THIRD TERM E-LEARNING NOTES CHEMISTRY SS 2 SCHEME OF WORK WEEK 2

 

WEEK 1

 

Revision of last term’s work.

 

WEEK 2

 

TOPIC: OXIDATION – REDUCTION (REDOX) REACTION

 

CONTENT: 1. OXIDATION AND REDUCTION

 

1. REDOX REACTIONS

 

1. OXIDATION NUMBERS OF CENTRAL ELEMENT IN SOME COMPOUNDS

 

1. CONNECTION OF OXIDATION NUMBER WITH IUPAC NAME

 

1. OXIDISING AND REDUCING AGENTS

 

1. REDOX EQUATION

 

PERIOD 1: OXIDATION AND REDUCTION

 

Redox is a short form for reduction and oxidation reactions. The two reactions are opposing and complementary and they occur simultaneously. Redox has the following definitions

 

Definition of Oxidation reactions with examples

In terms of addition of oxygen: Oxidation is the addition of oxygen or removal of hydrogen from a substance. Any reaction where there is addition of oxygen to a reactant is regarded as Oxidation.

Some examples are:

 

(i) 2Mg(s) + O2(g) 2MgO(s) oxidation of magnesium

 

(ii) C(s) + 2ZnO(s) CO2 (g) + 2Zn(s)

 

In this reaction, oxygen (O) was added to carbon (C) to form Carbon (iv) oxide (CO2). That is carbon was oxidized to Carbon (iv) oxide.

 

In terms of hydrogen removal: Oxidation is the removal of hydrogen from a compound e.g. (i) combustion of hydrogen sulphide

2H2S(g) + O2(g) 2H2O(l) + 2S(s)

 

H2S(g) + Cl2(g) 2HCl2(g) + S(s)

 

In both reactions, hydrogen sulphide was oxidized to Sulphur. The second reaction shows that oxidation can occur without the involvement of oxygen

 

In terms of loss of electron: Oxidation is the loss of electron.

4Na(s) + O2(g) 2Na2O(s)

 

Na Na+ + e- the loss of electron by sodium atom (Na) to form sodium ion (Na+) in sodium oxide

 

Definition of reduction with examples

Definitions of reduction are the opposites of the definitions of oxidation.

 

Removal of oxygen

E.g. Cu(s) + ZnO(s) CuO(s) + Zn(s)

 

In this reaction zinc oxide was reduced to zinc

 

Addition of hydrogen

e.g. H2S(g) +Cl2(g) 2HCl (g) + S(s)

 

chlorine is reduced to hydrogen chloride.

 

Gain of electrons

e.g. 4Na(s) + O2(g) 2Na2O(s) i.e oxygen gained electron

 

02 + 4e- 202-

 

In terms Of Decrease In Oxidation Number

Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)

 

The oxidation number of copper decreased from +2 in copper (ii) salt to zero in copper.

 

Reduction occurs when electro negative element is added

Redox Reaction And Electron Transfer: considering all the reactions used as examples so far, it would be seen that both oxidation and reactions occur together see illustration below:

 

 

Because 2Na 2Na+ + 2e- (loss of electron is oxidation)

 

O2 + 4e- 2O2- (gain of electron is reduction)

 

The above electron transfer equations can be written for most redox reactions.

 

More examples are

 

2KI + Fe2(SO4)3(aq) I2(aq) + K2SO4(aq) + 2FeSO4(aq)

Ionically

 

2I-(aq) I2 (aq) + 2e- (loss of electrons in oxidation)

 

The above equation is the redox reaction .Each half reaction cannot occur alone because an atom cannot receive or donate electron without another atom around that must donate or receive the electron.

 

Evidence of electron transfer in redox reactions can be proved using a simple electro chemical cell.

