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How to construct and balance redox equations?

How do we construct and balance redox equations?

The way to construct a redox equation is to write two balanced ionic half-reaction equations, one for the oxidising agent and the other one for the reducing agent, and combine them. The method to balance each half-reaction equation is slightly different from that of a normal stoichiometric equation and involves the following steps:-

    1. Balance elements other than O and H
    2. Balance O by adding H2O
    3. Balance H by adding H+
    4. Balance excess positive charges by adding electrons, e–
    5. (For reaction in a basic solution) Balance H+ by adding OH– on both sides, combine H+ and OH– to form H2O and cancel common terms

For example, the redox equation for the oxidation of acidified iron (II) sulphate by potassium dichromate (VI) is constructed as follows:

1st half-reaction equation

Fe2+ → Fe3+

Balancing the equation

Fe2+ → Fe3+ + e–            eq1

(note that steps 1 through 3 and step 5 do not apply)

2nd half-reaction equation

Cr2O72- → Cr3+

Balancing the equation

Cr2O72- → 2Cr3+

Cr2O72- → 2Cr3+ + 7H2O

Cr2O72-  + 14H+ → 2Cr3+ + 7H2O

Cr2O72-  + 14H+ + 6e–→ 2Cr3+ + 7H2O            eq2

(note that step 5 does not apply)

Next, combine the balanced half-reaction equations, cancel common terms and ensure that all electrons are eliminated in the final equation. For this example, we multiply eq1 by 6 throughout before adding to eq2, giving:

6Fe2+(aq) + Cr2O72- (aq) + 14H+ (aq) → 6Fe3+(aq) + 2Cr3+ (aq) + 7H2O (l)

 

Question

Write the redox equation for the oxidation of sodium formate by potassium manganite (VII) to form sodium carbonate and manganese dioxide in slightly basic conditions.

Answer

1st half-reaction equation

HCO2– → CO32-

Balancing the equation

HCO2–   + H2O → CO32-

HCO2–   + H2O → CO32- + 3H+

HCO2–   + H2O → CO32- + 3H+ + 2e–

HCO2–   + H2O + 3OH– → CO32- + 3H+ + 2e–+ 3OH–

HCO2–   + H2O + 3OH– → CO32- + 3H2O + 2e–

HCO2– + 3OH– → CO32- + 2H2O + 2e–

2nd half-reaction equation

MnO4– → MnO2

Balancing the equation

MnO4– → MnO2 + 2H2O

MnO4– + 4H+ → MnO2 + 2H2O

MnO4– + 4H+ + 3e– → MnO2 + 2H2O

MnO4– + 4H+ + 3e– + 4OH– → MnO2 + 2H2O + 4OH–

MnO4– + 4H2O + 3e– → MnO2 + 2H2O + 4OH–

MnO4– + 2H2O + 3e– → MnO2 + 4OH–

Combining the balanced half-reaction equations, we have,

3HCO2– (aq) + OH– (aq) + 2MnO4– (aq) → 3CO32- (aq) + 2MnO2 (s) + 2H2O (l)

 

 

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Disproportionation reaction

A disproportionation reaction is one in which an atom in a reactant undergoes both oxidation and reduction. For example,

The O in H2O2 (oxidation state of O = -1) is oxidised to the O in O2 (oxidation state of O = 0) and reduced to the O in H2O (oxidation state of O = -2).

 

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Redox reactions

A redox reaction is one in which oxidation and reduction occur concurrently. When two different reactants are involved, an atom, a molecule or an ion in one reactant undergoes oxidation, while an atom, a molecule or an ion in the other undergoes reduction. For example,

 

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Oxidising agent and reducing agent

An oxidising agent is a species that oxidises another species and a reducing agent is a species that reduces another species. For example,

Cu2+ in the above reaction is the oxidising agent as it oxidises Zn to Zn2+ while Zn is the reducing agent as it reduces Cu2+ to Cu.

 

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Oxidation state or oxidation number

The oxidation state/number of an atom or a monatomic ion is a number assigned to the atom or monatomic ion that corresponds to its degree of oxidation.

The rules for assigning an oxidation state/number to an atom or monatomic ion is as follows:-

    • The oxidation state/number of an uncombined atom or an atom in a neutral homonuclear molecule is zero, e.g.
 

Atom

Oxidation state/number

Cu

 Cu

0

Cl2

 Cl

0

C60

C

0
    • The oxidation state/number of a monatomic ion is equal to its charge, e.g.

Ion

Oxidation state/number

Fe3+

+3

Cl–

-1

Na+

+1
    • The oxidation state/number of an atom in a compound corresponds to the number of electrons the atom gains or loses when it hypothetically forms ionic bonds with its neighbours, e.g.

Compound

 Atom Assumed role

Oxidation state/number

MgCl2

 Mg Electron donor

+2

CO2

 C Electron donor

+4

KOH

 O Electron acceptor

-2

NH3

 N Electron acceptor

-3
    • The total oxidation state/number of all atoms in a species is equivalent to the overall charge of the species, e.g.

Species

 Overall charge of species

Oxidation state/number

NaCl

 0

+1 + (-1) = 0

NH4+

 +1

-3 + (+1 x 4) = +1

PO43-

 -3

+5 + (-2 x 4) = -3

 

Question

Determine the oxidation state/number of the underlined elements in AlH3, CH4 and Cr2O72-.

