Test: Redox Reactions (15 August) - JEE MCQ

The reaction of metallic cobalt in a nickel sulfate solution involves a competition for the release of electrons. This means that the cobalt metal can react with the nickel ions in the solution, or the nickel can deposit on the cobalt metal.

At equilibrium, the reaction has balanced out with no net change in the concentration of the reactants and products. The fact that both Ni⁺² (aq) and Co⁺² (aq) are present at moderate concentrations at equilibrium signifies that neither forward nor reverse reactions are greatly favoured.

• A: This option is incorrect because both reactants and products are present in moderate concentrations, indicating that neither is greatly favoured.
• B: This statement is not correct either. Even though Co (s) and Ni⁺² (aq) are part of the reaction, the fact that Co⁺² (aq) is also present at moderate concentrations shows that they are not the only favoured species.
• C: This option is also incorrect. Even though Co⁺² (aq) and Ni (s) are part of the reaction, the fact that Ni⁺² (aq) is also present at moderate concentrations shows that they are not the only favoured species.
• D: This is the correct answer. When a reaction is at equilibrium, it means that the rate of the forward reaction equals the rate of the reverse reaction. Therefore, neither the reactants nor the products are greatly favoured. In other words, the concentrations of the reactants and products remain constant over time

The flow of current in a Daniel cell or any other electrochemical cell is due to the movement of electrons from the anode to the cathode. This happens through an external circuit connecting the two electrodes.

• Electron Transfer: In a Daniel cell, the zinc (Zn) electrode is oxidized to zinc ions (Zn²⁺), releasing two electrons. These electrons then move through the external circuit to the copper (Cu) electrode, reducing the copper ions (Cu²⁺) to copper atoms. However, the transfer of electrons does not occur directly from Zn to Cu²⁺ in the solution. Instead, it happens via the external circuit. Therefore, option A is incorrect.

• Potential Difference: The potential difference between the two electrodes in a Daniel cell is what drives the flow of current. When zinc is oxidized at the anode, it leaves behind electrons, creating a negative charge. At the same time, the reduction of copper ions at the cathode absorbs electrons, creating a positive charge. This difference in charge creates a potential difference or voltage that causes the electrons to flow from the anode to the cathode, generating current. Therefore, option B is correct.

• Salt Bridge: A salt bridge is a device used to complete the circuit in a Daniel cell by allowing the flow of ions between the two half-cells. This maintains the electrical neutrality within the half-cells, enabling the reaction to continue. However, it does not directly facilitate the flow of current, which is driven by the potential difference between the electrodes. Therefore, option C is incorrect.
• Platinum Wire: While a wire, such as one made of platinum, is needed to connect the two electrodes and provide a path for the electrons to move, it is not the presence of the wire itself that allows for the flow of current. Rather, it is the potential difference between the electrodes. Therefore, option D is incorrect.

In conclusion, the flow of current in a Daniel cell is possible if there is a potential difference between the copper and zinc electrodes. This potential difference drives the movement of electrons from the anode to the cathode, generating current.

• Oxidation Numbers of Halogens: The halogens (Fluorine, Chlorine, Bromine, Iodine, and Astatine) are a group in the periodic table. They are known for their high electronegativity and hence, when they combine with almost all other elements, they tend to have an oxidation number of -1 as they gain one electron to achieve a stable electronic configuration.
• Exception in the case of Oxygen: However, oxygen is more electronegative than all halogens except fluorine. So, when halogens (chlorine, bromine, and iodine) combine with oxygen, they tend to lose electrons to oxygen and hence, they show positive oxidation states.
• Conclusion: Therefore, when chlorine, bromine, and iodine are combined with oxygen, they have an oxidation number of +1 or any positive number (depending on the number of oxygen atoms they are combined with). Hence, the correct answer is C: +1 or any positive number.

A disproportionation reaction is a particular type of redox reaction in which a single compound is transformed into two different compounds, one being reduced and the other being oxidized.

The decomposition of hydrogen peroxide to form water and oxygen is a classic example of a disproportionation reaction.

