SSS 2: REDOX REACTION

REDOX REACTION is a reaction involving both oxidation and reduction occuring at the same time. Most chemical reactions are redox reactions except neutralization reactions and precipitation reactions.

Redox reactions can be explained in four different ways:

1. Redox in terms of electron transfer

2. Redox in terms of change in oxidation number (O.N.)

3. Redox in terms of addition or removal of oxygen and hydrogen 

4. Redox in terms of addition or removal of electronegative and electropositive elements 

      Redox in terms of electron transfer 

(I) Oxidation is the process of electron loss

(II) Reduction is the process of electron gain

(a)   H₂(g) + Cu²⁺ (aq) → Cu(s) + 2H⁺(aq)

O.N.:  0             +2                0             +1 

(I) Cu²⁺ (aq) is undergoing reduction (is reduced). Hence Cu²⁺ (aq) is an oxidizing agent.

(II) H₂(g) is undergoing oxidation (is oxidized), hence H₂(g) is a reducing agent 

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

O.N.: +1 -2        0              0         +1 -1

 

(I) Cl₂(g) is undergoing reduction (is reduced); hence Cl₂(g) is an oxidizing agent.

(II) H₂S(g) is undergoing oxidation (is oxidized); hence, H₂S(g) is a reducing agent.

 Redox in terms of change in oxidation number

(I) Oxidation is Increase in oxidation number

(II) Reduction is decrease in oxidation number

(a) Zn(s) + Cu²⁺ (aq) →Zn²⁺(aq) + Cu(s)

O.N.: 0           +2                  +2              0 

(I) Zn(s) is undergoing oxidation (is increased from 0 to +2), hence Zn(s) is a reducing agent 

(II) Cu²⁺ (aq) is undergoing reduction (ON is reduced from +2 to 0), Hence Cu²⁺ (aq) is an oxidizing agent.

(b) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

O.N.:  0           +1 -1            +2 -1               0

 

(I) Mg(s) is undergoing oxidation (is oxidized from 0 to +2); hence, Mg(s) is a reducing agent.

(II) HCl(aq) is undergoing reduction (is reduced from +1 to 0); hence HCl(aq) is an oxidizing agent.

Redox in terms of addition or removal of oxygen and hydrogen

(I) Oxidation is the addition of oxygen to a substance and the removal of hydrogen 

(II) Reduction is the addition of hydrogen to a substance and the removal of oxygen 

(a) CuO(s) + H₂(g) → Cu(s) + H₂O(g)

(I) CuO is undergoing reduction (is reduced); hence CuO is an oxidizing agent

(II) H₂(g) is undergoing oxidation (is oxidized); Hence, H₂(g) is a reducing agent 

(b) 2Mg(s) + O₂(g) → 2MgO(s)

(I) Mg(s) is undergoing oxidation (is oxidized); hence, Mg(s) is a reducing agent 

(II) O₂(g) is undergoing reduction (is reduced); hence, O₂(g) is an oxidizing agent 

Redox in terms of addition or removal of electronegative and electropositive elements 

(I) Oxidation is the addition of electronegative element to a substance and the removal of electropositive element

(II) Reduction is the addition of electropositive element to a substance and the removal of electronegative element.

(a) 2FeCl₂(aq) + Cl₂(g)  → 2FeCl3 (aq)

(I) FeCl₂ is undergoing oxidation ( is oxidized); hence, FeCl₂ is a reducing agent

(II) Cl₂ is undergoing reduction (is reduced); hence, Cl₂ is an oxidizing agent

                      Oxidizing Agent

An oxidizing agent is a substance which accepts electrons or is an electron acceptor.e.g. Chlorine, Fluorine, Conc. H2SO4, Conc. HNO3, KMnO4, K2Cr2O7 etc

                       Reducing agent 

Reducing agent is a substance which donates electrons or is an electron donor.

Hydrogen, Hydrogen sulphide, active metals etc.

Example: Consider the following redox reactions. State the substance that is

(I) Oxidized (II) Reduced (III) Oxidizing agent (IV) Reducing agent 

(a) Ag+ (aq) + Cl-(aq) → AgCl(s)

(b) CuO(s) + CO(g) → Cu(s) + CO2(g)

(c) Cr2O72- (aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)

                             Solution

(a)    Ag+ (aq) + Cl-(aq) → AgCl(s)

O.N.: +1                -1                +1 -1

The O.N. of Ag+ (aq) = +1

The O.N. of Ag in AgCl (s) = +1

The O.N. of Cl- (aq) = -1

The O.N. of Cl in AgCl (s) = -1

The oxidation numbers of each substance remain unchanged, hence, the reaction is not a redox reaction.

