Preparation of azobenzene/olygoethyleneglycol-functionalized gold nanoparticles: AuNP-Gn@ AzoOEG/OEG

Oxidation is

(i) the gain of oxygen (ii) the loss of hydrogen

(iii) the loss of electrons (de-electronation) (iv) the increase of O.N.

Reduction is

(i) the loss of oxygen (ii) the gain of hydrogen (iii) the gain of electron (iv) the decrease in O.N.

2 O.N.Fe0 Fe2 ⁺ 2e⁻

= → + +

• Fe loses electrons

• There is increase in O.N.

• Hence Fe is said to be oxidized

• It is a source of electron hence it can act as a reducing agent (R.A.)

• Any species that can be oxidized is a reducing agent (R.A.)

• ²

O.N. 2 0

Cu ⁺ 2e Cu

+ + →

• Cu²⁺ gains electron

• There is decrease in O.N.

• Hence Cu²⁺ is said be reduced.

• Since it can gain electrons, hence it can act as an oxidizing agent (O.A.)

• Any species that can be reduced is an oxidizing agent (O.A.), or oxidant. Above examples represent half reactions.

A complete reaction showing oxidation and reduction together is called a Redox reaction.

R O

Fe + Cu²⁺ Cu + Fe²⁺

Now that you are clear on what is oxidation – reduction, we are now in the right position to know how to balance a redox (oxidation – reduction) reaction. This is important because many of the problems of stoichiometry would be based on such redox eractions.

Oxidation State/Oxidation Number

There are several chemical reactions in which oxidation – reduction takes place. Oxidation refers to a reaction in which electrons are removed from an atom and reduction refers to a reaction in which electrons are added to an atom. To describe these changes, the concept of oxidation state becomes necessary. For ionic species, the charge on each ion is said to be the oxidation state for that atom. For example in NaCl, Na exists as Na⁺ and Cl exists as Cl⁻. Therefore the oxidation state of Na in NaCl is +1 and that of Cl⁻ is –1. But in covalent molecules, the charge on an atom would be so small that sometimes it becomes impossible to calculate the exact charge on each atom of a molecule. Therefore, the Oxidation State (O.S.) or Oxidation Number (O.N.) is defined as the charge, an atom would have in a molecule if all the bonds associated with this atom in the molecule are considered to be completely ionic. For example in H2O there are two O–H bonds. If we assume both the O–H bonds to be completely ionic, then each H would possess a charge of + 1, while O possess a charge of –2. This is because oxygen is more electronegative than hydrogen. On the other hand, in H2O2 there are two OH bonds and one O–H bond (H–O–O–H). Considering each O–H bond to be ionic both the oxygen atoms acquire a charge of –1 and both the H, +1. This is because O – O bond can not be assumed to be ionic as both the atoms have the same electronegativity.

To calculate the oxidation state of an element in a molecule you need not always know the structure of the molecule. There are certain set of rules used to assign oxidation states in polyatomic molecules.

(i) The O.S. of all elements (in any allotropic form) is zero.

(ii) O.S. of oxygen is –2 in all its compounds except peroxides like H2O2 and Na2O2 where it is –1 and superoxides like KO² where it is –1/2. In OF² and O²F², oxygeneous +2, +1 respectively.

(iii) O.S. of hydrogen is +1 in all of its compounds except those with the metals where it is –1.

(iv) O.S. of all alkali metals is +1 and alkaline earth metals is +2 in all their compounds.

(v) O.S. of all the halogens is –1 in all their compounds except where they are combined with an element of higher electronegativity. Since fluorine is the most electronegative of all elements, its O.S. is always –1.

Illustra tions

Illustration 17

Calculate the O.S. of the atoms in the following species:

(i) In ClO¯ (ii) NO2¯ (iii) NO3¯ (iv) CCl4 (v) K2CrO4 (vi) KMnO4

Solution

(i) In ClO¯, the net charge on the species is –1 and therefore the sum of the oxidation states of Cl and O must be equal to –1. Oxygen will have an O.S. of –2 and if the O.S. of Cl is assumed to be ‘x’ then x – 2 should be equal to –1.

∴ x is + 1

(ii) NO2¯: (2 × –2) + x = –1 (where ‘x’ is O.S. of N)

∴ x = + 3

(iii) NO3¯^: : x + (3 × –2) = –1 (where ‘x’ is O.s. of N)

∴ x + 5

(iv) In CCl4, Cl has an O.S. of –1

∴ x + 4 × –1 = 0

∴ x = + 4 (where ‘x’ is O.S. of C)

(v) K2CrO4: K has O.S. of + 1 and O has O.S. of –2 and let Cr has O.S. ‘x’ then, 2 × + 1 + x + 4 × –2 = 0

∴ x = + 6

(vi) KMnO4 : + 1 + x + (4 × –2) = 0

∴ x = + 7 (where x is O.S. of Mn).

Balancing Redox Equations Some examples of redox reactions are

(a) R

O

Sn²⁺ + 2Hg²⁺ Hg₂²⁺ + Sn⁴⁺

(b)

R O

MnO₄^¯ + 5Fe²⁺ + 8H⁺ 5Fe³⁺ + Mn²⁺ + 4H₂O

(c)

R O

Cr₂O₇²⁺ + 6Fe²⁺ + 14H⁺ 6Fe³⁺ + 2Cr³⁺ + 7H₂O

(d) R

O

3Cu + 2NO₃¯ + 8H⁺ 2NO + 3Cu²⁺ + 4H₂O

If one of the half reactions does not take place, other half will also not take place. We can say oxidation and reduction go side by side.

