Gián án Chapter 18 Electrochemistry

Redox Reactionall single displacement, and combustion, some synthesis and decomposition split reaction into oxidation half-reaction and a reduction half-reaction half-reactions inclu

Chapter 18

Electrochemistry

Chemistry, Julia Burdge, 2st Ed.

McGraw Hill.

Redox Reaction

all single displacement, and combustion,

some synthesis and decomposition

split reaction into oxidation half-reaction and a

reduction half-reaction

half-reactions include electrons

• oxidizing agent is reactant molecule that causes oxidation

contains element reduced

• reducing agent is reactant molecule that causes reduction

contains the element oxidized

Oxidation & Reduction

• oxidation is the process that occurs when

oxidation number of an element increases

element loses electrons

compound adds oxygen

compound loses hydrogen

half-reaction has electrons as products

• reduction is the process that occurs when

oxidation number of an element decreases

element gains electrons

compound loses oxygen

Rules for Assigning Oxidation States

• rules are in order of priority

1 free elements have an oxidation state = 0

Rules for Assigning Oxidation States

polyatomic ion equals the charge on the ion

Rules for Assigning Oxidation States

5 in their compounds, nonmetals have oxidation

states according to the table below

 nonmetals higher on the table take priority

Nonmetal Oxidation State Example

Oxidation and Reduction

• oxidation occurs when an atom’s oxidation state increases during a reaction

• reduction occurs when an atom’s oxidation state decreases during a reaction

CH4 + 2 O2 → CO2 + 2 H2O

-4 +1 0 +4 –2 +1 -2

 if an atom loses electrons another atom must take them

 the reducing agent contains the element that is oxidized

 the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)

Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2 is the oxidizing agent

Identify the Oxidizing and Reducing Agents

in Each of the Following

3 H2S + 2 NO3– + 2 H+ → 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr → MnBr2 + Br2 + 2 H2O

Identify the Oxidizing and Reducing Agents

in Each of the Following

Common Oxidizing Agents

Oxidizing Agent Product when Reduced

HNO 3 (conc) or NO 3-1 NO 2 , or NO, or N 2 O, or N 2 , or NH 3

MnO 4-1 (base) MnO 2

MnO -1 (acid) Mn+2

Common Reducing Agents

Reducing Agent Product when Oxidized

Balancing Redox Reactions 1) assign oxidation numbers

a) determine element oxidized and element reduced

2) write ox & red half-reactions, including electrons

a) ox electrons on right, red electrons on left of arrow

3) balance half-reactions by mass

a) first balance elements other than H and O

b) add H 2 O where need O

c) add H +1 where need H

d) neutralize H + with OH - in base

4) balance half-reactions by charge

a) balance charge by adjusting electrons

5) balance electrons between half-reactions

Ex 18.3 – Balance the equation:

Ex 18.3 – Balance the equation:

Ex 18.3 – Balance the equation:

Ex 18.3 – Balance the equation:

Practice - Balance the Equation ClO3 -1 + Cl-1 → Cl2 (in acid)

Practice - Balance the Equation ClO3 -1 + Cl-1 → Cl2 (in acid)

+5 -2 -1 0

oxidation reduction

1 ClO3 -1 + 5 Cl-1 + 6 H+1 → 3 Cl2 + 3 H2O

Electrical Current

of a liquid in a stream, we are

discussing the amount of water

that passes by in a given period

of time

we are discussing the amount of

electric charge that passes a

point in a given period of time

 whether as electrons flowing

through a wire or ions flowing

through a solution

Redox Reactions & Current

• redox reactions involve the transfer of electrons from one substance to another

• therefore, redox reactions have the potential to generate an electric current

• in order to use that current, we need to separate the place where oxidation is occurring from the

Electric Current Flowing Directly Between Atoms

Electric Current Flowing Indirectly Between Atoms

Electrochemical Cells

• electrochemistry is the study of redox reactions

that produce or require an electric current

• the conversion between chemical energy and

electrical energy is carried out in an

electrochemical cell

• spontaneous redox reactions take place in a

voltaic cell

aka galvanic cells

• nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

Electrochemical Cells

 half-cells

through a solution constitutes an electric circuit

electrode to allow the transfer of electrons

 through external circuit

 electrolyte

• Anode

electrode where oxidation occurs

anions attracted to it

connected to positive end of battery in electrolytic cell

loses weight in electrolytic cell

gains weight in electrolytic cell

electrode where plating takes place in electroplating

Voltaic Cell

the salt bridge is required to complete the circuit and

maintain charge balance

Current and Voltage

 unit = Volt

 1 V of force = 1 J of energy/Coulomb of charge

 the voltage needed to drive electrons through the external circuit

 amount of force pushing the electrons through the wire is called the electromotive force, emf

