All questions of Oxidation, Reduction and Redox for BMAT Exam

Equivalent mass = molecular mass/(valency in case of elements and charge of cationic part in case of compounds)

For Cl2, valency and for HCHO, charge of cationic part are 1 in magnitude. Hence,
equivalent mass = molecular mass for A and B.

The oxidation number is a concept used in chemistry to keep track of the distribution of electrons in a compound or molecule. It is a measure of the charge that an atom would have if all the shared electrons were assigned to the more electronegative atom in a bond.

In the case of oxygen, its most common oxidation number is -2. However, in certain compounds, such as superoxides and peroxides, the oxidation number of oxygen deviates from -2.

Oxidation number of 1/2 is assigned to the oxygen atom in superoxides. Superoxides are a class of compounds that contain the superoxide ion, O2-. In this ion, each oxygen atom has an oxidation number of -1/2. This is because the oxygen-oxygen bond in the superoxide ion is a single bond with a bond order of 1/2. Therefore, each oxygen atom is assigned an oxidation number of -1/2 to account for the distribution of electrons in the bond.

In the case of peroxides, such as hydrogen peroxide (H2O2), the oxidation number of oxygen is -1. In peroxides, the oxygen-oxygen bond is a single bond with a bond order of 1. Each oxygen atom is assigned an oxidation number of -1 to account for the distribution of electrons in the bond.

When oxygen is bonded to fluorine, the oxidation number of oxygen is -1. Fluorine is the most electronegative element, and therefore, it attracts the shared electrons in the bond more strongly than oxygen. As a result, oxygen is assigned an oxidation number of -1 to account for the unequal distribution of electrons in the bond.

When oxygen is bonded to metals, the oxidation number of oxygen is typically -2. However, there are some exceptions to this rule, such as in certain metal peroxides or superoxides, where the oxidation number of oxygen deviates from -2.

In summary, the oxidation number of 1/2 is assigned to the oxygen atom in superoxides, where the oxygen-oxygen bond is a single bond with a bond order of 1/2. In other compounds, such as peroxides, fluorides, and most metal oxides, the oxidation number of oxygen is typically -1 or -2.

Yes because right side is deficient of oxygen.

In the ion-electron method (also called the half-reaction method), the redox equation is separated into two half-equations - one for oxidation and one for reduction. Each of these half-reactions is balanced separately and then combined to give the balanced redox equation.

To determine the number of moles of K2Cr2O7 needed to react completely with one mole of ferric sulphite in acidic medium, we need to first write and balance the chemical equation for the reaction.

The balanced equation for the reaction between K2Cr2O7 and Fe2(SO3)3 in acidic medium is as follows:

K2Cr2O7 + 14H+ + 6Fe2(SO3)3 → 2K+ + 2Cr3+ + 7H2O + 6Fe3+ + 6SO4^2-

From the balanced equation, we can see that the stoichiometric ratio between K2Cr2O7 and Fe2(SO3)3 is 1:6. This means that 1 mole of K2Cr2O7 reacts with 6 moles of Fe2(SO3)3.

Since we are given that we have 1 mole of Fe2(SO3)3, we can calculate the number of moles of K2Cr2O7 needed by using the stoichiometric ratio.

Number of moles of K2Cr2O7 = 1 mole of Fe2(SO3)3 × (1 mole of K2Cr2O7 / 6 moles of Fe2(SO3)3)

Simplifying the expression, we get:

Number of moles of K2Cr2O7 = 1/6 moles

Therefore, the correct answer is option 'D', 1.0 mole.

Oxidation number of oxygen in most compounds is -2.

Explanation:
- Oxidation number is the number assigned to an atom to indicate its degree of oxidation or loss/gain of electrons.
- Oxygen is a highly electronegative element, meaning it has a strong tendency to attract electrons.
- In most compounds, oxygen has an oxidation number of -2 because it tends to gain electrons to achieve a stable octet configuration (8 valence electrons).
- For example, in water (H2O), each hydrogen atom has an oxidation number of +1 and the oxygen atom has an oxidation number of -2, which balances out the charge to zero.
- There are some exceptions to this rule, such as in peroxides where oxygen has an oxidation number of -1, and in compounds with more electronegative elements where oxygen may have a positive oxidation number.
- Overall, the oxidation number of oxygen in most compounds is -2.

