Chapter 13, Preparation

• to work with cells reaction prediction is a necessary skill

• 1st step: Make a listGuidelines:• most ionics get listed as (aq) ions e.g. list NaNO3(aq) as Na+(aq), NO3‾(aq)

• list low solubility ionics as formula(s) e.g. MnO2(s), AgBr(s), AgI(s),

PbSO4(s)

• list molecular compounds as formula(SATP state) e.g. H2O(l),

NO(g), H2O2(l), N2O4(g), N2O(g)

Chapter 13, Preparation

• list pure elements as element formula(SATP state) e.g. F2(g), Cu(s), O2(g), etc

• list strong acids as H+(aq), @#‾(aq)• list weak acids as formula(aq) e.g. HNO2(aq), H2SO3(aq)

• if question says “acidified” list H+

(aq)• if question says “basic” list OH‾(aq)• list H2O(l)

Chapter 13, Preparation

• 2nd step: identify each entity as OA, RA, or \Use your table and the following generalizations:• Metal elements are always RA’s e.g. Ag(s), Zn(s), etc• Metal ions are always OA’s but a few, on your table are both OA and RA e.g. Sn2+(aq), Fe2+

(aq), Cr2+(aq)• Water is both OA and RA

Chapter 13, Preparation

• 3rd step: pick out SOA and SRA

• 4th step: write half-reactions for SOA and SRA and balance electrons to get net balanced equation

• 5th step: determine if predicted reaction is spontaneous or non-spontaneous

Chapter 13, Preparation

• Example 1:Nitric acid is painted onto a copper sheet to etch a design

List: H+(aq), NO3‾, Cu(s), H2O(l)

OA OA RA OA/RA

2 NO3‾(aq) + 4 H+(aq) + 2 e‾ N2O4(g) + 2 H2O(l) Cu(s) Cu2+(aq) + 2 e‾

S S

2 NO3‾(aq) + 4 H+(aq) + Cu(s) Cu2+(aq) + N2O4(g) + 2 H2O(l)

spont

Note: concentrated HCl(aq) would have no effect on Cu(s)

Chapter 13, Preparation

• Example 2:Engineers removed O2(g) from a solution by adding basic sodium sulfite solution

List: O2(g), OH‾(aq), Na+(aq), SO32‾(aq),

H2O(l)

OA RA OA RA OA/RA

O2(g) + 2 H2O(l) + 4 e‾ 4 OH‾(aq)

S S

2 x (SO32‾(aq) + 2 OH‾(aq) SO4

2‾(aq) + H2O(l) + 2 e‾)

2 SO32‾(aq) + O2(g) 2

SO42‾(aq)

2 SO32‾(aq) + 4 OH‾(aq) + O2(g) + 2 H2O(l) 2 SO4

2‾(aq) + 2 H2O(l) + 4 OH‾(aq)

spont

Do WS 54

Chemistry 30 – Unit 2 Part 2Cells and Batteries

To accompany Inquiry into Chemistry

PowerPoint Presentation prepared by Robert Schultzrobert.schultz@ei.educ.ab.c

a

Chapter 13, Section 13.1• Redox reactions – transfer of electrons

• In past, OA and RA were in contact and electrons flowed between the two

• In cells, OA and RA are kept away from each other so that electrons must flow through a wire to be transferred – this is an electric current

• Now the electrons can do work

Chapter 13, Section 13.1

• cell – transforms chemical potential energy to electrical energy via redox reaction

• cell (studied later in chapter) – transforms electrical energy into chemical potential energy

Chapter 13, Section 13.1Cell Definitions and Principles:

• electrode where oxidation occurs (where SRA reacts)

• electrode where reduction occurs (where SOA reacts)

• (salt bridge or porous cup) device to allow flow of ions

• (current flow): from anode to cathode

• anions to anode; cations to cathode

memory tool:

“An ox cared”

Chapter 13, Section 13.1• Daniell Cell

components: Zn(s) in ZnSO4(aq) solution;Cu(s) in CuSO4(aq) solution

To find out what will happen, list everything present in the cell and use your chart to identify the OA’s and RA’s