 

EVALUATION

 

1. Define reduction in terms of electron transfer.

 

1. Explain the experiment which proves that there is electron transfer in redox reaction

 

PERIO 2: OXIDATION NUMBERS OF CENTRAL ELEMENTS SOME COMPOUNDS

 

Oxidation number is the electrical change assigned to an atom in accordance with some prescribed set of rules

 

RULES FOR DETERMING OXIDATION NUMBER

 

The following sets of rules are used to determine the oxidation state or number of substances.

 

The oxidation number of an element in an un-combined state is zero, for example the oxidation number of Hydrogen, Oxygen and Sodium atom in a free state or un-combined with another element is zero.

In most compounds containing hydrogen, the oxidation number of hydrogen is +1 except in hybrids where it is -1

Electrons shared between two unlike atoms are counted with the more electro negative atom. For example, in water molecule the electron are regarded as being with the more electronegative oxygen. Thus in H2O, each hydrogen atom is in +1 oxidation state, and the oxygen atom is in -2 oxidation state.

In most compounds containing oxygen, the oxidation number of each oxygen atom is -2 except in peroxides where it is -1. E.g. hydrogen peroxides, (H2O2), sodium peroxide (Na2O2), barium peroxide (BaO2)

The oxidation number of each halogen is -1, except when bonded with fluorine which is the most electro negative. For instance in IF7, each fluorine atom is in oxidation state of -1 and iodine is in oxidation state of +7

The sum of all the oxidation numbers of elements in a compound is zero and with this simple relationship, the oxidation number of each element in a compound can be calculated.

In simple ions, i.e. ions containing one atom, the oxidation number is equal to the change on the ion. For example the ion Al3+ has on the oxidation number of +3, the ion Cu2+ has the oxidation number is -2

What about complex ion? In a complex ion (i.e. ion consisting of more than one element) the oxidation is the algebraic sum of all the oxidation numbers of all the elements in the ion. This will be the sign on the ion and of the same size. For example, in tetraoxosulphate (vi) ion (SO42-) the overall charge is -2 which its oxidation number is. It is obtained as follows:

• 4 =

 

+4 x (-2) = -2

 

For other ions such as OH-, NO3‑, NO2-3, SO2-3, PO3-4, NH4+ their oxidation numbers are -1, -1, -2, -3 and +1 respectively.

 

Calculation of oxidation number

 

With the rules for determining oxidation numbers in our memory, it is possible to calculate the oxidation number of any given element in an ion or a compound .

 

Worked examples

 

Calculate the oxidation number of chromium in

Solution

 

This can be solved by simple linear equation by making the unknown subject of formula. We have 4 oxygen and its oxidation is -2

 

Let the unknown (oxidation number of Cr be x)

 

But = -2 (because the sign on the ion is -2)

 

x + (-24) = -2

 

x – 8 = -2

 

x = -2 + 8

 

x= +6

 

Thus the oxidation number of Cr in is +6

 

Note: oxidation number is never written as a neutral number, i.e. it is either written as a neutral number, i.e .It is either written as a positive or negative number.

 

Calculate the oxidation number of sulphur in H2SO4

Solution

 

The oxidation number (O.N) of hydrogen is +1 and the two hydrogen atoms will give+2. The four oxygen atoms will give (-2 x4) = -8, since -2 is the oxidation number (O, N) of an atom of oxygen

 

2×(+1) + x + (4 × -2) =0

 

2 + x + -8 =0

 

=0 +8 -2

 

=+8-2

 

=+6

 

CONNECTION OF OXIDATION NUMBERS WITH IUPAC NAMES

 

From all the examples considered so far, it can be seen that the names of most inorganic compounds reflect.

 

The oxidation numbers of the elements with variable oxidation numbers e.g. iron (ii) ion, in this, oxidation number of Fe is +2

 

the number of oxygen atoms in the compound e.g.

CO32+ is trioxo (i.e. three oxygen atoms)

 

SO24- is tetraoxo (i.e. four oxygen atoms

 

Cr2O72- is heptaoxo (i.e. seven oxygen atoms)

 

Hence, if the formula of a substance is known, the IAUPAC name can be written by calculating the oxidation number of the central atom using the fixed oxidation numbers of the other atoms like oxygen, hydrogen and halogen.