A titration is a technique used to work out the concentration of an unknown solution based on its chemical reaction with a solution of known concentration. The process usually involves adding the known solution (the titrant) to a known quantity of the unknown solution (the analyte) until the reaction is complete.

To determine the number of moles of KMnO4 needed to react with one mole of ferrous sulphite in acidic solution, we need to balance the chemical equation and use stoichiometry.

The balanced chemical equation for the reaction between KMnO4 and ferrous sulphite in acidic solution is as follows:

5FeSO3 + 2KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O

From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of FeSO3. Therefore, the stoichiometric ratio is 2:5.

Now, let's calculate the number of moles of KMnO4 needed.

Given that we have 1 mole of FeSO3, we can set up the following proportion:

2 moles KMnO4 / 5 moles FeSO3 = x moles KMnO4 / 1 mole FeSO3

Cross-multiplying and solving for x, we get:

2 moles KMnO4 = 5 moles FeSO3 * x moles KMnO4
x moles KMnO4 = (2 moles KMnO4 * 1 mole FeSO3) / 5 moles FeSO3
x moles KMnO4 = 0.4 moles KMnO4

Therefore, the number of moles of KMnO4 that will be needed to react with one mole of ferrous sulphite in acidic solution is 0.4 moles, which corresponds to option B.

Understanding the Compound Cl2O7
To find the oxidation state of chlorine (Cl) in the compound Cl2O7, we need to analyze its structure and the general rules for oxidation states.
Oxidation State of Oxygen
- In most compounds, the oxidation state of oxygen is -2.
- Since there are 7 oxygen atoms in Cl2O7, the total contribution from oxygen is:
- 7 atoms × (-2) = -14
Calculating the Total Oxidation State
- Let the oxidation state of chlorine be x.
- Since there are 2 chlorine atoms, their total contribution to the oxidation state is 2x.
- The overall charge of the compound is neutral (0), leading to the equation:
- 2x + (-14) = 0
Solve for x
- Rearranging the equation gives:
- 2x = 14
- x = 14 / 2
- x = +7
Conclusion
- Therefore, the oxidation number of Cl in Cl2O7 is +7, which corresponds to option 'C'.
This analysis shows that chlorine exhibits a +7 oxidation state, a characteristic of its higher oxidation states in oxoacids and related compounds.

Understanding the Half-Reaction
In this half-reaction, we are looking at the transformation from cyanide ion (CN-) to the cyanate ion (CNO-). To balance this reaction, we need to account for both charge and mass.
Analyzing the Charges
- Cyanide Ion (CN-): The charge is -1.
- Cyanate Ion (CNO-): The charge is -1.
These ions have the same charge, but we also need to consider the atoms involved.
Balancing the Atoms
- Atoms in CN-: 1 Carbon (C) and 1 Nitrogen (N).
- Atoms in CNO-: 1 Carbon (C), 1 Nitrogen (N), and 1 Oxygen (O).
In this case, we see that while the number of Nitrogen and Carbon atoms remains the same, we need to add an Oxygen atom to balance the transformation from CN- to CNO-.
Adding Electrons
To balance the change in oxidation state and ensure that both sides of the equation have the same charge:
- Oxygen Addition: When adding oxygen, we must also add electrons to maintain charge balance.
- Oxygen typically requires 2 electrons to form. Thus, we add 2 electrons to the right side of the half-reaction to balance the charges.
Final Balanced Half-Reaction
The balanced half-reaction can be expressed as follows:
CN- + 2 electrons → CNO-
This shows that 2 electrons must be added to the right side to balance the reaction.
Conclusion
The correct answer is indeed option 'D': 2 electrons on the right. This is necessary to account for the addition of the oxygen atom and to maintain charge balance in the half-reaction.