ListList:Zn(s), Zn2+(aq), Cu(s), Cu2+(aq), SO4

2-(aq),

H2O(l) RA OA RA OA OA* OA/RA

Chapter 13, Section 13.1

Zn(s), Zn2+(aq), Cu(s), Cu2+(aq), SO42-(aq),

H2O(l) RA OA RA OA OA* OA/RA

Identify the strongest oxidizing agent, SOA, strongest reducing agent, SRA, and write the half-reactions

S S

Cu2+(aq) + 2 e– Cu(s)

Zn(s) Zn2+(aq) + 2 e–

By definition

@ cathode@ anode

Chapter 13, Section 13.1

anode – cathode +

electron flow

anion flow

cation flow

Zn(s) Cu(s

)

Zn2+

(aq)Cu2+

(aq)

KNO3(aq)

V

SO42-(aq) SO4

2-(aq)

Always use red meter lead for (+) and black for (–)

Chapter 13, Section 13.1• Why do ions flow? Look at the half-

reactions

Zn(s) Cu(s

)

Zn2+

(aq)Cu2+

(aq)

KNO3(aq)

V

SO42-

(aq)

SO42-

(aq)

Cu2+(aq) changes to Cu(s)

SO42-(aq) concentration

unchanged – build up of negative charge around cathode. Cations flow to neutralize charge

Zn(s) changes to Zn2+(aq)

build up of positive charge around anode. Anions flow to neutralize charge

anode cathode

Cu2+(aq) + 2 e− Cu(s)

Zn(s) Zn2+(aq) + 2 e−

Chapter 13, Section 13.1• Daniell

Cell in action!

• Look at the cell half-reactions again. What would you expect to happen to copper electrode? zinc electrode?

• Discuss

Chapter 13, Section 13.1• The voltmeter reading in the

animation was +1.10 V or -1.10 V

• Do you see any connection to electrical potential numbers from your redox chart in Data Booklet? more to come on this

• Any idea about the meaning of the (+) or (-)?

Chapter 13, Section 13.1• Inert electrodes: many of the half-

reactions on your chart do not include a metal element that can function as an electrode

• Examples:2 H+(aq) + 2 e– H2(g)

Fe3+(aq) + e– Fe2+(aq) Cr2O7

2-(aq) + 14 H+(aq) + 6 e– 2 Cr3+(aq) + 7 H2O(l)

To use these half-reactions in a cell, inert electrodes are used. Common inert electrodes: C(s), Pt(s)

Chapter 13, Section 13.1• Cell Notation:

• For the Daniell Cell:Zn(s) | Zn2+(aq), SO4

2-(aq) || Cu2+(aq), SO42-(aq) |

Cu(s)

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

common to leave out spectators:

anode | anode solution || cathode solution | cathode

oxidation half-cell

reduction half-cell

porous barrier

Chapter 13, Section 13.1Oxidation (@ anode) produces electrons making anode negativeReduction (@cathode) consumes electrons making cathode positive

Electrons at anode have lots of potential energy (they are repelled through the wire away from the anode)

Electrons at cathode have zero potential energy

Chapter 13, Section 13.1• Potential difference (commonly

called voltage) = difference in Ep per unit charge between anode and cathode

• Eºcell = Eº

net = ΔEº = Eºcathode– Eº

anodeStandard cell potential*Cell potentialCell voltage

(V)Standard reduction potential for anode ½ -reaction

(V)

Standard reduction potential for cathode ½ -reaction

(V)**Important note: never multiply the Eº values by balancing coefficients!

Chapter 13, Section 13.1• For Daniell cell

Eºcell=+0.34 V – – 0.76 V = +1.10 V

same as on animation!• Recall the spontaneity rule from

chapter 12:OA

RA

Note that spontaneous reactions will have a + Eº

cell

Chapter 13, Section 13.1Reference half-cell:

…….|| H+(aq), H2(g) | Pt(s)

or

Pt(s) | H+(aq), H2(g) || …..

