Data Page 8 Chemistry 12

F g e F

S O e SO

H O H e H O

MnO H e Mn H O

Au e Au s

2

2 82

42

2 2 2

42

23

2 2 2 87

2 2 2 0 1

2 2 2 1 78

8 5 4 1 51

3 1 50

( ) + →← ++ →← +

+ + →← ++ + →← + +

+ →← ( ) +

− −

− − −

+ −

− + − +

+ −

.

.

.

.

.

BrO H e Br H O

ClO H e Cl H O

Cl g e Cl

Cr O H e Cr H O

O g H e

312 2 2

4 2

2

2 72 3

2

12 2

6 5 3 1 48

8 8 4 1 39

2 2 1 36

14 6 2 7 1 23

2 2

− + −

− + − −

− −

− + − +

+ −

+ + →← ( ) + ++ + →← + +

( ) + →← ++ + →← + +

( ) + + →

ℓ .

.

.

.

←← +H O2 1 23.

MnO s H e Mn H O

IO H e I s H O

Br e Br

AuCl e Au s Cl

NO H e NO g H

22

2

312 2 2

2

4

3 2

4 2 2 1 22

6 5 3 1 20

2 2 1 0 9

3 4 1 0 0

4 3 2

( ) + + →← + ++ + →← ( ) + +

( ) + →← ++ →← ( ) + +

+ + →← ( ) +

+ − +

− + −

− −

− − −

− + −

.

.

.

.

ℓ

OO +0 9 6.

Fe e Fe

O g H e H O

Mn O H O e Mn O s OH

I s e I

Cu e Cu s

3 2

2 2 2

4 2 2

2

0 7 7

2 2 0 7 0

2 3 4 0 60

2 2 0 5 4

0 5 2

+ − +

+ −

− − −

− −

+ −

+ →← +( ) + + →← ++ + →← ( ) + +

( ) + →← ++ →← ( ) +

.

.

.

.

.

H SO H e S s H O

Cu e Cu s

SO H e H SO H O

Cu e Cu

Sn e Sn

2 3 22

42

2 3 22

4 2

4 4 3 0 45

2 0 34

4 2 0 17

0 15

2 0 15

+ + →← ( ) + ++ →← ( ) +

+ + →← + ++ →← +

+ →← +

+ −

+ −

− + −

+ − +

+ − +

.

.

.

.

.

S s H e H S g

H e H g

Pb e Pb s

Sn e Sn s

Ni e Ni s

( ) + + →← ( ) ++ →← ( ) ++ →← ( ) −+ →← ( ) −+ →← ( ) −

+ −

+ −

+ −

+ −

+ −

2 2 0 14

2 2 0 00

2 0 13

2 0 14

2 0 26

2

22

2

2

.

.

.

.

.

H PO H e H PO H O

Co e Co s

Se s H e H Se

Cr e Cr

H O e H OH M

3 4 3 3 22

23 2

2 27

2 2 0 28

2 0 28

2 2 0 40

0 41

2 2 2 10 0 41

+ + →← + −+ →← ( ) −

( ) + + →← −+ →← −

+ →← + ( ) −

+ −

+ −

+ −

+ − +

− − −

.

.

.

.

.

Fe e Fe s

Ag S s e Ag s S

Cr e Cr s

Zn e Zn s

Te s H e H Te

2

22

3

2

2

2 0 45

2 2 0 69

3 0 74

2 0 76

2 2 0 79

+ −

− −

+ −

+ −

+ −

+ →← ( ) −( ) + →← ( ) + −

+ →← ( ) −+ →← ( ) −

( ) + + →← −

.

.

.

.

.

2 2 2 0 83

2 1 19

3 1 66

2 2 37

2 71

2 22

3

2

H O e H g OH

Mn e Mn s

Al e Al s

Mg e Mg s

Na e Na s

+ →← ( ) + −+ →← ( ) −+ →← ( ) −+ →← ( ) −+ →← ( ) −

− −

+ −

+ −

+ −

+ −

.

.

.

.

.

Ca e Ca s

Sr e Sr s

Ba e Ba s

K e K s

Rb e Rb s

2

2

2

2 2 87

2 2 89

2 2 91

2 93

2 98

+ −

+ −

+ −

+ −

+ −

+ →← ( ) −+ →← ( ) −+ →← ( ) −+ →← ( ) −+ →← ( ) −

.

.

.

.

.

Cs e Cs s

Li e Li s

+ −

+ −

+ →← ( ) −+ →← ( ) −

3 03

3 04

.

.

Hg e Hg

O g H M e H O

NO H e N O H O

Ag e Ag s

Hg e Hg

2

12 2

72

3 2 4 2

12 2

2

2 0 85

2 1 0 2 0 82

2 4 2 2 0 80

0 80

0 80

+ −

+ − −

− + −

+ −

+ −

+ →← ( ) +( ) + ( ) + →← +

+ + →← + ++ →← ( ) ++ →← ( ) +

ℓ

ℓ

.

.

.

.

.

STANDARD REDUCTION POTENTIALS OF HALF-CELLSIonic concentrations are at 1M in water at 25 °C.

