Reduction-Oxidation (redox) reactions

Redox reactions are those reactions involving the transfer of electrons from a chemical specie to an other, for example:

Zn + 2H⁺ = Zn²⁺ + H₂

in this reactions Zn losses two electrons to yield Zn²⁺ and H⁺ gaines two electrons to yield H₂. A molecular specie is oxidized when it losses electrons and reduced when it gains electrons, to an oxidation process always corresponds a reduction process.

From the reaction of oxidation of Zn, described above, it is immediate to define the oxidation state or oxidation number of an atom as its net charge. Thus Zn²⁺, Zn, H⁺, H₂, has oxidation numers 2, 0, 1, 0 respectively.

Therefore, in the reaction above, Zn increases its oxidation state (from 0 to +2) it results oxidised. H decreases its oxidation state ( from +1 to 0) and thus it results reduced.

It should be evident that in order to deal with redox reactions it is necessary to define the oxidation state of an atom in a molecule.

For pure ionic substances (such as NaCl) it is easy to assign a positive charge to Na (oxidation number +1) and a negative charge to Cl (oxidation number -1). In other words, for pure ionic substance the oxidation state corresponds to the real charges on the atoms.

In the case of covalent bond among two differents atoms, the oxidation number of an atom is defined as the charge resulting when electrons are assigned to the more electronegative atom. As an example, consider a molecule of hydrochloric acid

H - Cl

Chlorine atom is more electronegative than hydrogen, if we assign the two bonding electrons to Cl (as in a pure ionic bond) we can write:

H⁺¹ Cl^-1

i.e., H brings one positive charge while Cl brings one negative charge. From our definition the oxidation number of hydrogen in the molecule of HCl is +1 while the oxidation number of Cl is -1.

Here follows a set of rules which helps in the calculation of the oxidation state

0. The oxidation state of elements in any allotropic form is zero. This is true also for molecules such as Cl₂ , H₂, etc.

1. In bynary compounds involving differents atoms the element with greater electronegativity is assigned a negative oxidation number equal to its charge in simple ionic compounds of the element (e.g. in the compound PCl₃ the chlorine is more electronegative than the phosphorous. In simple ionic compounds Cl has an ionic charge of 1-, thus, its oxidation state is -1)

2. The oxidation state of O₂ is -2 in almost all its compounds, significant exceptions are peroxides (e.g., H₂O₂ and Na₂O₂) in which oxidation state is  -1.

3. The oxidation state of hydrogen is +1 in all its compounds with the exception of hydrides (e.g., NaH, LiH)  in which oxidation state is  -1.

4. The algebraic sum of the oxidation states of all the elements in a compound is equal to the net charge of the molecule or ion.

5. Alkali metals exhibit only an oxidation state of +1 in compounds

6. Alkaline earth metals exhibit only an oxidation state of +2 in compounds

By using these rules it is possible to determine the oxidation state of other elements in the molecule considered. However, certain elements almost always display a unique oxidation number. For example, the oxidation state of the alkali metals (Group I) in compounds is normally +1 and the oxidation state of the alkaline earths (Group II) is normally +2. The oxidation state of the halogens in compounds is normally -1 except when they are directly combined with oxygen (e.g., NaClO₄).

Example I. Determine the oxidation states (OS) of Cl in the ion ClO^-

According to rule 2 we assign -2 to oxygen and since the algebraic sum of the oxidation states of all the elements in ClO^- must be -1 (rule 4) we write

-2 + OS Cl = -1

OS Cl = -1 + 2 = 1

Example II. Determine the oxidation states (OS) of N in NO₂

By assigning -2 to oxygen (rule 2), according to rule 4 it must be

-2 (2) + OS N = -1

OS N = 4 - 1 = 3

Example III. Determine the oxidation states (OS) of N in NO₃^-

By assigning -2 to oxygen (rule 1), according to rule 4 it must be

-2 (3) + OS N = -1

OS N = 6 - 1 = 5

Example IV. Normally when there is the introduction of oxygen atoms in a carbon compound we think to a redox reaction but this is not always true, consider the following reaction:

2CCl₄ + K₂CrO₄ = 2Cl₂CO + CrO₂Cl₂ + 2KCl

In CCl₄ OS _C is +4 (by assigning -1 to each chlorine atom) in Cl₂CO (by assigning -1 to each chlorine atom and +2 to the oxygen atom) OS _C is still +4. You can verify by applyng the rule above that oxidation states of chromium and chlorine also remain unchanged. Theerefore, the reaction is not a redox reaction.

