Redox Reactions - Free MCQ Practice Test with solutions, NEET Chemistry

Cl show different oxidation state as -1 to +7 due to vacant d orbital.

As oxygen is more electronegative than Cl.

Oxygen size is small hence its more electronegative and show -2 oxidation states.

2x + 7 (-2) = 0

x = +7

When zinc is placed in copper nitrate solution the intensity of the blue colour is produced and copper iron is deposited on zinc. This is a Redox reaction between zinc and an aqueous solution of copper nitrate occurring in a beaker.

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

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.

• 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.

Oxidation Number of Oxygen

• Oxygen is a highly electronegative element with a strong tendency to attract shared electrons towards itself. It almost always has an oxidation number of -2 in its compounds, except in peroxides like H₂O₂ or Na₂O₂ where it has an oxidation number of -1.
• The oxidation number of oxygen in its elemental form (O₂) is 0.
• The reason why the oxidation number of oxygen is usually -2 is due to its electron configuration. Oxygen has 6 valence electrons and needs 2 more electrons to achieve the stable electron configuration of the nearest noble gas, neon. Therefore, in most of its compounds, oxygen gains 2 electrons, giving it an oxidation number of -2.
• So, the correct answer to the question is D: -2 because the oxidation state of oxygen in most of its compounds is -2. This is due to the fact that oxygen is highly electronegative and tends to gain 2 electrons to achieve a stable electron configuration.

• Let's denote the oxidation number of P as x.
• NaH₂PO₄ is a neutral compound. Therefore, the sum of oxidation numbers is zero.
• The sum of oxidation numbers would be: +1 (for Na) + 2*(+1) (for two H atoms) + x (for P) + 4*(-2) (for four O atoms) = 0.
• Solving this equation: 1 + 2 + x - 8 = 0, we get x = 5.

Therefore, the oxidation number of P in NaH₂PO₄ is +5 (option D).

When a copper rod is dipped in silver nitrate solution, a redox reaction occurs between Copper and an aqueous solution of silver nitrate.

• So the intensity of blue colour increases gradually as silver deposits on the rod.

Zn is oxidized to Zn⁺² while H+ is reduced to H₂. So H⁺ is acting as oxidant and Zn is acting as reductant.

Option A is correct: F.

Fluorine (F) always has oxidation state -1 in its compounds because it is the most electronegative element and always attracts electron density toward itself; it does not exhibit positive oxidation states in known stable compounds.

Cesium (Cs) is an alkali metal and characteristically loses its single valence electron to give oxidation state +1; it does not show negative oxidation states under normal chemistry.

Iodine (I) can show both negative and positive oxidation states. Examples:

• -1 in iodide ions, e.g., I^- (as in NaI).
• +1, +3, +5, +7 in interhalogen and oxyanion species, e.g., ICl (I +1), IO₃^- (iodate, I +5), IO₄^- (periodate, I +7).

Neon (Ne) is a noble gas with a filled shell and shows oxidation state 0 in its elemental form; it forms no stable compounds under normal conditions.

Therefore, among the given elements only F exhibits exclusively negative oxidation state; choose A.

To determine the decreasing order of oxidation states of nitrogen in the given compounds, we need to find the oxidation state of nitrogen in each compound:

1. HNO3 (Nitric acid): Oxidation state of nitrogen: +5

2. NO (Nitric oxide): Oxidation state of nitrogen: +2

3. N2 (Dinitrogen): Oxidation state of nitrogen: 0

4. NH4Cl (Ammonium chloride): Oxidation state of nitrogen: -3

Now, let's arrange these compounds in decreasing order of oxidation states:

• HNO3: +5
• NO: +2
• N2: 0
• NH4Cl: -3

So, the correct order in decreasing oxidation state is:

HNO3, NO, N2, NH4Cl