Class 11 Chemistry (Chemistry Part-II) Chapter 7 Redox Reactions

Key Concept: Electron Flow

c) Zinc electrode to copper electrode

[Solution Description] Electrons in a Daniell cell flow from the zinc electrode, which acts as the anode where oxidation occurs, to the copper electrode, which acts as the cathode where reduction takes place.

Key Concept: Medium Identification

b) Basic

[Solution Description] The given reaction involves hydroxide ions $(OH^-)$, which indicates that it is balanced in a basic medium.

Key Concept: Basic Reaction Identification

a) $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$

[Solution Description] Decomposition reactions involve breaking down a single compound into two or more products, at least one of which is in an elemental state. Here, option a) represents such a breakdown.

Key Concept: Reaction Classification

b) Decomposition

[Solution Description] The given reaction represents a decomposition reaction where $KClO_3$ breaks down into $KCl$ and $O_2$. Decomposition reactions involve breaking down a compound into simpler substances, often seen when a compound separates into its components.

Key Concept: Alkali Metals

b) +1

[Solution Description] Alkali metals like potassium have a common oxidation number of +1 in compounds. Potassium in KBr follows this rule and thus has an oxidation number of +1.

Key Concept: Comparative Questions

b) $Mg(s) + Cl_2(g) \rightarrow MgCl_2(s)$

[Solution Description] The correct answer is the reaction where magnesium (Mg) reacts with chlorine $(Cl_2)$: $Mg(s) + Cl_2(g) \rightarrow MgCl_2(s)$. This reaction involves the addition of the electronegative element chlorine to magnesium.

Key Concept: Element Requirement

c) Chlorine

[Solution Description] Chlorine can undergo disproportionation reactions because it can exist in multiple oxidation states including negative, zero, and positive states. For instance, in the reaction $Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O$, chlorine changes from zero to both -1 and +1.

Key Concept: Fractional Oxidation States

b) +2.67

[Solution Description] The compound $Fe_3O_4$ can be represented as a mixture of $FeO$ and $Fe_2O_3$. In $FeO$, iron has an oxidation state of +2, while in $Fe_2O_3$, iron has an oxidation state of +3. Therefore, for $Fe_3O_4$, we calculate the average oxidation state:

$\text{Average oxidation state} = \frac{(1 \times 2) + (2 \times 3)}{3} = \frac{2 + 6}{3} = \frac{8}{3} = 2.67$

Thus, the average oxidation state of iron in $Fe_3O_4$ is 2.67.

Key Concept: Half-Reaction Balancing

c) $\text{MnO}_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4H_2O(l)$

[Solution Description] In acidic solution, we balance the oxygen by adding water and hydrogen by adding hydrogen ions, but since it’s already balanced for atoms, we only focus on balancing charges. The net charge on the left side initially considering one electron added is $1- + 8 = +7$. On the right, it is $2+$.

To balance the charge, five more electrons must be added to the left: $\text{MnO}_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4H_2O(l)$

Key Concept: Peroxide vs Superoxide

d) $-\frac{1}{2}$

[Solution Description] In superoxides, such as $KO_2$ or $RbO_2$, the oxidation number of oxygen is $-\frac{1}{2}$. This is because the overall negative charge is split over two oxygen atoms, resulting in each having an oxidation state of $-\frac{1}{2}$.

Key Concept: Comparative Reactivity

b) Chlorine

[Solution Description] According to the reactivity series of halogens, chlorine is more reactive than bromine. Therefore, chlorine can displace bromine from its compounds. The reaction is:

$Cl_2(g) + 2KBr(aq) \rightarrow 2KCl(aq) + Br_2(l)$

Hence, chlorine will displace bromine from potassium bromide.

Key Concept: Broadened Definition

b) $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$

[Solution Description] Oxidation can be defined as the removal of hydrogen from a compound. In this context, we need to identify a reaction where hydrogen is removed from one of the reactants. The reaction $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ involves methane $(CH_4)$, where hydrogen atoms are being removed to form carbon dioxide $(CO_2)$ and water $(H_2O)$. This matches the definition of oxidation through hydrogen removal.

