16. Redox Chemistry

and Corrosion

Everyday redoxRedox reactions include:• the respiration reaction that is

the source of energy in almost all living things

• photosynthesis in green plants• burning of fuels to propel our

cars• Often associated with rusting of

materials– Australia spends about $3 billion

each year in an effort to prevent metals from corroding and replacing metal that has corroded

Redox

A redox reaction is a reaction where oxidation and reduction simultaneously occur.

• Oxidation is loss of electrons (e- are products)• Reduction is gain of electrons (e- are reactants)

Reduction Oxidation

Of electrons

Redox half equations

Terminology

• OXIDANT– Causes oxidation to occur. The oxidant itself will

be reduced– Also called oxidising agent

• REDUCTANT– Causes reduction to occur. The reductant itself will

be oxidised– Also called reducing agent

Writing redox half equations

Zn(s) → Zn2+(aq)

Cu2+(aq) → Cu(s)

Overall chemical equationZn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

+ 2e–

+ 2e–

oxidation

reduction

Redox half equations

Au+ (aq) → Au3+(aq) • Au+ (aq) → Au3+(aq) + 2e- • Oxidation

Br2 (l) → Br− (l)

• Br2 (l) + 2e- → 2Br− (l) • ReductionF− (aq) → F2 (g)

• F− (aq) → F2 (g) + e- • Oxidation

H2 (g) → H+ (aq)

• H2 (g) + 2e- → 2H+ (aq) • Reduction

Complete and balance the following half equations, AND classify as either Oxidation of Reduction.

Steps for writing complex redox half equations: KOHES

• K = balance the key element (undergoing oxidation or reduction)

• O = balance oxygen by adding H2O to the opposite side of the equation.

• H = balance hydrogen by adding H+ to the opposite side of the equation.

• E = balance the charge on both sides of the equation by adding electrons to the more positive side

• S = add states

NOTE: Common error students have is where they locate the electrons REMEMBER to balance for atoms AND charge for half equations!

Over to you..

Permanganate ions MnO4- reduced to Mn2+ ions

K: MnO4

- Mn2+ (Manganese Mn is the Key Element)O: MnO4

- Mn2+ + 4H2O (balance Oxygen)

H: MnO4- + 8H+ Mn2+ + 4H2O (Balance hydrogen)

E: MnO4- + 8H+ + 5e- Mn2+ + 4H2O (balance charge)

S: MnO4-(aq) + 8H+

(aq) + 5e- Mn2+(aq) + 4H2O(l) (States)

Oxidation Numbers (states)

Sometimes it can be hard to distinguish how electrons have been transferred.

Oxidation numbers also help to identify whether a redox reaction has occurred.

• Increase in ON means the element has been Oxidized • Decrease in ON means the element has been Reduced

Species Rule for oxidation number Examples

Elements Oxidation number of zero Cl2, Mg , C, K, P4

Ions

Oxidation number is equal to (or adds up to) their charge

Na+ (+1), Cl- (-1),

NH4+ : overall +1 (N -3,

4H (4x +1) = +1)

Oxygen compounds

Oxygen in compounds = -2

(except H2O2 and BaO2 = -1)

H2O : O = -2,

H2O2: O = -1

Hydrogen compounds

Hydrogen compounds = +1

(except in metal hydrides = -1)

H2O : H = +1HCl : H = +1,

LiH: H=-1

Molecular ions and molecules

The sum of ON = charge on molecular ion or 0 in case of neutral molecule.

The most Electroneg elements have a negative ON

CH4: C (-4) + 4H (4 x +1) = 0

Using oxidation numbers to identify a redox reaction determine whether the following reaction is a redox reaction.

H2(g) + I2(g) 2HI(g)

• The oxidation number of hydrogen has changed from 0 to +1, so the hydrogen has been oxidized.

• The oxidation number of iodine has changed from 0 to –1, so the iodine has been reduced.

• This is therefore a redox reaction.

Although tungsten, W, is a rare element, it is very common in light globes. The metal is obtained from tungsten(VI) oxide by heating it with hydrogen, according to the equation:

WO3(s) + 3H2(g) W(s) + 3H2O(g)

Using oxidation numbers, determine whether this equation represents a redox reaction and, if so, identify the oxidant and the reductant.

• The oxidation number of tungsten decreases, it must belong to the compound that is reduced.

• The oxidation number of hydrogen increases, indicating oxidation. • Therefore the reaction is redox: WO3 (not just W) is the oxidant and H2 is the

reductant.

