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Oxidation–Reduction (Redox)

Reactions

Dr. F. Akinwunmi

CHM 101

• Oxidation–reduction reactions are reactions involving a transfer

of electrons from one species to another.

• Example: When an iron nail is dipped into a blue solution of copper (II)

sulphate (Figure 1).

Fig 1. [Reaction of Fe with Cu2+(aq)]

• The iron nail becomes coated with a reddish-brown tinge of metallic copper.

Oxidation Numbers

• The oxidation number (or oxidation state ) of an atom in a substance is defined as the actual charge of the atom if it exists as a monatomic ion, or a hypothetical charge assigned to the atom in the substance by simple rules.
• An oxidation–reduction reaction is one in which one or more atoms change oxidation number, implying that there has been a transfer of electrons.
• Consider the combustion of calcium metal in oxygen gas (Figure 2).

2Ca( s ) + O 2 ( g ) → 2CaO( s )

• This is an oxidation–reduction reaction
• Ca and O in O 2 have oxidation numbers of zero
• The oxidation number of Ca in CaO is +2 (the charge on Ca2+), and that of O

in CaO is -2 (the charge on O2-)

i.e

 Calcium atom in the metal loses two electrons to form Ca2+^ ions, and each oxygen atom

in O 2 gains two electrons to form O2-^ ions.

 The net result is a transfer of electrons from calcium to oxygen, so this reaction is an

Figure 2 oxidation–reduction reaction

• Calcium has increased in oxidation number from 0 to +2. (Each calcium atom loses two electrons.) calcium is said to have been oxidized.
• Oxygen, has decreased in oxidation number from 0 to -2. (Each oxygen atom gains two electrons.) oxygen is said to have been reduced.
• An oxidation–reduction reaction always involves both oxidation ( the loss of electrons ) and reduction ( the gain of electrons ).
• Note: Oxidation numbers are represented as +n or – n while charges are represented as n+^ or n-.
• Another example is Burning of Calcium in Chlorine (Fig 3)

Fig 3: Burning of Calcium in Chlorine

The chemical equation is

 The calcium atom is oxidized, because it increases in oxidation number (from 0 to +2).

 Chlorine is reduced; it decreases in oxidation number from 0 to -1.

• This is clearly an oxidation–reduction reaction that does not involve oxygen.

o Formerly, the term oxidation meant “reaction with oxygen.”

DETERMINATION OF OXIDATION NUMBERS

PRACTICE QUESTION

• 1. Determine the O.N of N in i) N 2 O 4 ii) NO 3 - iii) NH Worked Examples
• 1. Determine the O.N of Cl in HClO
  • 1 i) N 2 O 4 = 2x + 4(-2)= Let the O.N of N be x
    • = 2x - 8 =
      • 2x = + - X = +8/2 = +
  • ii) NO 3 - = x + 3(-2) = - - x – 6 = - - x = -1 + - x = +
  • iii) NH 3 = x + 3(1) = - x + 3 = - x = - - 2 HClO - Let Cl be y, HClO 4 +1 + y + 4(-2) = - +1 + y - 8 = - y + 1 – 8 = - y – 7 = - y = + - a) Ca(ClO 3 ) Calculate O.N of the underline elements in each of the following: - c) H 3 PO b) Cr 2 O 7 2-

OXIDIZING AND REDUCING AGENTS

• Oxidizing Agents [O.A] are species that

 Oxidize other substances

 Contains atoms that are reduced

 Gain or appear to gain electrons.

• Reducing Agents [R.A] are species that

 reduce other substances

 Contains atoms that are oxidized

 lose or appear to lose electrons.

 There is always a reduction for every oxidation.

 The two species involved in a redox reaction acts as agents (O. A and R. A) for each other.

• Therefore,

 any spp that undergoes oxidation is a reducing agent or reductant and

 any spp that undergoes reduction is a oxidizing agent or oxidant

 Thus, a redox reaction may also be regarded as an interaction between an O.A and R.A.

An oxidizing agent is an electron acceptor which

causes a co-reactant to be oxidized in a reaction e.g O2,

Cl2, Br2, acidified KMnO4 and K2Cr2O

A reducing agent is an electron donor which causes a

co-reactant to be reduced in a reaction e.g metals,

metallic ions, Fe2+, Pb2+^ e.t.c, LiAlH4, NaBH4.