Eor= 0.00 V

Chapter 13, Section 13.1Cell analysis:• When you look at cell notation it’s

good idea to check that left hand side of cell notation really is anode ½ cell (where SRA reacts)

Example:

Ni(s) | Ni2+(aq) || Cr2O72-(aq), H+(aq) |

C(s) RA OA OA OA \S S

inert electrode

Chapter 13, Section 13.1Half-reactions:3 x (Ni(s) Ni2+(aq) + 2 e–)

(anode)Cr2O72-(aq) + 14 H+(aq) + 6 e– 2 Cr3+(aq) + 7

H2O(l) (cathode)Cr2O7

2-(aq) + 14 H+(aq) + 3 Ni(s) 2 Cr3+(aq) + 7 H2O(l) + 3 Ni2+(aq)

1.23 0.26 1.49

o o ocell cathode anodeE E E

V V V

Spontaneous: why are standard voltaic cells always spontaneous?

Chapter 13, Section 13.1

Interruption:

Another half-cell separator (instead of salt bridge)

Porous cup:

unglazed pottery – holds water but ions are small enough to flow through

Chapter 13, Section 13.1Back to the cell:

electron flow

anode–cathod

e

+

V

Ni(s)

C(s)

Ni2+

(aq)

Cr2O72-

(aq)

H+(aq)

anions

cations

Chapter 13, Section 13.1• Other solutions can be used, but

KNO3(aq) is particularly good for use in a salt bridge

• Why?

Both K+(aq) and NO3–(aq) have high

solubility with all other ions – no precipitates in salt bridge

K+(aq) and NO3–(aq) without H+(aq)

are weak OA’s and unlikely to interfere with the desired cell reaction

Chapter 13, Section 13.1Investigation 13.A page 488

Prelab:Make predictions for the following cells:A Ag(s) in Ag+(aq) with Cu(s) in Cu2+(aq)

B Zn(s) in Zn2+(aq)C Mg(s) in Mg2+(aq)D Mg(s) in Mg2+(aq) with Zn(s) in Zn2+(aq)E Cu(s) in Cu2+(aq)F Zn(s) in Zn2+(aq) with Cu(s) in Cu2+(aq)

""" "

"" " "

" " " "

We’ll do the first together

Chapter 13, Section 13.2Write cell notation (in proper order)

• Draw cell diagrams showing electrodes labeled with identity and also as + or -, anode or cathode

• Label electrolyte solution in each half-cell

• Show direction of electron flow, direction of ion flow, predicted E°cell, anode and cathode half-reactions

Chapter 13, Section 13.2Commercial voltaic cells:• Dry Cells, also called carbon-zinc

cells or zinc chloride cells (Eveready “cat”)zinc anode (case), carbon cathode, electrolyte paste made up of MnO2(s), starch, and a very small amount of waterHalf-reactions:Zn(s) Zn2+(aq) + 2 e–2 MnO2(s) + H2O(l) + 2 e– Mn2O3(s) + 2

OH–(aq)1.5 V, primary cell (non-rechargeable)

Chapter 13, Section 13.2• alkaline cells (Duracell, Energizer,

etc)

• more expensive version of dry cellSignificant differences between these and regular dry cells

Chapter 13, Section 13.2• batteries – multiple cells e.g. a 9 V

alkaline is made up of six 1.5 V alkaline cells

Button batteries – read page 492-3

Chapter 13, Section 13.2• Fuel Cells: batteries where a “fuel” in

the anode half-reaction is consumed and products are released from the cell

• Battery is renewed by replacing the fuel

• Common fuel cell fuels: hydrogen, methanol, hydrocarbons

• Conversion of fuels to energy in fuel cells more efficient than combustion(also much cleaner than combustion)

Chapter 13, Section 13.2• The hydrogen fuel cell used in

the space shuttle provides electricity and water for the astronauts as net reaction is 2 H2(g) + O2(g) 2 H2O(l)

• Your text has interesting pictures and reading on pages 493-6 on fuel cells

Chapter 13, Section 13.2• Discuss questions 2, 6, 7, 10 page

501

Chapter 13, Section 13.2• Rusting (oxidation/corrosion of iron)

Many metals such as copper, aluminium, and zinc, when they react with O2 in air, get covered with a passive oxide coat that protects them from further oxidation

• Not true for iron and steelRust forms on the surface and continues to form below the surfaceOnce it starts . . . . . . .