The half-reaction concept

An important feature of redox reactions is that they can take place also when reactants are separated in space. Let's consider the following reaction:

Zn (s) + Cu⁺⁺ (aq) ===> Cu (s) Zn⁺⁺ (aq)

This reaction can be carried out in a galvanic cell which consist of

As first let's say that no reaction can take place when connections beetween the two solutions and the two metallic rods are missing.

When the connections are working we can observe:

1) the ammeter indicates that electrons are flowing from the zinc rod to the copper rod;

2) Zinc rod dissolves

3) Copper is deposited on the copper rod

These observations can be explained as follows

at zinc rod

Zn = Zn++ + 2e

Zinc rod dissolves and electrons flows through the circuit

at copper rod

2e + Cu++ = Cu

Copper is reduced by electrons coming from the zinc rod and copper is deposited on the copper rod.

It should be evident that the reactions above tend to produce a net positive and negative charges in the left and right beaker respectively. The purpose of the salt bridge is the prevention of the accumulation of charge by allowing the diffusion of negative ions in the left beaker and positive ions in the right beaker. Without salt bridge, the charge accumulating in the beakers immediately stop the reaction.

By concluding the overal reaction above can be separated into two half-reactions:

Zn = Zn⁺⁺ + 2e (oxidation)
2e + Cu⁺⁺ = Cu (reduction)
----------------------------------------------------
Zn (s) + Cu⁺⁺ (aq) ===> Cu (s) Zn⁺⁺ (aq)

Although not all redox reactions can be carried out successfully in a galvanic cell, all redox reactions can be conceptually separated into half-reactions.

Balancing redox reactions

We illustrate here the half-reaction method of balancing redox reactions taking as an example the following reaction:

H₂O₂ + I^- = I₂ + H₂O (occurs in acidic aqueous solution)

Step 1: identify species being oxidized or reduced.

Iodide is oxidized from the -1 state to the zero state while oxygen is reduced from state -2 (in H₂O₂) to state -1 (H₂O).

Step 2. write half-reactions.

I^- = I₂ (oxidation)

H₂O₂ = H₂O (reduction)

Step 3 . balance half-reactions with respect to atoms.

2I^- =I₂ (oxidation)

the reaction of reduction can be balanced with respect to oxygen by writing H₂O₂ = 2H₂O but hydrogen can not be balanced withouth the addition of some form of hydrogen. Since the reaction take place in acidic solution we can write:

2H⁺ + H₂O₂ = 2H₂O

Step 4 . balance half-reactions with respect to charges.

2I^- = I₂ + 2e (oxidation)

2H⁺ + 2e + H₂O₂ = 2H₂O (reduction)

Step 5 . Combine half-reactions to obtain the overal balanced equation. By adding the two half-reactions we get:

2I ^- = I₂ + 2e (oxidation)
2H⁺ + 2e + H₂O₂ = 2H₂O (reduction)
----------------------------------------------------------
2H⁺ + H₂O₂ + 2I^- = 2H₂O + I₂

It should be noted that the effect of adding the two half-reactions is the elimination of electrons from the overal reaction.

Example 1.

C₆H₅CHO + Cr₂O₇⁼ = C₆H₅COOH + Cr³⁺

we will show here that our knowledges on oxidation states are not necessary to balance the reaction but can serves to check our work. We just need to identify and balance the two half-reactions. Let's start with

C₆H₅CHO = C₆H₅COOH

considering that the reaction requires an acidic enviroment the mass balanced half-reaction is

C₆H₅CHO + H₂O = C₆H₅COOH + 2H ⁺

to balance the charges we write

C₆H₅CHO + H₂O = C₆H₅COOH + 2H ⁺ + 2e

This should mean that the oxidized specie (carbon in this case) loss two electrons, let's check:

Benzaldehyde contains six atoms of hydrogen in the state +1 and one atom of oxygen in the state -2 for a total of

6(1) + (- 2) = +4

then the oxidation number of seven carbon atoms is -4.

Benzoic acid contains six atoms of hydrogen in the state +1 and two atoms of oxygen in the state -2 for a total of

6(1) + 2(-2) = +2

then the oxidation number of seven carbon atoms is -2.

Therefore the oxidation of benzaldeyde to benzoic acid involves the loss of two electrons as result from the balanced half-reaction.