 

For example:

[mediator_tech]

KMnO4 has oxidation number K= +1 and O4= 4 × -2

 

i.e. +1+x + -8 = 0 where x is the oxidation number of Mn, therefore= 0 + 8 – 1 = +7. Therefore, the name is potassium tetraoxomaganate(VII)

 

The Following Prefixes and Suffixed Should Be Noted

 

PREFIX SUFFIX

 

1= mono- = sulphate (IV)

 

2= di- = sulphate (VI)

 

3.= tri – =maganate (VII)

 

1. =tetra- = phosphate(V)

 

5= penta =Chlorate(V)

 

6= hexa- = Carbonate(IV)

 

7= hepta – = dichromate (vi)

 

1. =octa- = nitrate (v)

 

9.= nona – BrO- = bromate (i)

 

1. = deca IO- =iodate (i)

 

• iodate (V)

 

. ClO– = Chlorate (V)

 

Worked examples

 

Calculate the oxidation number of manganese in KMnO4 and hence name the compound in IUPAC nomenclature

Solution

 

Let unknown (oxidation number of Mn) = x

 

K= +1, O = -2

 

KMnO4= 0

 

1 + x + (-2 x4) =0

 

1 + x -8 =0

 

X – 8 = -1

 

X= -1 +8

 

X= +7

 

The oxidation number of manganese is +7 and the name of the compound is potassium tetraoxomanganate(vii) . The tetraoxo means there are 4 oxygen atoms in the compound.

 

Calculate the oxidation state of chromium in and give its IUPAC name

Let the unknown = x

 

O= -2 and = -2

 

2x + (-2 x 7) = -2

 

2x – 14 =-2

 

2x = -2 +14

 

2x =+ 12

 

X=+6

 

The oxidation state of chromium is +6

 

There are two chromium atoms and there are seven oxygen atoms, there is a charge on the compound making it an ion. Thus, the name is heptaoxodichromate(VI) ion.

[mediator_tech]

EVALUATION

 

What is the oxidation number of manganese in each of the following species? (i) MnCl2 (b) MnO2 (c) MnO-4

Determine the oxidation number of chromium in Cr2O72-, then name in IUPAC.

PERIOD 3: OXIDIZING AND REDUCING AGENTS

 

Oxidising agent: This is a substance which adds oxygen or removes hydrogen form a substance or a substance which accepts electron, i.e. electron acceptor. Consider the following reactions.

Oxidation in terms of addition of oxygen: oxidizing agent will add oxygen to a substance e.g.

C(s) +2ZnO(s) CO2(g) + 2Zn (s)

 

ZnO is the oxidising agent because it added oxygen to (i.e. oxidise) carbon to form CO2

 

Oxidation in term of removal of hydrogen: oxidizing agent will add oxygen to a substance e.g

2H2S (g) + O2 (g) 2H2O (l) + 2S(s) or

 

H2S (g) + Cl2 (g) 2HCl (g) + S(s)

 

Oxygen and chlorine are the oxidizing agents because they removed hydrogen from H2S to form sulphur

 

Oxidation in term of loss of electrons: oxidizing agent will gain electrons e.g

2FeCl2(s) + Cl2 (g) 2FeCl3(s)

 

Fe2+ Fe3+ + 2e-

 

Cl2 + 2e- 2Cl-

[mediator_tech]

The oxidizing agent is chlorine (Cl2) because it gained electrons lost by Fe2+

 

Other means of identifying an oxidising agent is the substance.

 

(i) Which is reduced

 

(ii) Whose oxidation number has decreased

 

Reducing agent: This is a substance which adds hydrogen to another substance or removes oxygen from that substance or this substance which donates election, i.e. an electron donor.