Identify the oxidation numbers of all elements in the equation and then deduce which element is reduced.

Decrease in ON means the element has been Reduced

Therefore the molecular oxygen has been reduced

Oxidation Numbers and Vanadium

Today we will be reacting vanadate ions (V) with Zinc to form vanadium (IV), vanadium(III) and vanadium (II):

The reactions:VO2

+ (aq)

+ 2H+(aq)

+ e- VO2+ (aq) + H2O(l)

VO2+ (aq) + 2H+

(aq) + e- V3+

(aq) + H2O(l)

V3+(aq) + e- V2+

(aq)

TASK1. Identify the change of oxidation number for vanadium in each

reaction

2. Analyse whether vanadium is being oxidized or reduced and therefore whether it is an oxidant or reductant

Oxidation Numbers and Vanadium

VO2+

+5

VO2+

+4

V3+

+3

V2+

+2

Homework

• Read pages 273 – 283

• Complete questions: 2, 4, 7, 8, 9, 10ace

• W/s Booklets due this Monday complete and corrected (with DIFFERENT coloured pen)

Revision Quiz (6min/test conditions)1. Distinguish between:

a. Oxidation and reduction

b. Oxidant and reductant

2. Write the half equation for Cu2+ solution forming Cu solid.

Identify whether oxidation or reduction has occurred.

3. Write the half equation for CO gas forming CO2 gas.

Identify whether oxidation or reduction has occurred.

4. Assign oxidation numbers for the underlined species:a. Vb. V2+

c. V2O5

d. O2

e. MnO4-

0+2+50+7

PREDICTING REDOX REACTIONSThe Electrochemical Series

The Electrochemical

Series

Strongest oxidant

Strongest reductant

Incr

easi

ng O

xidi

zing

str

engt

h Increasing reducing strength

This table can be used to predict whether a spontaneous redox reaction will occur

The electron transfer (note the negative gradient)

The oxidant must be higher in position to the reductant

Top-left /bottom right

The Electrochemical Series as it appears in your textbook (p287)

Relative strength of reductant (causes reduction)

Relative strength of oxidant (causes oxidation)

How can we predict this?

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Oxidation

Reduction

Stronger oxidant

Stronger reductant

Will a reaction occur?

The Electrochemical Series (p287)

Oxidation

Reduction

Stronger oxidant

Stronger reductant

Will a reaction occur?

Zn (s) Zn2+(aq) + 2e-

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

• Conjugate Redox pairs– Zn2+(aq)/Zn(s) and Cu2+(aq)/Cu(s)

OxidationZn(s) Zn2+(aq) + 2e-

ReductionCu2+ (aq) + 2e- Cu(s)

• Conjugate Redox pairs– Zn2+(aq)/Zn(s) and Cu2+(aq)/Cu(s)

OxidationZn(s) Zn2+(aq) + 2e-

ReductionCu2+ (aq) + 2e- Cu(s)

reductantOxidant

Oxidant and Reductant

• What is undergoing oxidation?• What is undergoing reduction?

• What is the oxidant? (Causing oxidation)• What is the reductant? (causing reduction)

Na

O2

Na

O2

Will a reaction occur?

Ni(s)

Al(NO)3(aq)

Will a reaction occur? NO.

Ni(s)

Al(NO)3(aq)

Use an electrochemical series to predict whether the following combinations will

spontaneously react

a Fe is added to a solution containing Ni2+ ions

b Ag is added to a solution containing Co2+ ions

c Cr is added to a solution containing Sn2+ ions.

Incr

easi

ng O

xidi

zing

str

engt

h

Homework

• Questions 13 and 14 should be completed today in class

• On Monday we will be predicting whether redox reactions will occur and construction galvanic cells. pre reading required: p 283-286

Revision from Last class

Y Pb(s) + Cu2+(aq) Cu(s) + Pb2+

(aq) NNNY Zn(s) + Pb2+(aq) Pb(s) + Zn2+

(aq) NY Fe(s) + Cu2+(aq) Cu(s) + Fe2+

(aq) NY Zn(s) + 2Ag+(aq) 2Ag(s) + Zn2+(aq) N

Using the electrons:Instead of reacting the oxidant and reductant directly, you can separate the reactants and have the electron transfer happen through an external wire.