Cell Potential

• the difference in potential energy between the anode the cathode in a voltaic cell is called the

cell potential

• the cell potential depends on the relative ease

with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at

the anode

• the cell potential under standard conditions is

called the standard emf, E°

Cell Notation

• shorthand description of Voltaic cell

• electrode | electrolyte || electrolyte | electrode

• oxidation half-cell on left, reduction half-cell on the right

• single | = phase barrier

if multiple electrolytes in same phase, a comma is

used rather than |

• double line || = salt bridge

Standard Reduction Potential

occur has a large + half-cell potential

electrons will flow so that the half-reaction

with the stronger tendency will occur

a half-reaction, we can only measure it

relative to another half-reaction

conditions, which we assign a potential

difference = 0 v

 standard hydrogen electrode, SHE

Standard Cell

Half-Cell Potentials

• E°cell = E°oxidation + E°reduction

 E°oxidation = − E°reduction

Ex 18.4 – Calculate E°cell for the reaction at 25°C

Ex 18.4a – Predict if the following reaction is spontaneous

under standard conditions

sketch the cell and

label the parts –

oxidation occurs at

the anode; electrons

flow from anode to

cathode

Practice - Sketch and Label the Voltaic Cell

Half-Reactions and Overall Reaction, and Determine

the Cell Potential under Standard Conditions.

ox: Fe(s) → Fe 2+(aq) + 2 e − E° = +0.45 V

Predicting Whether a Metal Will

Dissolve in an Acid

• acids dissolve in metals if the

reduction of the metal ion is

easier than the reduction of H+

(aq)

• metals whose ion reduction

reaction lies below H+ reduction

on the table will dissolve in

E°cell, ∆ G° and K

• for a spontaneous reaction

one the proceeds in the forward direction with the

chemicals in their standard states

E° > 1 (positive)

K > 1

• ∆ G° = −RTlnK = −nFE°cell

Example 18.6- Calculate ∆ G° for the reaction

I2(s) + 2 Br−

(aq) → Br2(l) + 2 I−

(aq)

the forward direction under standard conditions

.1G

55.0485

,96mol

2G

5

C

J mol

C

×+

( ) ( )

12 5

.

11 3 2 10 10

5 1

1 V

0592

0

mol

2 V 34 0 log

(aq) → H2(g) + Cu2+

(aq)

the left under standard conditions

cell E E

V 0592

0

Ecell =

E ° at Nonstandard Conditions

V 41 1 E

.0]

1 [ 0]

2 [

] 010 0 [ log 6

V 0592 0 V 34 1 E

] [H ] [MnO

] Cu [ log V 0592 0 E E

cell

8 2

3 cell

8 2 4

3 2 cell

cell E E

n log

V 0592 0 E

Ecell = cell −

ox: Cu(s) → Cu 2+

(aq) + 2 e − }x3 E° = −0.34 v red: MnO4−

0 E

Ecell = cell −

Concentration Cells

reduction reactions are the same, as long as the electrolyte

concentrations are different

solutions

 the more concentrated solution has lower entropy than the less

concentrated

solution to the electrode in the more concentrated solution

 oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution – the less concentrated solution has the anode

when the cell concentrations are equal there is no difference in

energy between the half-cells and

no electrons flow

Concentration Cell

when the cell concentrations

are different, electrons flow

from the side with the less

concentrated solution

(anode) to the side with the

more concentrated solution

(cathode)

Cu(s) Cu 2+

(aq) (0.010 M)  Cu 2+

(aq) (2.0 M)  Cu(s)

LeClanche’ Acidic Dry Cell

Alkaline Dry Cell

electrolyte is alkaline KOH paste

rechargeable, little corrosion of zinc

Lead Storage Battery

• rechargeable, long life, light – however

recharging incorrectly can lead to battery

breakdown

Ni-MH Battery

Lithium Ion Battery

solution

ions

 reduction of transition metal

to cathode causing a corresponding

migration of electrons from anode to

cathode

more environmentally friendly, greater

energy density

Fuel Cells

• like batteries in which

reactants are constantly

being added

 so it never runs down!