Consider the following reactions

 

Adds hydrogen e.g hydrogen sulphide reduces chloride to hydrogen chloride in

H2S(g) + Cl2(g) 2HCl (g) + 2S(s)

 

Removes oxygen e.g. Carbon in

C(s) + ZnO(s) CO2(g) + 2Zn(s)

 

Has its oxidation number increases e.g. in

Zn(s) + CuSO4 (aq) ZnSO4 (a q) + Cu(s)

 

Zinc is the reducing agent because its oxidation number increased from zero in zinc to +2 in zinc tetraoxosulphate (vi) that is, it was oxidized.

 

The Table below give some common reducing agent and oxidizing agents.

 

OXIDIZING AGENTS

 

Oxygen

 

O2

 

Hydrogen peroxide

 

H2O2

 

Chlorine (and other halogen)

 

Cl2

 

Acidified potassium tetraoxomanganate (viii)

 

KMnO4

 

Acidified potassium heptaoxchromate (vi)

 

K2Cr2O7

 

Hot concentrated tetraoxsulphate (vi) acid

 

H2SO4

 

Concentrated trioxonitrate(v) acid

 

HNO3

 

Silver salt (and metals low in the series e.g cu)

 

Ag+

 

REDUCING AGENTS

 

Hydrogen

 

H2

 

Carbon

 

C

 

Potassium

 

KI

 

Hydrogen sulphide

 

H2S

 

Sulphur(iv) oxide

 

SO2

 

Iron(II) salt

 

SO2

 

Sodium (and other reactive metals)

 

Na

 

Ammonia

 

NH3

 

Generally, metals that are highly electro positive e.g. Na, K and Mg lose electrons easily and so are good reducing agents (they release their electrons to another thus reducing it). Non-metals that are highly electro negative e.g. O, Cl and F are good oxidizing agents (they easily accept electrons from another thus oxidizing it).

 

TEST FOR OXIDIZING AGENTS

 

Strong reducing agents like iron (ii) salt and hydrogen sulphide are used.

 

Green aqueous solution of iron (ii) salt turns brown in the presence of an oxidising agent .

Fe2+ Fe3+ +e-

 

Blue brown

 

Note: Freshly prepared iron(II) salt is used.

 

(ii) Hydrogen sulphide forms deposit of sulphur when passed through the solution of an oxidising agent.

 

S2- 2S(s) + 2e-

 

TEST FOR REDUCING AGENTS

 

Strong oxidizing agents like acidified potassium tetraoxomagnate(viii) and acidified potassium heptaoxo chromate (vi) are used

 

The purple colour of acidified potassium tetraoxomanganate (vii) is turned colourless in the presence of a reducing agent.

 

 

The orange colour of acidified potassium heptaoxodichromate(VI) is turned green in the presence of a reducing agent.

 

 

 

EVALUATION:

 

1. define reducing agent in terms of oxidation number.

 

1. Identify the oxidizing and reducing agent in the reaction below:

 

Zn(s) + CuSO4 (aq) ZnSO4 (a q) + Cu(s)

 

PERIOD 4: BALANCING IONIC (REDOX) EQUATION

 

In balancing ionic equations, the following steps can be taken.

 

Examine the given equation carefully

Balance the number of atoms first by adding the correct number of H+ and H2O on the appropriate sides of the equation when an acid is involved.

Next, balance the number of charges by adding appropriate number of electrons on the right side of the equation.

Lastly, count the atoms on both sides of the equation. If they are equal, then the equation is balanced in atoms (or molecules); also check the charge; if equal, then it is balanced ironically too.

WORKED EXAMPLES

 

What is the value of n in the following equation

• 8H+ + ne- X2+ + 4H2O

 

Solution

 

• 8H+ + ne- X2+ + 4H2O

 

In the given equation, the atoms are balanced but the charges are not. On the left of the equation, we have a net charge of

 

-1 on XO-4 + 8H+ + ne-, i.e. -1 +8 =+7

 

On the right hand side, we have just +2 which is on X2+. Then, for these charges to be equal on both sides of the equation, we have to add 5electrons which will neutralize 5 positive charges from the +7, thereby reducing it to +2, which balances the equation below

 

The balanced equation is

 

• 8H+ + 5e- X2+ + 4H2O

 

WORKED EXAMPLE 2

 

Balance the following Redox equation:

 

• H+(aq) Cr3+(aq) + H2O(l)

 

Solution

 

The first step is to balance the atoms before the charges.

