Category: Basic Chemistry

Some of the reactions of metals with water and acid are summarised in the table below.

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The reactivity of metals broadly refers to the relative extent of change that metals undergo with respect to certain reactions. Some of these reactions are:

1. 1. Metal reaction with water
   2. Metal reaction with acid
   3. Metal displacement reactions
   4. Extraction of metals from metal oxides (or other ores)
   5. Heating of metal hydroxides, metal carbonates and metal nitrates

In general, the greater the extent to which a metal changes from its elemental form to a compound, the more reactive it is. For example, sodium is more reactive than copper, as elemental sodium reacts violently with cold water, while elemental copper does not react with water and steam. The reactivities of metals for reactions 1 to 5 are summarised in the following table, which is known as the metal reactivity series:

The two non-metals, C and H are included for reference.

 

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What are the effects of pH on redox reactions?

pH affects the strength of an oxidising agent. For example, the oxidising strength of potassium manganate (VII), as measured by its standard electrode potential, is as follows:

Acidic medium

MnO₄^– + 8H⁺+ 5e^– → Mn²⁺ + 4H₂O               E^o_red  = +1.51 V

Neutral/slightly basic medium

MnO₄^– + 2H₂O + 3e^– → MnO₂ + 4OH^–              E^o_red  = +1.23 V

Strongly basic medium

MnO₄^– + e^– → MnO₄^2-               E^o_red  = +0.56 V

Hence, potassium manganate (VII) is most strongly oxidising in an acidic medium and least oxidising in a strongly basic medium.

The behaviour of MnO₄^– in different pH conditions is best explained using the the reaction MnO₄^– + 2H₂O + 3e^– → MnO₂ + 4OH^– as a reference. If the pH of this reaction is lowered, excess H⁺ allows manganese dioxide to be further reduced to the more stable +2 oxidation state ion:

MnO₂ + 4H⁺+ 2e^– → Mn²⁺ + 2H₂O

Combining the half reactions MnO₄^– + 2H₂O + 3e^– → MnO₂ + 4OH^– and MnO₂ + 4H⁺+ 2e^– → Mn²⁺ + 2H₂O gives the overall reduction reaction of MnO₄^– to Mn²⁺ in acidic medium.

On the other hand, if the pH of the same reaction MnO₄^– + 2H₂O + 3e^– → MnO₂ + 4OH^– is increased, excess OH^– will shift the equilibrium towards the left. As the solution becomes more strongly basic, the manganate (VII) ion is instead reduced to the manganate (VI) ion. This implies that the manganate (VI) ion is stable only in very basic conditions. It disproportionates in acidic, neutral and slightly basic conditions as follows:-

3MnO₄^2- + 4H⁺ → 2MnO₄^– + MnO₂ + 2H₂O

Incidentally,

MnO4–	purple
MnO42-	green
MnO2	brown solid
Mn2+	pale pink

 

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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. 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^– to both sides, then combine H⁺ and OH^– to form H₂O and cancel out any 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

Cr₂O₇^2- → Cr³⁺

Balancing the equation

Cr₂O₇^2- → 2Cr³⁺

Cr₂O₇^2- → 2Cr³⁺ + 7H₂O

Cr₂O₇^2-  + 14H⁺ → 2Cr³⁺ + 7H₂O

Cr₂O₇^2-  + 14H⁺ + 6e^–→ 2Cr³⁺ + 7H₂O            eq2

(note that step 5 does not apply)

Next, combine the balanced half-reaction equations, cancel any 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) + Cr₂O₇^2- (aq) + 14H⁺ (aq) → 6Fe³⁺(aq) + 2Cr³⁺ (aq) + 7H₂O (l)

 

 

 

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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).

A disproportionation reaction happens when the oxidation state of an element in a compound is unstable and can achieve greater stability by splitting into two species — one at a higher oxidation state and one at a lower. This is often driven by the products having a lower total Gibbs energy than the reactant. It may also occur due to favourable changes in enthalpy and entropy.

 

 

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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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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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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:-

1. • 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

 

 

 

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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.

1. • 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 is a technique in which a chromatogram is first developed using a particular solvent (either by ascending or descending paper chromatography), and then the paper is rotated 90 degrees for a second run with a different 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.

 

 

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