Preparation of Hydrogen from H₂O with Ca (Calcium)

• Hydrogen can be prepared from water (H₂O) through a chemical reaction with calcium (Ca).
• In this reaction, calcium acts as a reducing agent. This is because calcium has the ability to donate electrons during the reaction, which reduces the other reactant, in this case, water.
• The overall chemical reaction that occurs is: Ca + 2H₂O → Ca(OH)₂ + H₂
• In this reaction, calcium (Ca) reacts with water (H₂O) to form calcium hydroxide (Ca(OH)₂) and hydrogen gas (H₂).
• The hydrogen gas that is produced can be collected and used for various purposes.
• Thus, out of the given options, it is calcium that can act as a reducing agent to prepare hydrogen from water.

Why Other Options are Incorrect

• Ag (Silver) and Au (Gold) are noble metals and do not readily participate in chemical reactions. They cannot act as reducing agents to produce hydrogen from water.
• Al (Aluminium) can act as a reducing agent, but it does not react with water under normal conditions to produce hydrogen.

Conclusion

• Therefore, the correct answer is option B: 'Ca, which acts as reducing agent'.

• Molecules of most electronegative elements, compounds having an element in the highest oxidation state and oxides of metals and nonmetals are examples of oxidising agents.
• But sodium hydride is a metallic hydride, so it is not an oxidizing agent.

Oxidation Numbers in Oxygen Difluoride (OF₂) and Dioxygen Difluoride (O₂F₂)

Oxidation numbers are assigned to elements in compounds according to several rules. Here, we'll use the rule that states the oxidation number of fluorine (F) in a compound is always -1, and the rule that states the sum of oxidation numbers of all atoms in a compound is 0.

Oxidation Number in OF₂:

• In Oxygen Difluoride (OF₂), there are two fluorine atoms, each with an oxidation number of -1. So, the total oxidation number contributed by fluorine is -2.
• To mak e the total oxidation number 0, the oxygen atom must have an oxidation number of +2.

Oxidation Number in O₂F₂:

• In Dioxygen Difluoride (O₂F₂), there are two fluorine atoms, each with an oxidation number of -1. So, the total oxidation number contributed by fluorine is -2.
• There are also two oxygen atoms. To make the total oxidation number 0, the combined oxidation number of the two oxygen atoms must be +2.
• Therefore, the oxidation number of each oxygen atom in O₂F₂ is +1 (because +2 divided by 2 equals +1).

So, in OF₂ and O₂F₂, the oxygen is assigned an oxidation number of +2 and +1, respectively. Therefore, the correct answer is B: +2 and +1.

as oxygen is more electronegative than Cl,Br and I. So they have positive oxidation state.

P4 is undergoing oxidation as well as reduction. As oxidation number of P₄ is 0 in reactant and it increases to +1 in H₂PO₂^- and decreases to -3 in PH₃

Oxidation Number Transition in the Conversion of Br₂ to BrO₃

• The oxidation number of an element in its free or uncombined state is zero. This rule applies to bromine in Br₂, so the initial oxidation state of Br in Br₂ is zero.
• In BrO₃^-, the combined state of Bromine, Oxygen has an oxidation state of -2. However, the overall charge of the ion is -1. Since we have three Oxygen atoms, the total contribution of Oxygen is -6. To balance this, the Bromine must have an oxidation state of +5.
• Therefore, in the conversion of Br₂ to BrO₃^-, the oxidation state of Bromine changes from 0 in Br₂ to +5 in BrO₃^-
• Hence, the correct answer is Option A: zero to +5.
   

Explanation of Highest Value of Oxidation Number Changes from 1 to 7

The highest value of oxidation number changes from 1 to 7 across the third period in the periodic table. This is due to the following reasons:

• The Atoms of Transition Elements:Transition metals are those elements located in the d-block of the periodic table. They have varying oxidation states, but they do not usually reach an oxidation state of 7. Their oxidation states primarily range between +2 and +3, although some can reach states of +4 or +5.
• The First Three Groups:The first three groups of the periodic table include alkali metals, alkaline earth metals, and boron group elements. These groups generally have oxidation states of +1, +2, and +3 respectively. They do not reach an oxidation state of 7.
• In Alkaline Earth Metals:Alkaline earth metals are the elements in the second group of the periodic table. These elements generally have an oxidation state of +2 due to the presence of two valence electrons which are readily lost in chemical reactions.
• Across the Third Period in the Periodic Table:The elements in the third period of the periodic table show a wider range of oxidation states, which can vary from +1 to +7. This is due to the presence of both s and p orbitals in their valence shell, which allows the elements to lose or gain more electrons in chemical reactions. Therefore, across the third period in the periodic table, the highest value of oxidation number changes from 1 to 7.