Now let's balance the other half-reaction:

Cr₂O₇⁼ = Cr³⁺

considering that the reaction requires an acidic enviroment the mass balanced half-reaction is

14H ⁺ + Cr₂O₇⁼= 2Cr⁺³ + 7H₂O

to balance the charges we write

6e + 14H ⁺ + Cr₂O₇⁼= 2Cr⁺³ + 7H₂O

the oxidation state of chromium in Cr₂O₇⁼ is +6 (check) while that of chromium in Cr³⁺ is +3 thus we need three electrons to reduce one atom of chromium and six electrons to reduce two atoms. Therefore the two balamced half reactions are:

C₆H₅CHO + H₂O = C₆H₅COOH + 2H ⁺ + 2e

6e + 14H ⁺ + Cr₂O₇⁼= 2Cr⁺³ + 7H₂O

It should be evident that in order to eliminate electrons from the overall reaction, before of adding the two half-reactions, we need to multiply with 3 the half-reaction of oxidation of benzaldeyde

3C₆H₅CHO + 3H₂O = 3C₆H₅COOH + 6H ⁺ + 6e
6e + 14H ⁺ + Cr₂O₇⁼= 2Cr³⁺ + 7H₂O
---------------------------------------------------------------
3C₆H₅CHO + Cr₂O₇⁼ + 8H ⁺ = 3C₆H₅COOH + 2Cr⁺³ + 4H₂O

The multiplication with 3 of the half-reaction of oxidation of benzaldeyde, is consistent with the fact that the oxidation yields only two electrons thus 3 moles of benzaldehyde should be oxidized to reduce on mole of Cr₂O₇⁼ .

Example 2 .

ClO ^- + CrO₂^- = CrO₄⁼ + Cl ^- (in basic solution)

Let's balance the half-reaction involving chlorine:

ClO ^- = Cl ^-

since we are dealing with a basic reaction we might balance oxigen by introducing either H₂O or OH^-. This can result in a more complicated procedure of material balancing which can be simplified by first deciding how many electrons must appear in the reaction. The reduction of chlorine from state +1 in ClO ^- to the state - 1 in Cl ^- requires two electrons:

2e + ClO ^- = Cl ^-

and, in order to balance charge, we must introduce 2 OH^- to the right side of the reaction

2e + ClO ^- = Cl ^- + 2OH^-

Eventually, the material balance is completed by adding water

2e + H₂O + ClO ^- = Cl ^- + 2OH^-

Now let's balance the other half-reaction:

CrO₂^- = CrO₄⁼

The oxidation of chromium from state +3 in CrO₂^- to the state +6 in CrO₄⁼ requires three electrons:

CrO₂^- = CrO₄⁼ + 3e

therefore charge balance require the introduction of 4 OH^- on the left side of the equation:

4OH^- + CrO₂^- = CrO₄⁼ + 3e

Eventually, the material balance is completed by adding water

4OH^- + CrO₂^- = CrO₄⁼ + 3e + 2H₂O

By multiplying with 3 the half-reaction of reduction and with 2 the half-reaction of oxidation we get

6e + 3H₂O + 3ClO ^- = 3Cl ^- + 6OH^-
8OH^- + 2CrO₂^- = 2CrO₄⁼ + 6e + 4 H₂O
-------------------------------------------------------------------
3ClO ^- + 2CrO₂^- + 2OH^- = 3Cl ^- + 2CrO₄⁼ + H₂O

Example 3.

P₄ + OH^- = PH₃ + H₂PO₂^-

In this reaction phosphorus is both oxidized and reduced. Reactions in which a substance is both oxidized and reduced are said disproportionation reactions.

Let's balance the half-reaction of oxidation:

P₄ = 4H₂PO₂^-

Since the oxidation state of P change from 0 to +1 and four atoms of P are involved then 4 electrons must be produced

P₄ = 4H₂PO₂^- + 4e

For the charge balance we need to introduce 8 negative charges as OH^- to the left side of the equation

8OH^- + P₄ = 4H₂PO₂^- + 4e

Now let's balance the reduction reaction:

P₄ = 4PH₃

Since the oxidation state of P change from 0 to -1 and four atoms of P are involved then 12 electrons are needed for the process of reduction:

12 e + P₄ = 4PH₃

For the charge balance we need to introduce 12 negative charges as OH^- to the right side of the equation

12e + P₄ = 4PH₃ + 12OH^-

eventually we add H₂O to the left side for the balance of H and 0:

12e + P₄ + 12H₂O = 4PH₃ + 12OH^-

By multiplying with 3 the half-reaction of oxidation we get

24OH^- + 3P₄ = 12H₂PO₂^- + 12e
12e + P₄ + 12H₂O = 4PH₃ + 12OH^-
----------------------------------------------------------------
12OH^- + 4P₄+ 12H₂O = 12H₂PO₂^- + 4PH₃