Electrochemistry - SS2 Chemistry Past Questions and Answers - page 1

In the given reaction, copper (Cu) reacts with silver nitrate (AgNO3) to form copper nitrate (Cu(NO3)2) and silver (Ag). Copper starts with an oxidation state of 0 and ends with an oxidation state of +2 in Cu(NO3)2. Therefore, copper undergoes oxidation, losing electrons.

Oxygen typically has an oxidation state of -2 in most of its compounds. This is due to its high electronegativity, resulting in a strong attraction for electrons and a tendency to gain two electrons to achieve a stable octet configuration.

In the given reaction, iron(III) oxide (Fe2O3) reacts with carbon monoxide (CO) to form iron (Fe) and carbon dioxide (CO2). Carbon starts with an oxidation state of +2 in CO and ends with an oxidation state of +4 in CO2. Therefore, carbon undergoes reduction, gaining electrons.

In potassium permanganate (KMnO4), manganese (Mn) exhibits a variable oxidation state. In this compound, manganese has an oxidation state of +7. The presence of oxygen (O) and the -2 charge on each oxygen atom necessitate the +7 oxidation state for manganese to balance the compound's charge.

In most compounds, hydrogen (H) has an oxidation state of +1. However, hydrogen can also have an oxidation state of -1 when it forms a hydride ion (H-) in compounds with highly electropositive elements.

Redox (reduction-oxidation) reactions involve the transfer of electrons between species. In a redox reaction, one species loses electrons (undergoes oxidation) while another species gains those electrons (undergoes reduction). Oxidation and reduction always occur together and are interconnected processes.

Oxidation numbers are assigned to atoms in a compound or ion to indicate the distribution of electrons within the molecule. They provide a way to track electron transfer in redox reactions. In general, oxidation numbers can be determined based on a set of rules:

●     The oxidation number of an element in its elemental form is zero (e.g., O2, Na, Cl2).

●     The oxidation number of a monatomic ion is equal to its charge (e.g., Na+ has an oxidation number of +1).

●     Oxygen typically has an oxidation number of -2, except in peroxides (e.g., H2O2) where it has an oxidation number of -1.

●     Hydrogen usually has an oxidation number of +1, except in metal hydrides (e.g., NaH) where it has an oxidation number of -1.

●     The sum of the oxidation numbers in a neutral compound is zero, and the sum in a polyatomic ion is equal to the ion's charge.

During a redox reaction, the oxidation numbers of certain elements change. The element that increases its oxidation number is oxidised, and the one that decreases its oxidation number is reduced. For example:

In the reaction: Zn + CuSO4 → ZnSO4 + Cu

●     Zinc (Zn) goes from an oxidation number of 0 to +2. It is oxidised (loses electrons) and is the reducing agent.

●     Copper (Cu) goes from an oxidation number of +2 to 0. It is reduced (gains electrons) and is the oxidising agent.

Balancing redox reactions is crucial to ensure that the number of electrons gained and lost in the oxidation and reduction processes is equal. This balancing is achieved by adjusting the coefficients of the reactants and products in the chemical equation. Oxidation numbers play a significant role in balancing redox reactions by tracking electron transfers.

To balance a redox reaction, the following steps are generally followed:

1.    Assign oxidation numbers to all atoms in the reaction.

2.    Identify the species that undergo oxidation and reduction.

3.    Write half-reactions for the oxidation and reduction processes, showing the changes in oxidation numbers.

4.    Balance the atoms (except hydrogen and oxygen) in each half-reaction by adding water molecules (H2O) or hydroxide ions (OH-) as needed.

5.    Balance the charges in each half-reaction by adding electrons (e-).

6.    Multiply each half-reaction by appropriate coefficients to equalise the number of electrons gained and lost.

7.    Combine the balanced half-reactions to obtain the balanced overall redox equation.

Example of a balanced redox reaction using oxidation numbers:

In the reaction: Cl2 + 2NaBr → 2NaCl + Br2

●     Chlorine (Cl) has an oxidation number of 0 in Cl2 and -1 in NaCl.

●     Bromine (Br) has an oxidation number of -1 in NaBr and 0 in Br2.

Half-reactions:

Oxidation: Cl2 → 2Cl- (loses 2 electrons)

Reduction: 2Br- → Br2 + 2e- (gains 2 electrons)

Balanced equation:

Cl2 + 2NaBr → 2NaCl + Br2

By assigning oxidation numbers and following the steps for balancing redox reactions, we can achieve a balanced equation that satisfies the conservation of mass and charge.