R

3Cl₂ + 6OH¯ 5Cl¯ + ClO₃^¯ + 3H₂O

0 –1 +5

O.N. =

In this we find the Cl2 has been oxidized as well as reduced. Such type of redox reaction is called Disproportionation reaction. Examples are

R O

2Cu⁺ Disproportionates

Cu + Cu²⁺

R

2HCHO + OH¯ CH₃OH + HCOO¯

O

R O

4ClO₃¯ 3ClO₄¯ + Cl¯

R O

3MnO₄^2- + 2H₂O MnO₂ + 2MnO₄^- + 4OH

-How to Balance a Redox Reaction Ion Electron Method

This method involves the following steps

(i) Divide the complete equation into two half reactions, one representing oxidation and the other reduction.

(ii) Balance the atoms in each half reaction separately according to the following steps (a) First of all balance the atoms other than H and O.

(b) In a reaction taking place in acidic or neutral medium, oxygen atoms are balanced by adding molecules of water to the side deficient in oxygen atoms while hydrogen atoms are balanced by adding H⁺ ions to the other side deficient in hydrogen atoms.

On the other hand, in alkaline medium (OH¯), for every excess of oxygen atom on one side is balanced by adding one H2O to the same side and 2OH¯ to the other side. In case hydrogen is still unbalanced, then balance by adding one OH¯, for every excess of H atom on the same side and one H²O on the other side.

(c) Equalize the charge on both sides by adding suitable number of electrons to the side deficient in negative charge.

(iii) Multiply the two half reactions by suitable integers so that the total number of electrons gained in one half reaction is equal to the number of electrons lost in the other half reaction.

(iv) Add the two balanced half equations and cancel any term common to both sides.

Illustra tions

Illustration 18

(a) NO₃⁻ +H S₂ acidic medium^H+ →HSO₄⁻ +NH₄⁺ (b) Fe+N H2 4 →alkaline medium^OH− Fe(OH)2+NH3

Solution

(a) Step 1 N⁵⁺ + 8e → N^3–

Step 2 S^2– → S⁶⁺ + 8e

Step 3 NO3¯ + H2S → NH4+ + HSO4¯

Step 4 No other atom (except H and O) is unbalanced therefore no need for this step.

Step 5 Balancing O atom. This is made by using H²O and H⁺ io ns. Add desired molecules of H2O on the side deficient in O atom and double H⁺ on opposite side. Therefore

3 2 2 4 4

NO⁻ +H S+H O →NH⁺ +HSO⁻ +2H⁺ Step 6 Balancing charge by H⁺

3 2 2 4 4

NO⁻+H S+H O+3H⁺ →NH⁺ +HSO⁻ +2H⁺

∴ Balanced equation is

3 2 2 4 4

NO⁻+H S+H O+H⁺ →NH⁺ +HSO⁻

(b) Step 1 Fe → Fe²⁺ + 2e Step 2 N^22– + 2e → 2N^3–

Step 3 Fe + N²H⁴ → Fe(OH)² + 2NH³

Step 4 No other atom (except H and O) is unbalanced and therefore no need for this step.

Step 5 Fe+N H2 4 +4OH⁻ →Fe(OH)2 +2NH3 +2H O2

Step 6 Balance charge by H⁺

∴ Fe + N²H⁴ + 4OH¯ + 4 H⁺ → Fe(OH)² + 2NH³ + 2H²O Finally Fe + N2H4 + 2H2O → Fe(OH)² + 2NH3

Practice Exercise

11. Balance the redox equation, HNO3 + H2S → NO + S 12. Balance the following redox equation,

FeC2O4 + KMnO4 + H2SO4 → Fe2(SO4)3 + CO2 + MnSO4 + K2SO4. Answers

11. 2HNO3 + 3H2S → 3S + 2NO + 4H2O

12. 10FeC2O4 + 6KMnO4 + 24H2SO4 → 5Fe2(SO4)3 + 6MnSO4 + 3K2SO4 + 20CO2 + 24H2O Oxidation Number Method

This method is based on the principle that the number of electrons lost in oxidation must be equal to the number of electrons gained in reduction. The steps to be followed are

(i) Write the equation (if it is not complete, then complete it) representing the chemical changes.

(ii) By knowing oxidation number of elements, identify which atom(s) is(are) undergoing oxidation and reduction. Write down separate equations for oxidation and reduction.

(iii) Add required electrons on the right hand side of oxidation reaction and on the left hand side of reduction reaction. Care must be taken to ensure that the net charge on both the sides of the equation is same.

(iv) Multiply the oxidation and reduction reactions by suitable numbers to make the number of electrons lost in oxidation reactions equal to the number of electrons gained in reduction reactions.

(v) Transfer the coefficient of the oxidizing and reducing agents and their products to the main equation. By inspection, arrive at the co-efficient of the species not undergoing oxidation or reduction.

Illustra tions

Step (vi) Making provision of KCl and H2O in the product: Since the reactant has 7 oxygen atoms in the product 7H2O must be present. For accounting 14 hydrogen atoms of water in the product, the reactants must have 12 HCl molecules (the only H containing species). For accounting the 2K atoms and 14 – 12 = 2 additional Cl atoms in the reactant, the product must have 2KCl. Hence the balanced equation is K2Cr2O7 + 14HCl → 2KCl + 2CrCl³ + 7H2O + 3Cl2

Practice Exercise

13. Balance the following oxidation-reduction equationm KMnO4 + KCl + H2SO4 → MnSO4 + K2SO4 + H2O + Cl2

14. Balance the following redox equation K2Cr2O7 + HCl → KCl + CrCl3 + Cl2 + H2O Answers

13. 2KMnO4 + 10KCl + 8H2SO4 → 2MnSO4 + 6K2SO4 + 8H2O + 5Cl2

14. K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 3Cl2 + 7H2O

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