1. 2FeBr3 + 3Cl2 → 2FeCl3 + 3Br
   • Ionically, 2[Fe3+^ + 3Br-] + 3Cl2 → 2[Fe3+^ + 3Cl-] + 3Br
   • Fe3+ acts as spectator ions and do not take part in the electron transfer. Hence its cancellation allows for easy identification of the O.A (Cl2) and R.A (Br-) Thus, the ionic equation becomes: 2Br- + Cl2 → 2Cl- + Br

PRACTICE QUESTIONS

Write each of the following formular units equations as a net equation if the two differ. Which ones are redox reactions? For the redox reactions, identify the O.A, R.A, Spp oxidized and reduced:

a. 2AgNO3 + Cu → Cu(NO3)2 + 2Ag b. 4KClO3 → KCl + 3KClO c. 3AgNO3 + K3PO4 → Ag3PO4 + 3KNO

BALANCING REDOX EQUATIONS

• The basis for balancing redox equations is that the total increase in O.N for oxidation must be equal to the total O.N for reduction.
• All balanced equations must satisfy two criteria namely:

 There must be mass balance i.e the same no of atoms of each kind must appear in reactants and products  There must be charge balance. i.e the sum of actual charges on the LHS and RHS of the equation must be equal to each other.

Two methods are employed viz:

 Half reaction method

 Change in O.N method

Worked Examples

1. Write a balanced net ionic equation for the reaction between I 2 and sodium thiosulphate Na 2 S 2 O 3
   • Iodine I 2 oxidises thiosulphate S 2 O 3 2-^ to tetrathionate S 4 O 6 2-^ ions but it is reduced to iodide I-. Thus, I2 is the O.A while the thiosulphate is the R.A.
   • The unbalanced equation: I 2 + S 2 O 3 2-^ → I-^ + S 4 O 6 2- I 2 → I-^ (Reduction half reaction) Balance the no of atoms: I 2 → 2I- Balance the charges: I 2 + 2e- → 2I-^ (Balanced reduction half reaction) S 2 O 3 2-^ → S 4 O 6 2-^ (Oxidation half reaction) Balance the no of atoms: 2S 2 O 3 2-^ → S 4 O 6 2- Balance the charges: 2S 2 O 3 2-^ → S 4 O 6 2-^ + 2e- (Balanced oxidation half reaction) Add the balanced half-reactions algebraically to obtain the overall equation: I 2 + 2e- → 2I- 2S 2 O 3 2-^ → S 4 O 6 2-^ + 2e-

Overall balanced equation: I 2 + 2S 2 O 3 2-^ → 2I-^ + S 4 O 6 2-

1. Permanganate ions oxidize iron II to iron III in sulphuric acid solution but it is reduced to manganese II ions. Write the net ionic equation for this reaction. Fe2+^ + MnO 4 -^ → Fe3+^ + Mn2+ Atoms are balanced Fe2+^ → Fe3+^ (Oxidation half reaction) Balance the charges: Fe2+^ → Fe3+^ + e- (Balanced oxidation half reaction) MnO 4 -^ → Mn2+^ (reduction half reaction) Balance the no of atoms: MnO 4 -^ + 8H+^ → Mn2+^ + 4H 2 O Balance the charges: MnO 4 -^ + 8H+^ + 5e- → Mn2+^ + 4H 2 O (Balanced oxidation half reaction) Since redox reactions are complimentary , the no of electrons transferred must be equal. Thus, balance the electron transfer by multiplying the balanced oxidation half reaction by 5 and balanced reduction half reaction by 1. Net ionic equation by algebraic addition: 5Fe2+^ → 5Fe3+^ + 5e- MnO 4 -^ + 8H+^ + 5e- → Mn2+^ + 4H 2 O 5Fe2+^ + MnO 4 -^ + 8H+^ → 5Fe3+^ + Mn2+^ + 4H 2 O

PRACTICE QUESTIONS

1. Balance the following equations in acidic medium , state the O.A, R.A, species oxidized and reduced a. Fe2+^ + Cr 2 O 7 2-^ → Fe3+^ + Cr3+

b. Cr 2 O 7 2-^ + SO 2 → Cr3+^ + SO 4 2- c. MnO 4 -^ + SO 3 2-^ + H+^ → Mn2+^ + SO 4 2-^ + H 2 O d. Zn + NO 3 -^ → Zn2+^ + NH 4 +

1. Balance the following equations in basic medium , state the O.A, R.A, species oxidized and reduced a. MnO 4 -^ + H 2 O + NO 2 → MnO 2 + NO 3 -^ + OH- b. CrO 4 2-^ + H 2 O + HSnO 2 -^ CrO 2 -^ + OH-^ + HSnO 3 -

CHANGE IN OXIDATION NUMBER METHOD

This method is based on equal total increases and decreases in Oxidation Number.

General Procedure:

1. Write as much of the overall unbalanced equation as possible
2. Assign O. N to find the elements that undergo changes in O.N
3. Draw a bracket to connect atoms of the elements that is oxidized and reduced
4. Insert coefficients into the equation to make the total increase and decrease in O.N equal
5. Balance the other atoms by inspection