Key Concept: Oxidation Number Calculation

c) +7

[Solution Description] To find the oxidation number of chlorine in $KClO_4$, consider that:

– Potassium (K) has an oxidation number of +1.

– Oxygen (O) typically has an oxidation number of -2.

The overall charge of the compound is 0, so we set up the equation: $(+1) + x + 4(-2) = 0$

Simplifying gives: $1 + x – 8 = 0 \implies x = +7$

Thus, the oxidation number of chlorine in $KClO_4$ is +7.

Key Concept: Real-World Application

c) Hall-Héroult process

[Solution Description] Metallurgical processes like the extraction of aluminum involve electrolytic decomposition. During the Hall-Héroult process, aluminum oxide is decomposed to obtain pure aluminum.

Key Concept: Half-Reaction Identification

b) $Cl_2(g) + 2e^- \rightarrow 2Cl^-(g)$

[Solution Description] Reduction involves the gain of electrons. For the formation of $NaCl$, chlorine gas $(Cl_2)$ undergoes reduction by gaining electrons to form chloride ions (Cl^-). The correct half-reaction is: $Cl_2(g) + 2e^- \rightarrow 2Cl^-(g)$

Key Concept: Basic Concept

d) 0.00 volts

[Solution Description] By definition, the standard electrode potential of the hydrogen electrode is taken as 0.00 volts for reference in electrochemistry.

Key Concept: Basic Definition

c) A reaction where oxidation and reduction occur simultaneously

[Solution Description] A redox reaction is one where oxidation and reduction occur simultaneously. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons.

Key Concept: Ionic Form

a) $Br_2(l) + 2I^-(aq) \rightarrow 2Br^-(aq) + I_2(s)$

[Solution Description] In this reaction, bromine $(Br_2)$ displaces iodine from potassium iodide (KI). The ionic form highlights that bromine molecules gain electrons donated by iodide ions, forming bromide ions and elemental iodine.

$Br_2(l) + 2I^-(aq) \rightarrow 2Br^-(aq) + I_2(s)$

Key Concept: Rule Application

c) $CaH_2$

[Solution Description] Among the given options, $CaH_2$ is a metal hydride. In metal hydrides like $CaH_2$, the oxidation state of hydrogen is -1 due to its combination with a metal.

Key Concept: Complex Reaction Analysis, Reaction Balancing

c) Chlorine (Cl)

[Solution Description] The balanced equation is already given, so we need to identify which element’s oxidation state increases during the reaction.

– Determine the oxidation states of elements in reactants and products:

– In $\text{KMnO}_4$, Mn has an oxidation state of +7.

– In $\text{MnCl}_2$, Mn has an oxidation state of +2.

– Thus, Mn is reduced from +7 to +2.

– In $\text{HCl}$, Cl has an oxidation state of -1.

– In $\text{Cl}_2$, Cl has an oxidation state of 0.

– Thus, Cl is oxidized from -1 to 0.

Since we are identifying the element being oxidized, it is Chlorine.

Key Concept: Reaction Type Identification

a) $3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)$

[Solution Description] The definition of a combination reaction states that two or more substances combine to form one product. Let’s analyze each option:

– Option A: The reactants are $3Mg(s)$ and $N_2(g)$, combining to form $Mg_3N_2(s)$. This is a single product from multiple reactants, so it is a combination reaction.

– Option B: The equation shows decomposition into separate elements, not a combination.

– Option C: There is no chemical reaction depicted in this option.

– Option D: Shows a decomposition reaction breaking down into simpler substances.

Therefore, option A is a combination reaction as it involves multiple reactants forming one product.

Key Concept: Reaction Classification

c) Displacement

[Solution Description] First, identify the species being oxidized and reduced:

– Copper (Cu) changes from $0$ to $+2$ in $Cu(NO_3)_2$, hence it is oxidized.

– Silver $(Ag^+ in AgNO_3)$ changes from $+1$ to $0$ in $Ag$, hence it is reduced.

This reaction involves the transfer of electrons between copper and silver ions, indicating it is a displacement reaction where one element displaces another in a compound.

Key Concept: Colour Change

c) Purple to colourless

[Solution Description] During a redox titration using $MnO_4^-$ as a self-indicator, the solution starts out purple. As the reaction proceeds and the $MnO_4^-$ is reduced, the solution turns colourless, indicating the endpoint of the titration.