 

On the left hand sides (LHS) there are 7 oxygen atoms on

 

CrO2-7,, but on the right hand side (RHS), there is only one oxygen atom, the oxygen is balanced by writing 7 in front (coefficient) of H2O

 

On the same LHS, there are 2 atoms of chromium, but one on RHS; write 2 in front of the Cr3+ to balance it.

 

The 7 in front of H2O (i.e. 7H2O) now on RHS, gives 14 atoms of hydrogen, but only one on the LHS. Then write 14 as hydrogen, but only one on the LHS. Then write 14 as a coefficient of H+, this gives the equation below which is balanced in terms of atom, but not in charges yet

 

• 14H+(aq) 2Cr3+(aq) + H2O(l)

 

Now to balance the charges

 

LHS: -2 on Cr2O2-7, +14 on 14H+

 

i.e. -2 +14= +12 net charge

 

RHS: +3×2= +16 net change

 

Add 6 electrons to neutralize the 6 excess positive charges on LHS.

 

This gives the balanced equation below:

 

• 14H+(aq) + 6e- 2Cr3+(aq) + H2O(l)

 

EVALUATION

 

1. Define reduction in terms of election transfer

 

1. Using appropriate examples, explain (i) oxidizing agent (ii) reducing agent

 

1. What is the oxidation number of iron in FeO3?

 

1. What is the oxidation number of nitrogen in Al(NO3)3?

 

1. What is the value of x in the following equation?

 

Cr2O72- + 14H+ + Xe- 2Cr3+ + 7H2O

 

GENERAL EVALUATION

 

OBJECTIVE TEST:

 

The oxidation number of manganese in the reaction: MnO-4 + 8H+ + 5e- Mn2+(aq) +4H2O(l)

+3 to +5 (b) +7 to +2 (c) +3 to +2 (d) +7 to +5

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) the oxidizing agent is (a) Fe2O3 (b) CO (c) Fe (d) CO2

In which of the following is oxidation number of sulphur +2? (a) H2SO4 (b) H2SO3 (c) SO2 (d) H2S

Which of the following can act as both oxidizing and reducing agent? (a) KMnO4 (b) H2O2 (c) SO2 (d) H2S

Which of these does not define reduction? (a) it is loss of electron (b) it is addition of hydrogen (c) it is addition of electrons (d) decrease in oxidation number

ESSAY QUESTIONS

 

List the oxidising agent in the following

(i) 2H2O(l) + 2F2(g) 4HF(aq) + O2(g)

 

(ii) 2FeCl2(s) + Cl2(g) 2FeCl3(s)

 

(iii) H2S(g) + Cl2(g) 2HCl(g) +S(s)

 

Consider the following equation:

MnO-4 + 8H+ + Xe- Mn2+ + YH2O

 

State the (i) Values of x and y (ii) oxidation state of Mn in MnO-4

 

Calculate the oxidation number or state of sulphur in the following compounds and give their IUPAC names.

(i) CS2 (ii) H2SO4 (iii) H2S2O7

 

1. Cu2S(s) +O2(g) 2Cu(s) +SO2(g)

 

What is the change in oxidation number of copper in reaction?

 

1. What is the oxidation number of manganese in each of the following species? (i) MnCl2 (ii) MnO2 (iii) MnO4-

 

WEEKEND ASSIGNMENT

 

Read comprehensive chemistry by G N C. Ohia et al

 

PRE READING ASSIGNMENT

 

Read about ionic theory from Melrose chemistry for senior secondary 2 pages 111 to 116

 

WEEKEND ACTIVITY

 

Explain the experiment which proves that there is election transfer in redox reactions

 

REFRENCE TEXTS

 

Comprehensive Chemistry by G N C Ohia et al