Reducing AgentsOxidizing AgentsS

TR

ON

GW

EA

KS

TR

ON

GW

EA

K

Overpotential Effect

Overpotential Effect

ST

RE

NG

TH

OF

OX

IDIZ

ING

AG

EN

TS

TR

EN

GT

H O

F R

ED

UC

ING

AG

EN

TE° Volts( )

264

Jequilibrium arrow

reduction

✓←Oxidation

Fzcg) is the F-

is the weakest RA

strongest OA =F-

is least likely" to lose:

Fug ) is mostlikely to game?

Zncs)

cult 2¥-

@

Lit is the weakestOA Lies) is the strongest RA

= Lit is least likely = Lies) is most

to be reduced likely to be oxidized .

O O

6

On the table…

1. Left hand side of the table is increasing strength as an oxidizing agent…Think Reduction!!!

Strongest oxidizing agent is F2

2. Right hand side, increases going down the table, is increasing reducing agent…Think Oxidation!!!

Strongest reducing agent is Li

3. Most metals are located at the bottom right. Metals tend to become positively charged by undergoing oxidation. (Except Cu, Ag, Hg and Au)

4. Halogens / oxyanions are located in the top left. They tend to become negatively charged by undergoing reduction.

5. Some metals have more than one oxidation number. Therefore, there is more than one one-half reaction on the table. This can either be an oxidation or reduction. Example: Cu+, Sn2+ and Fe2+.

• Isolated half-reactions can go forward or backwards. Use equilibrium arrows.

Example: Au3+ + 3e- ⇌ A(s)

• If a half reaction is part of a redox reaction and MADE to go as either an oxidation or reduction, then use a one-way arrow.

• Oxidations, like bases in the previous section, run backwards on the table.

• How to predict if there is an oxidation or reduction

1. Determine available substance and identify them in the Table of Reduction Potential.

2. The strongest oxidizing agent (upper left) will undergo reduction.

3. The strongest reducing agent (bottom right) will undergo oxidation.

4. Unless there is a combination of 2 and 3, the redox won’t be spontaneous. a. Reactant to be reduced (left side) is HIGHER on the table than the reactant to be oxidized (right side). In this case, the

reaction will be SPONTANEOUS. b. Reactant to be reduced (left side) is LOWER on the table than the reactant to be oxidized (right side). In the case,

there will be Non-SPONTANEOUS reaction. c. If BOTH reactants are found on the left OR right side of the table, then, NO REACTION is possible because

there aren’t enough chemicals to allow full exchange of e-.

265

FZ

tics)

positivelyoxidation ( by losing e-)

negativelyreduction (by gaining e-)

forward backward For

4 #a ⑥

→or←part

RE OX

backwards

→

←

⇐* spontaneous rxn

It; ⇒ D Non- spontaneousrxn

=

⇐ ing,

7

Example: You have: Solid Zn(s) and

Cu+2 solution

Possible half-reactions are…

Notice that the only species that can be reduced is Cu2+. Therefore, Cu+2 will act as the oxidizing agent.

On the other hand, the only species that can be oxidized is Zn(s). Hence, Zn(s) will act as the reducing agent.

The combination of a strong oxidizing agent and a reducing agent causes a spontaneous redox reaction.

To write a redox equation, each half reaction must be written in the appropriate direction.

The reduction reaction is written such that the e- appears on the reactant side.

Cu+2 is reduced…therefore the reduction reaction is:

The oxidation reaction must be written in “reverse”. Remember to have e- on the product side. Zn(s) is oxidized…therefore the oxidation reaction is:

Add the half reactions together to show the redox equation.

**If only one species in a half reaction is present, you CAN NOT assume the is also present UNLESS you have been explicitly told.**

Example: The reduction half reaction will not take place unless SO4

2-, H+(acidc solution), and electrons are all available.

SO4

2- + 4H+ + 2e- ! H2SO3 + H2O

Examples: Write a redox equation if the reaction is spontaneous. If not, explain why. 1. Reactants are Zn(s) and Cu(s)

2. Reactants are K+ and Ca(s)

266OA RA

* tze- FF cues )e e

-

=

znzt + 2e-

IE Ents ) AtOA

RA

a redox ( spontaneous )order

-- This reduction rxn Ts

Cutt +,

Ze- BEEN Cues) the one that is most likely

Zncs) EBAN Znztt 2¥t

CU" t Zncs )- Cues , t znzt

Cancel out

electrons .

MMothers

There is no redox bleCutt e- E.cno reduction is possible -

cutter- 5Cwe only have RA ( can

zn2tt2e- 5Zn④ undergo oxidation )

Catt 25K€ There is non- spontaneous rxn

④ e- gy Kes ,

b/c the reduction to oxidation

orientation is incorrect .

8

3. Reactants are Mg+2 and Mn2+

4. Reactants are Br2(l)

and Zn(s) 5. Reactants are Br2(l)

and Cl- (H2O is not available)

267

There is no redox ble

⇒t 2e-

5 Mncs) no oxidation is possible .

④+ 2e-

G luges ) Both Mg't te Mn

't will

want to gain e: there isn't

any species willing to losee:

t 2e-

§pg2B§R :

q zincs )O'

.ZnHt2e-*nrn

ZncS)→znzt+#

Redox Brue) t Zncs) → 2Br-

t Zn't ⇐ Spontaneous Redox