Answer

For AlH3, we assume that Al is the electron donor as it is more electropositive than H, and so, +3 + (H x 3) = 0. Therefore, H = -1.

For CH4, we assume that H is the electron donor as it is more electropositive than C, and so, C + (+1 x 4) = 0. Hence, C = -4.

For Cr2O72-, we assume that Cr is the electron donor as it is more electropositive than O, and so, (Cr x2) + (-2 x 7) = -2. Therefore, Cr = +6.

 

 

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Oxidation and reduction: introduction

Oxidation is the gain in oxygen, loss of hydrogen, loss of electron or increase in oxidation number, while reduction is the loss of oxygen, gain in hydrogen, gain in electron or decrease in oxidation number.

    • Gain/loss of oxygen, e.g.

The reactant H2 is oxidised as it gains an oxygen atom to form H2O, while O2 is reduced as it loses an oxygen to form H2O.

    • Gain/loss of hydrogen, e.g.

The reactant H2S is oxidised as it loses hydrogen atoms to form S, while Cl2 is reduced as it gains hydrogen atoms to form HCl.

    • Gain/loss of electron, e.g.

The reactant Zn is oxidised as it loses two electrons to form Zn2+, while Cu2+ is reduced as it gains two electrons to form Cu.

    • Increase in oxidation state or oxidation number (see next article for details), e.g.

The reactant Zn is oxidised as its oxidation state/number increases from 0 in Zn to +2 in Zn2+, while Cu2+ is reduced as its oxidation state/number decreases from +2 in Cu2+ to 0 in Cu.

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2-dimensional paper chromatography

2-dimensional paper chromatography is a paper chromatographic technique in which a chromatogram is first developed using a particular solvent (either through ascending or descending paper chromatography) and then rotated by 90 degrees for a second run with another solvent.

2-dimensional paper chromatography allows components that are not separated by the first solvent (due to their insolubility in that solvent) to be separated by the second solvent, thereby producing a chromatogram with better resolution than a 1-dimensional paper chromatogram.

 

Question

With reference to the above diagram, what is the minimum number of chemical components in mixture X?

Answer

5

 

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Locating agent

Some components that are separated from a mixture by paper chromatography are colourless. A locating agent is sprayed on the chromatograph to make these colourless components visible. An example of a locating agent is ninhydrin (2,2-dihydroxyindane-1,3-dione), which is used to identify colourless amino acids in the separation of an amino acid mixture.

Ninhydrin reacts with colourless amino acids (with primary amine groups) to form the ninhydrin chromophore, which is purplish blue.

Iodine and UV light are other examples of locating agents.

 

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\(R_f\) value

To analyse paper chromatography results, we determine the retention factor, Rf, where

R_f=\frac{distance\: travelled\: by\: substance\, from\: origin}{distance\: travelled\: by\: solvent\: front\: from \: origin}

 

For example (with reference to the above diagram),

R_{f,orange\: substance}=\frac{a}{c}

R_{f,yellow\: substance}=\frac{b}{c}

 

Question
    1. Based on the diagram below, what are the Rf values for samples x and y?
    2. With reference to the calculated Rf values, describe the likely solubilities of samples x and y in the mobile and/or stationary phases.
    3. What could be the composition of sample z?

Answer
    1. Rf,x = 11.2/13.4 = 0.84; Rf,y = 3.1/13.4 = 0.23
    2. Since Rf,x > Rf,y , sample x is relatively more soluble in the mobile phase than sample y (or sample y is relatively more soluble in the stationary phase than sample x), assuming partitioning is the main mechanism of separation.
    3. Sample z could be a mixture of x and y.

 

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Principles of separation

What are the principles of separation in paper chromatography?

The distances travelled by the chemical components of a mixture in paper chromatography are affected by propelling and retardation forces.

Propelling force

Solvent-flow along the length of the paper by capillary action

Retardation forces

 Partitioning of chemical components between the mobile phase and the stationary phase

Adsorption of chemical components onto cellulose

Ion-exchange bonding of chemical components with cellulose

Filter paper consists of almost pure cellulose, which is a polymer of D-glucose. However, some functional groups on the polymeric chain may be oxidised to the carboxyl form as shown in the diagram below.

The hydroxyl, carboxyl and ether groups on cellulose interact with water molecules in the atmosphere via hydrogen bonds and van der Waals forces to form a multilayer of water that is known as the stationary phase.

The solvent, known as the mobile phase, is usually selected to be relatively less polar than water so that the components to be separated dissolve in different amounts in each phase. As the chemical components are propelled by the solvent along the paper via capillary action, they are partitioned between the stationary and mobile phases; that is, the chemical components equilibrate and distribute themselves between the two phases according to their solubilities in each phase.

The moving components are slowed down by the partitioning process because they are held back by stationary polar water molecules (via van der Waals forces and hydrogen bonds) when they dissolve and equilibrate in the stationary phase. Eventually, the partition process may bring the components to a stop over a certain length of the paper.

The more soluble a component is in the mobile phase (relatively less polar) versus the stationary phase (relatively more polar), the further it is propelled along the paper by the mobile phase. Furthermore, direct adsorption on to the cellulose functional groups via hydrogen bonding and ion-exchange bonding with the carboxyl and possibly hydroxyl groups also serve to slow down and separate the chemical components according to their affinity with these functional groups.

 

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