Chapter 13, Section 13.2• in corrosion of iron, iron acts as anode

and is oxidized• the OA is oxygen in presence of water• cathode, where OA reacts, is inert

conductor on surface of iron – soot in some casesHalf-reactions:

Fe(s) Fe2+(aq) + 2 e–

O2(g) + 2 H2O(l) + 4 e– 4 OH–(aq)

Products: Fe2+(aq) and OH–(aq) react to form Fe(OH)2(s), a precursor of rust

Chapter 13, Section 13.4

soot less soot

Chapter 13, Section 13.2• Fe(OH)2(s) further reacts with air to

form Fe(OH)3(s) and Fe2O3•x H2O(s)

• This is rust

Rust prevention:Physical - cover the iron with paint, tar, or grease, plate it with chromium, alloy it with chromium (stainless steel)

Chapter 13, Section 13.2Rust prevention continued

• Chemical – attach, by some means, a metal that is a stronger RA than Fe to the Fe

• Under oxidizing conditions the other metal (sacrificial anode) will be oxidized instead of the Fe

Chapter 13, Section 13.2• iron becomes inert cathode where

O2(g) + 2 H2O(l) + 4 e− 4 OH−(aq)

takes place

• Process called cathodic protection since iron becomes inert cathode

• Examples of cathodic protection: galvanizing, sheradizing, contact with blocks of Zn or Mg

• Discuss questions 13, 14, 16 page 500

Chapter 13, Section 13.2

Chapter 13, Section 13.2• Review sample diploma exam

questions on cathodic protection

Chapter 13, Section 13.3• Electrolytic Cells: Cells that convert

electrical energy to chemical potential energy

Chapter 13, Section 13.3Voltaic Cell

+ – oxidation

electrons produced

reduction

electrons consumed

e-e-

Electrolytic Cell

cathodeanode

reductionelectrons consumed

oxidationelectrons produced

spontaneous

–

anode

cathode

e.g.

C(s)|Cu2+(aq)|C(s)SOA Cu2+

(aq)

Cu2+(aq) + 2 e– Cu(s)

SRA H2O(l)

2 H2O(l) O2(g) + 4 H+(aq) + 4 e–

Non-spontaneous

Eºnet=+0.34 V – +1.23 V = –0.89 V

Chapter 13, Section 13.3Observations:Anode (+) in electrolysis; cathode (–) in electrolysis

Electrons in electrolysis circuit flow from (+) to (-) but still from anode to cathode!

Anions in electrolytic cell flow towards (+); cations flow towards (-) but this is still ……….Anions to anode; cations to cathode

Voltage in electrolytic cells is (-): non-spontaneous

Text has good chart on page 503 – turn to it

Chapter 13, Section 13.3• What do you think the meaning of

the negative Eºcell is?

• Predict anode and cathode half-reactions for following electrolysis reactions:

• Na2SO4(aq) with Pt(s) electrodes• NaCl(aq) with Pt(s) electrodes• KI(aq) with C(s) electrodes• CuSO4(aq) with C(s) electrodes

Use the handout provided

Be sure to list H2O(l) when finding SOA & SRA

Chapter 13, Section 13.3All of these electrolysis reactions will be demonstrated for you next day

Chapter 13, Section 13.3• When you calculate Eºnet for

electrolysis of H2O(l):2 H2O(l) + 2 e– H2(g) + 2 OH–(aq) Eºr=–

0.83 V2 H2O(l) O2(g) + 4 H+(aq) + 4 e– Eºr=

+1.23 V

2 x ( )

4 H2O(l)

6 H2O(l) 2 H2(g) + O2(g) + 4 H+(aq) + 4 OH–

(aq)2

Eºnet=−0.83 V − +1.23 V= −2.06 V : predicted minimum voltage 2.06 V

But this is way too high! Why?Nowhere near standard conditions!! Under these conditions Eºr’s much differentEºnet = −0.42 V − +0.82 V= −1.24 V

−0.42 V

+0.82 V

overpotential

Chapter 13, Section 13.3• If you did this electrolysis and applied a

voltage of 1.24 V (new predicted minimum voltage) you would find that the reaction would not occur

• Eºr’s are measured with respect to the reference half-cell (H2(g), H+(aq)|Pt(s))