Superoxides: Superoxides are a class of compounds that contain the superoxide anion, which has the chemical formula O2-. In superoxides, the oxidation number of oxygen is -1/2.

• The reason the oxidation number of oxygen in superoxides is -1/2 is because of the unique way in which the electrons are distributed in the superoxide ion. Each superoxide ion is composed of two oxygen atoms. However, these two oxygen atoms share one electron, resulting in an overall charge of -1 for the entire ion. Therefore, each oxygen atom effectively has a charge of -1/2.
• Some common examples of superoxides include potassium superoxide (KO₂) and rubidium superoxide (RbO₂). These are often used in life-support systems to generate oxygen and absorb carbon dioxide.
• It's important to note that the oxidation number of oxygen can vary depending on the specific compound in which it is found. For instance, in peroxides (like hydrogen peroxide, H₂O₂), the oxidation state of oxygen is -1. In most other compounds, including water, the oxidation state of oxygen is typically -2.
• When oxygen is bonded to fluorine, the oxidation state of oxygen is positive since fluorine is more electronegative. When oxygen is bonded to metals, the oxidation state is typically -2, as metals are usually less electronegative than oxygen.

Key Point: In superoxides, the oxidation state of each oxygen atom is -1/2 due to the sharing of one electron between two oxygen atoms in the superoxide ion.

The standard electrode potential, E₀, is a measure of the tendency of a chemical species to be reduced. The higher the E₀ value, the greater the tendency for reduction.

In a redox reaction, the chemical species that gets reduced acts as the oxidising agent. Therefore, the species with the highest E₀ value will be the strongest oxidising agent.

Given the E0 values:

• Fe³⁺ / Fe²⁺ = +0.77V
• I₂(s) / I^- = +0.54V
• Cu²⁺ / Cu = +0.34V
• Ag⁺ / Ag = +0.80V

From the given values, it is clear that Ag+ / Ag has the highest E₀ value (+0.80V). Therefore, Ag+ is the strongest oxidising agent among the given options.

Conclusion:

• Ag+ is the strongest oxidising agent because it has the highest standard electrode potential (E₀) value among the given species.
• Remember, the higher the E₀ value, the greater the tendency of the species to be reduced, and hence, stronger is its oxidising power.

Metal activity series or electrochemical series is a series in the decreasing order of metals which are active during a chemical reaction comparatively with each other.

• Here, Zinc’s activity is greater than Copper’s activity and Copper’s activity is greater than that of silver.

In the given reaction, 2H₂O₂ (Hydrogen Peroxide) decomposes to form 2H₂O (Water) and O₂ (Oxygen). This reaction is a redox reaction, meaning it involves the transfer of electrons.

Understanding Oxidation and Reduction:
• Oxidation is the process in which an atom, ion, or molecule loses electrons.
• Reduction is the process in which an atom, ion, or molecule gains electrons.

Oxygen's Role in the Reaction:
• In H₂O₂, the oxidation state of Oxygen is -1.
• In H₂O, the oxidation state of Oxygen is -2.
• In O₂, the oxidation state of Oxygen is 0.

Applying the Concept of Oxidation and Reduction:
• The change in oxidation state from -1 in H₂O₂ to -2 in H₂O indicates a gain of electrons, hence, Oxygen is reduced.
• The change in oxidation state from -1 in H₂O₂ to 0 in O₂ indicates a loss of electrons, hence, Oxygen is oxidised.

Hence, in this reaction, Oxygen is both oxidised and reduced. Therefore, the correct answer is C: Oxygen is both oxidised and reduced.

Please note that in such reactions where the same element is both oxidised and reduced, it is referred to as a disproportionation reaction.This is a dispropotionation reaction.