• Anode and Cathode

both Pt coated metal

Electrolytic Cell

cause a non-spontaneous reaction

 must be DC source

reduced, anions release electrons to the anode and are

oxidized

electroplating

In electroplating, the work is

at the cathode

Cations are reduced at the

cathode and plate to the

surface of the work piece

The anode is made of the

plate metal The anode

oxidizes and replaces the

metal cations in the

Electrochemical Cells

anode, reduction occurs at the cathode

 anode is the source of electrons and has a (−) charge

 cathode draws electrons and has a (+) charge

 electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery

 electrons are forced toward the anode, so it must have a source

of electrons, the − terminal of the battery

• electrolysis is the process of using

electricity to break a compound

apart

cell

separate elements from their

compounds

Electrolysis of Water

Electrolysis of Pure Compounds

• must be in molten (liquid) state

• electrodes normally graphite

• cations are reduced at the cathode to metal

element

• anions oxidized at anode to nonmetal element

Electrolysis of NaCl(l)

Mixtures of Ions

• when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode

• when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode

Electrolysis of Aqueous Solutions

• Complicated by more than one possible oxidation and reduction

 reduction of cation to metal

2 H2O + 2 e -1 → H2 + 2 OH -1 E° = -0.83 v @ stand cond.

E° = -0.41 v @ pH 7

 oxidation of anion to element

graphite doesn’t oxidize

Electrolysis of NaI(aq) with Inert Electrodes

Faraday’s Law

• the amount of metal deposited during

electrolysis is directly proportional to the charge

on the cation, the current, and the length of time the cell runs

charge that flows through the cell = current x time

Example 18.10- Calculate the mass of Au that can be plated in

25 min using 5.5 A for the half-reaction

C 5 5

Au g

6 5

Au mol

1

g

196.97 mol

3

Au mol

1 C

96,485

mol

1 s

1

C

5.5 min

1

s

60 min

C 6,485 9

Au mol

1

Au mol 1

g 196.97

• corrosion is the spontaneous oxidation of a metal by chemicals in the environment

• since many materials we use are active

metals, corrosion can be a very big problem

• rust is hydrated iron(III) oxide

• moisture must be present

water is a reactant

required for flow between cathode and anode

• electrolytes promote rusting

enhances current flow

• acids promote rusting

some metals, like Al, form an oxide that strongly

attaches to the metal surface, preventing the rest from corroding

• another method to protect one metal is to attach

it to a more reactive metal that is cheap

Sacrificial Anode

Sample Calculations

COMMERCIAL PRODUCTION

OF CHEMICALS

basic units of joules/second

P = V I, (J/C) (C/S) = J/S

• A kilowatt-hour is an energy unit and is equal to 3.60x106J:

3600 S/hr times 1000 W/kW

• These unit factor problems change: current x time

==> coulombs ==> moles e- ==>

moles oxidized or reduced species ==> mass

• The balanced half-reaction provides the conversion from moles electrons to

moles of species sought

at the cathode if a current of 4.40 A

flows for 1.50 hours.

Let’s do an example first.

COUNTING ELECTRONS

Quantitative Aspects of

Electrochemistry Consider electrolysis of aqueous silver ion.

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

could know the quantity of Ag formed.

Consider electrolysis of aqueous silver ion.

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

1 mol e- -> 1 mol Ag

If we could measure the moles of e-, we could know the quantity of Ag formed.

But how to measure moles of e-?

Current = charge passing time

I (amps) = coulombs

Quantitative Aspects of

Electrochemistry

But how is charge related to moles of electrons?

1.50 amps flow thru a Ag+(aq) solution for 15.0 min What mass of Ag metal is deposited?

Solution (a) Calculate the charge

Coulombs = amps x time

= (1.5 amps)(15.0 min)(60 s/min) = 1350 C

I (amps) = coulombs

seconds

Quantitative Aspects of

Electrochemistry

1.50 amps flow thru a Ag+(aq) solution for 15.0 min What mass of Ag metal is deposited?

The anode reaction in a lead storage battery is

If a battery delivers 1.50 amp, and you have 454 g of

Pb, how long will the battery last?

Solutiona) 454 g Pb = 2.19 mol Pb

• Remember:

The balanced half-reaction

provides the conversion from

moles electrons to moles of

species sought

Now solve the problem:

at the cathode if a current of 4.40 A

flows for 1.50 hours.

COUNTING ELECTRONS

• Remember:

The balanced half-reaction provides the conversion from moles electrons to

Now solve the problem:

• Calculate the grams of aluminum formed at the cathode if a current of 4.40 A flows for 1.50 hours.

COUNTING ELECTRONS