(b)   CuO(s) + CO(g) → Cu(s) + CO2(g)

O.N.: +2 -2       +2 -2         0        +4 -2

(I) CO is oxidized; oxidation number of C in CO increases from +2 to +4

(II) CuO is reduced; oxidation number of Cu in CuO is reduced from +2 to 0

(III) CuO is an oxidizing agent; it undergoes reduction 

(IV) CO is a reducing agent; it undergoes oxidation

(c) Cr2O72- (aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)

O.N.: +6  -2       +1  →   +3           +1  -2

(I) H+ is oxidized; it undergoes oxidation 

(II) Cr2O72- (aq) is reduced

(III)  Cr2O72- (aq) is an oxidizing agent; it undergoes reduction 

(IV) H+  is a reducing agent; it undergoes oxidation

                 Oxidation number 

Oxidation number, also known as oxidation state, is the electrical charge it appears to have as determined by a set of arbitrary rules.

Rules for determining oxidation number 

The set of rules used for the determination of oxidation number are as follows:

1. The oxidation number of all elements in the free state is zero. Eg. Na, Cl2, Mg, P4, H2 etc

2. The oxidation number of an ion represents the charge it carries. For example, the oxidation numbers of the following ions are:

K⁺ = +1, Ca²⁺ = +2, Al³⁺ = +3, Cl⁻ = –1, and S²⁻ = –2.

For compound ions (radicals), the oxidation number is the algebraic sum of the oxidation numbers of all the atoms present in the ion.

Examples include: NH₄⁺ = +1, OH⁻ = –1, NO₃⁻ = –1, SO₄²⁻ = –2, PO₄³⁻ = –3, and MnO₄⁻ = –1.

NOTE:

1. For the hydroxide ion (OH⁻):

Let the oxidation number of oxygen be –2 and that of hydrogen be +1.

Sum of oxidation numbers = (+1) + (–2) = –1,

which corresponds to the charge on the ion.

2. The algebraic sum of the oxidation numbers of all the elements in a compound is always zero.

For example, in AlCl₃:

(O.N. of Al) + 3 × (O.N. of Cl) = O.N. of AlCl₃

(+3) + 3(–1) = 0

3. The oxidation number of oxygen in most compounds is –2, but in peroxides, 

it is –1.

4. The oxidation number of hydrogen is generally +1, but in metal hydrides such as KH, NaH, and CaH₂, it is –1.

Here’s the rewritten explanation with step-by-step solutions for each example:

Example:

Determine the oxidation number of the underlined element(s) in the following compounds or ions:

(I) NH₄⁺ (II) NO₃⁻ (III) HNO₃ 

(IV) SO₄²⁻ (V) H₂SO₄ (VI) KMnO₄ 

(VII) MnO₄⁻ (VIII) K₂Cr₂O₇ (IX) KClO₃ (X) H₂SO₃ (XI) NaMnO₄ (XII) Na₂ZnO₂ (XIII) HCO₃⁻ (XIV) K₃Fe(CN)₆ (where C = –4, N = +3)

Solutions:

(I) NH₄⁺

Let the oxidation number of N = x

x + 4(+1) = +1

x + 4 = +1

x = +1 – 4 = –3

O.N. of N = –3

(II) NO₃⁻

Let O.N. of N = x

x + 3(–2) = –1

x – 6 = –1

x = +5

O.N. of N = +5

(III) HNO₃

Let O.N. of N = x

(+1) + x + 3(–2) = 0

x – 5 = 0

x = +5

O.N. of N = +5

(IV) SO₄²⁻

Let O.N. of S = x

x + 4(–2) = –2

x – 8 = –2

x = +6

O.N. of S = +6

(V) H₂SO₄

2(+1) + x + 4(–2) = 0

2 + x – 8 = 0

x = +6

O.N. of S = +6

(VI) KMnO₄

(+1) + x + 4(–2) = 0

x – 7 = 0

x = +7

O.N. of Mn = +7

(VII) MnO₄⁻

x + 4(–2) = –1

x – 8 = –1

x = +7

O.N. of Mn = +7

(VIII) K₂Cr₂O₇

2(+1) + 2x + 7(–2) = 0

2 + 2x – 14 = 0

2x – 12 = 0

x = +6

O.N. of Cr = +6

(IX) KClO₃

(+1) + x + 3(–2) = 0

x – 5 = 0

x = +5

O.N. of Cl = +5

(X) H₂SO₃

2(+1) + x + 3(–2) = 0

2 + x – 6 = 0

x = +4

O.N. of S = +4

(XI) NaMnO₄

(+1) + x + 4(–2) = 0

x – 7 = 0

x = +7

O.N. of Mn = +7

(XII) Na₂ZnO₂

2(+1) + x + 2(–2) = 0

2 + x – 4 = 0

x = +2

O.N. of Zn = +2

(XIII) HCO₃⁻

Let O.N. of C = x

(+1) + x + 3(–2) = –1

1 + x – 6 = –1

x – 5 = –1

x = +4

O.N. of C = +4

(XIV) K₃Fe(CN)₆

3(+1) + x + 6(–1) = 0 (because [CN]⁻ = –1 for each)

3 + x – 6 = 0

x – 3 = 0

x = +3

O.N. of Fe = +3

                     Class activities

Consider the following redox reactions and identify the substances that are:

(I) Oxidized (II) Reduced (III) Oxidizing agent (IV) Reducing agent

(a) Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

(b) MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g)