Galvanic Cells

Components of a Galvanic Cell

Anode Cathodeexternal

internal

The Galvanic Cell aka Electrochemical CellBeaker A:Copper strip Dilute Copper Sulphate

Beaker B:Zinc stripDilute Zinc Chloride

Galvanometer – detects flow of e-

salt bridge – contains a soluble ionic substance that allows the circuit to be completed.

e- e-

Zn(s) —> Zn2+(aq) + 2e−

Cu2+(aq) + 2e−—> Cu(s)

Zinc is losing electrons…Oxidation!

Copper is gaining electrons… Reduction!

e- e-e- e-e- e-

A Closer Look at OxidationZn(s) —> Zn2+

(aq) + 2e−

Zn Zn

Zn

Zn

Zn

Zn

Zn 2+ Zn 2+

Zn 2+

Zn2+

Zn 2+

Zn 2+

2e−

2e−

2e−

2e−

2e−

2e−`

(aq)

(aq)

(aq) (aq)(aq)

(aq)

• The electrons lost by zinc travel through the wire to the copper solution

• The electrons are lost from zinc atoms, forming ions.

• These ions have to return to solution in the form of Zn2+ ions (aq)

• The metal gradually loses mass as metal atoms form metal ions in solution.

Given the zinc strip is the source of the negatively charged electrons we assign this half cell as the anode.

The negative anode is the cell which undergoes oxidation! This is dependent on which metals are part of the electrochemical cell – not always zinc

A closer look at reduction

Cu2+(aq) + 2e−—> Cu(s)

• The electrons lost by zinc travel through the wire to the copper solution.

• In the copper solution are Cu2+ ions (aq)

• The ions receive (gain) the electrons and as a result form Cu(s) atoms.

• The atoms then combine with the copper strip to add further mass.

Cu2+

Cu2+

Cu2+

Cu2+

e- flow

Cu

Cu

Cu

Cu

The Salt Bridge• Anions migrate towards the half cell that contains the anode• Cations migrate towards the half cell that contains the

cathode

K+ NO3-

Cu2+

2e−

Zn 2+

Ions in salt bridge balance the

change of ion concentration in

each half cell

Positively charged cathode

Ni2+/Ni half-equation: Ni(s) → Ni2+(aq) + 2e–

OxidationPb2+/Pb half-equation: Pb2+(aq) + 2e– → Pb(s)

Reduction

e- (from Ni to Pb)

Negatively charged anode

+

Ni2+

Stronger oxidant (will undergo reduction)

Stronger reductant (will undergo oxidation)

Complete this diagram of galvanic cell for Ni(s) and

Pb2+(aq)

A galvanic cell is set up by combining the A2+(aq)/A(s) and B2+(aq)/B(s) half- cells via a conducting wire and a salt bridge. After some time, the A and B

electrodes are taken out of the half-cells and reweighed. The mass of the A electrode has increased, whereas the mass of the B electrode has decreased. Complete the labeling of the galvanic cell shown below including the overall

redox equation for the reaction occurring in the cell.

Overall Equation: A2+(aq) + B(s) → A(s) + B2+(aq)

A2+/A half-equation: A2+(aq) + 2e–→ A(s)Reduction

B2+/B half-equation: B(s) → B2+(aq) + 2e–

Oxidation

e-

B is the negatively charged anodeA is the positively charged cathode

To be completed:

Now: • Worksheet • Textbook question 15 (13 and 14 should be

complete from last class for those who were absent)

Homework: • Textbook questions: 22, 25, 27, 29, 30, 31, 32, 36,

37, 38, 40, 42, 43, 46, 47, 50• W/S booklet: worksheet 14.1 and 14.2

Prac: Galvanic cellsAIM: To construct and observe several galvanic cells and to use experimental data to create an electrochemical series

MATERIALS:Zn(s)/Zn2+

(aq) Mg(s)/Mg2+(aq) Fe(s)/Fe2+

(aq) Cu(s)/Cu2+(aq)

RESULTS:

DISCUSSION:1. Create an electrochemical series from your data (order reduction equations for each

conjgate redox pair from strongest oxidant to weakest) 2. Compare with the electrochemical series in the textbook and comment.

Cell Cell Voltage Half cell equations Oxidant Reductant

Fe(s)/Fe2+(aq) 0.56V Cathode (+): Cu2+

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

Cu2+(aq)

Cu(s)/Cu2+(aq)

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

Holiday homework

• Textbook questions for redox chem: 22, 25, 27, 29, 30, 31, 32, 36, 37, 38, 40, 42, 43, 46, 47, 50

• W/S booklet: worksheet 7 – 10 complete and corrected