• Particularly in half-reactions where gases are formed, a slightly higher voltage called the overpotential is required

• For formation of H2(g) and O2(g) overpotential is approximately 0.6 V

Chapter 13, Section 13.3• Overpotential needed to

understand NaCl(aq) electrolysis

• In electrolysis of NaCl(aq), predicted reaction didn’t occur at anode

• Prediction: H2O(l) both SRA and SOAanode: 2 H2O(l) O2(g) + 4 H+(aq) + 4 e– Eºr =

+1.23 V

cathode: 2 H2O(l) + 4 e– H2(g) + 2 OH–(aq) Eºr = - 0.83 V

Predicted Products:

O2(g) & H2(g)

Eºnet= –0.83 V – +1.23 V = –2.06 V

Chapter 13, Section 13.3• Actual:

• Products:

anode: 2 Cl–(aq) Cl2(g & aq) + 2 e– Eºr = +1.36 V

cathode: 2 H2O(l) + 4 e– H2(g) + 2 OH–(aq) Eºr = - 0.83 V

Cl2(g), H2(g), OH–(aq)

Eºnet= –0.83 V – +1.36 V = –2.19 V

• Even though production of O2(g) seems to be favoured, the overpotential for O2(g) is greater than that for Cl2 making production of Cl2 favoured• called “the chloride anomaly”

Chapter 13, Section 13.3Chlor-Alkali Process – industrial

electrolysis of NaCl(aq)Figure 13.25, page 506

Chapter 13, Section 13.3• Electrolysis of aqueous ionics to

produce elements has several problems:

• Some ionics have low solubility in water

• H2O(l) is a stronger OA than some metal ions

• Answer: Use molten ionics (molten salts)

Chapter 13, Section 13.3

Down’s Cell

anode & cathode both C(s)Figure 13.26,

page 510

Chapter 13, Section 13.3• Rechargeable cells and batteries

(secondary cells and batteries)

• Example: lead-acid battery

Figure 13.28, page 511

• while battery discharges it’s voltaic; while it recharges it’s electrolytic

Chapter 13, Section 13.3• You should know the half-reactions

– they are in your Data Booklet

• Example: nicad – discussKnow ecological concerns

• Read pages 511-2

Chapter 13, Section 13.4Cell Stoichiometry

Necessary definitions: rate of flow of electric

charge; 1 ampere of current = 1 Coulomb (unit of charge) per second

• Charge of 1 mol of electrons called 9.65 x 104 C/mol

• No need to memorize; page 3 Data Booklet

qI

t units: 1 1CA s

Chapter 13, Section 13.4

e

I tn

F

e

n = # mol of electrons (mol)I = current (A) (must be

amps)t = time (s) (must be seconds) 4' 9.65 10 CF Faraday s number mol

Chapter 13, Section 13.4Example: question 9, page 516

Zn2+(aq) + 2 e− Zn(s)n1

750 mA

3.25 h

n2

m=?

current, time info always related to

en

1 4

0.750 3.25 36000.0909

9.65 10

sA hI t hn molCF mol

2

10.0909 0.0455

2n mol mol

0.0455 65.41 2.97gm mol gmol

Chapter 13, Section 13.4Example: question 11, page 516Ni(s) Ni2+(aq) + 2 e−

n1

1.20 g

n2

35.5 minI=?

1

1.200.0204

58.69

gn mol

gmol

2

20.0204 0.0409

1n mol mol

40.0409 9.65 101.85

35.5min 60 min

e

e

I tn

FCmoln F molI A

st

Chapter 13, Section 13.4• Do worksheet 60

• Answers:1. 51 min, also indicate electrode2. 6.35 kg, note this is Cl2 not Cl-

also indicate electrode3. 5.96 h4. 274 A5. 9.35 g

Chapter 13, Section 13.4Electroplating –• The process of plating a metal

coating on top of another metal by electrolysis

• Sounds easy, but there are numerous problems

• What are some?

Chapter 13, Section 13.4Extraction and refining of metalsExtraction –Extracting the metal from its ore;

oftendone by electrolysis of solutions ormoltens

Refining –Purification of metalsafter extraction

− +

?? anode: impure Cu

cathode: purified Cu

Chapter 13, Section 13.4