USE THIS EQUATION IF YOU ARE USING THE FLIPPING METHOD


Student Question


Calculate E° of the following cell

Cr(s)|Cr3+(aq)||Br-(aq)|Br2(l)|Pt(s) Helpful Information: Cr3+ + 3e-

Cr E° = -0.73 V

Br2 + 2e- 2Br- E° = 1.09 V

a) 2.55 V

b) 1.82 V

c) -1.82 V

d) -2.55 V


Bonus: Determine the balanced reaction for this cell.

14



Calculate the solubility product of

Hg2Cl2 at 298.15 K.

Solubility Product Ksp:

Reactions of Interest:

Hg2Cl2 + 2e- 2Hg + 2Cl- E˚= 0.27 V

Hg22+ + 2e- 2Hg E˚= 0.79 V

15


Student Question


The equilibrium constant for the following

general reaction is 10. at 25°C. Calculate E° for

the cell.X2(s) + Y+(aq) X2+(aq) + Y(s)

(unbalanced)

a) 0.015V

b) 0.030 V

c) 0.060 V

d) 0.045 V


16


Student Question

Nernst Equation

Calculate the emf, at 25ºC, of the cell

Zn(s)|Zn2+(aq,1.5 M)||Fe3+(aq, 0.0010 M)|Fe(s)

Helpful Information: Fe3+ + 3e- Fe(s) Eo = -0.04 V

Zn2+ + 2e- Zn(s) Eo = -0.76 V

a) 0.72V

b) 0.66 V

c) 2.1 V

d) 0.69 V


17


Batteries/ Fuel Cells

Primary Cells: Galvanic cell with the reactants

sealed inside at the manufacturer. They can not

be recharged.

18

Type: Primary Cell/Dry Cell

Anode Electrode: Zn

Cathode Electrode: CCathode Material: MnO2,

Electrolyte: Carbon black, and NH4Cl

Voltage :1.5 V

(A) Zn Zn2+ + 2e-

(C) 2MnO2 + 2NH4+ + 2e-

Mn2O3 + 2NH3 + H2O

Zn+2NH4++2MnO2 Zn2++Mn2O3+2NH3+ H2O

Zinc Carbon



Secondary Cell: Galvanic cell that must be

charged before it can be used.

19

Type: Secondary Cell

Anode Electrode: Pb

Cathode Electrode: Pbcovered with PbO2

Electrolyte: H2SO4

Voltage : 12 V

(A) Pb + HSO4– PbSO4 + H+ + 2e–

(C) PbO2 + 3H+ + 2HSO4– + 2e– PbSO4 + 2H2O

Pb + PbO2 + 2H+ + 2HSO4– 2PbSO4 + 2H2O

Lead Acid



Fuel Cell: A Galvanic cell in which the reactants

are continuously supplied.

20

Type: Fuel Cell

Anode Electrode: Platinum

Cathode Electrode: Platinum

Electrolyte: KOH

Voltage : 1.2 V

(A) 2H2 +4OH- 4H2O + 4e-

(C) 4e- + O2 +2H2O 4OH-

2H2 + O2 2H2O

Alkaline


Electrolytic Cell

General Difference between Galvanic and

Electrolytic Cells

Electroplating: The deposition of a thin film of

metal on an object.

21

Galvanic Cell Electrolytic Cell

E > 0 E < 0

No external power External power

2 Compartments 1 Compartment

2 Electrolytes 1 Electrolyte

Room temp. & press. High temp. & press.

Note: Metal or graphite coated plastic is placed at the cathode.

Note: An aqueous solution of the salt of the plating material is the electrolyte.


Electrolytic Cell

How much voltage must be supply to plate Zinc

onto copper?

Zn2+(aq) + 2e- Zn(s) E˚ = -0.76 V

Cu2+(aq) + 2e- Cu(s) E˚ = 0.34 V

Overpotential: Additional voltage over the emf

required to run a reaction.

22


Electrolytic Cell

Faraday’s law of

Electrolysis: The

amount of product

formed or reactant

consumed by an

electrical current is

stoichiometrically

equivalent to the

amount of electrons

supplied.

23


Student Question

Electrolytic Cell

What time is required to plate a metal tray

(24.0 cm×12.0 cm) with a coating (thickness 0.00200 cm) of silver (density =10.54 𝑔

𝑐𝑚3) using a

current 7.65 A? Neglect the amount of silver

required to coat the edges.

Helpful Information: The reaction of interest is

Ag+ + e- Ag, and 𝑀𝐴𝑔 = 107.9 𝑔

𝑚𝑜𝑙

a) 1420 s b) 912 s

c) 1130 s d) 708 s


24


Take Away From Chapter 11

Big Idea: Electron transfer in a chemical reaction

is both material and concentration specific. If

the process is spontaneous the transfer of

electrons can be used to produce a current and

drive electrical devices.


Be able to assign oxidation numbers

Be able to identify element oxidized/reduced

OIL RIG

Be able to identify oxidizing/reducing agent

Be able to balance redox equations (Ch. 4 - 83)

Acidic conditions (no OH-)

Basic conditions (no H+)

25

Numbers correspond to end of chapter questions.



Galvanic Cell

Know that a galvanic cell is a spontaneous reaction (3)

Be able to draw a galvanic cell (19&21)

Identify anode

Identify cathode

Identify electron flow

Identify ion migration through salt bridge


Know that for galvanic cell Ecell is positive

Be able to calculate E°cell for a system (22&24)

Be able to select the best oxidizing/reducing agent from a list

of standard potentials (25,26,27,28,29,&30)

Be able to calculate ΔG, wmax , and K of a cell (42,43,44,45,

47,&100)

∆𝐺 = −𝑛𝐹𝐸 = 𝑤𝑚𝑎𝑥

𝐸° =𝑅𝑇

𝑛𝐹ln(𝐾)

26




Nernst Equation

Be able to calculate E at conditions other than the standard

state (10,55,56,59,65,&98)

𝐸 = 𝐸° −𝑅𝑇

𝑛𝐹ln(𝑄)

Batteries / Fuel Cell

Electrolytic Cell

Know that an electrolytic cell is a non spontaneous reaction

Be able to calculate the products of electrolysis.

(72,73,74,75,&84)

𝑛 =𝐼𝑡

𝐹

Be able to determine when competitive reactions (water) will

effect electrolytic cell (81)

27


chapter 11: phenomena · chapter 11: electrochemistry redox reaction review assigning oxidation...

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