Key Concept: Ion Charge Identification

b) +2

[Solution Description] The oxidation number of a monoatomic ion equals its charge. Therefore, for the $\text{Ca}^{2+}$ ion, the oxidation number is +2.

Key Concept: Observation of Color Change

a) Blue to colorless

[Solution Description] The reaction between zinc and copper nitrate involves the reduction of $\text{Cu}^{2+}$ ions to copper metal $(\text{Cu(s)})$, which deposits on the zinc strip, and the oxidation of zinc to $\text{Zn}^{2+}$. The blue color of the copper nitrate solution (due to $\text{Cu}^{2+}$ ions) fades as copper metal forms.

Key Concept: Reaction Medium

a) $6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$

[Solution Description] To balance the redox equation in an acidic medium, we will use the half-reaction method: Half-Reaction for $Fe^{2+} \rightarrow Fe^{3+}$:

– Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$

Half-Reaction for $Cr_2O_7^{2-} \rightarrow Cr^{3+}$:

– Reduction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

Now, to balance the electrons, multiply the oxidation half-reaction by 6: $6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$

Combine both half-reactions:

$6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$

The balanced equation is option a).

Key Concept: Complex Reaction Analysis, Balancing Equations

a) $5H_2O_2 + 2MnO_4^- + 16H^+ \rightarrow 5O_2 + 2Mn^{2+} + 8H_2O$

[Solution Description] To balance the redox reaction between $H_2O_2$ and $MnO_4^-$ in acidic conditions, follow these steps:

– **Write the unbalanced equation:**

$$H_2O_2 + MnO_4^- \rightarrow O_2 + Mn^{2+}$$

– **Separate into half-reactions:**

Oxidation half-reaction for $H_2O_2$:

$$H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$$

Reduction half-reaction for $MnO_4^-$:

$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

– **Equalize electrons by multiplying oxidation half-reaction by 5 and reduction by 2:**

$$5(H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-)$$

$$2(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O)$$

– **Combine and simplify:**

$$5H_2O_2 + 2MnO_4^- + 16H^+ \rightarrow 5O_2 + 2Mn^{2+} + 8H_2O$$

This is the balanced equation.

Key Concept: Complex Reaction Analysis, Advanced Oxidation Number

c) Iron, changing from $+\frac{8}{3}$ to $0$

[Solution Description] The balanced chemical equation for the reaction is: $Fe_3O_4 + 4CO \rightarrow 3Fe + 4CO_2$

Now, determining the oxidation numbers: – In $Fe_3O_4$, each $Fe$ has an average oxidation number of $+8/3$. This is because $3x + 4(-2) = 0 \Rightarrow 3x – 8 = 0 \Rightarrow x = +8/3$.

– In $CO$, carbon is in oxidation state $+2$ and oxygen is $-2$.

– In $CO_2$, carbon’s oxidation state changes to $+4$.

The iron changes from $+\frac{8}{3}$ in $Fe_3O_4$ to $0$ in $Fe$, so it is reduced. Hence, the answer is Iron (Fe), changing from $+\frac{8}{3}$
to $0$.

Key Concept: Half-Reaction Analysis

b) $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

[Solution Description] In the given reaction, zinc is oxidized and copper ions are reduced. The reduction half-reaction involves the gain of electrons by copper ions:

$\text{Reduction half-reaction: } Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

Key Concept: Electron Transfer

b) 2

[Solution Description] In this redox reaction:

– Zinc loses two electrons to become $\text{Zn}^{2+}$, hence zinc undergoes oxidation.

– These two electrons are gained by $\text{Cu}^{2+}$ ions, reducing them to copper metal.

Thus, the number of electrons transferred is 2.

Key Concept: Reaction Balancing

a) $MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$

[Solution Description] In acidic solution, the first step is to write the oxidation and reduction half-reactions separately:

Reducing half-reaction: $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

Oxidizing half-reaction: $$Fe^{2+} \rightarrow Fe^{3+} + e^-$$

Next, we need to multiply the oxidizing half-reaction by 5 to balance the electrons: $$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$$

Now combine both half-reactions: $$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$$

This gives us the balanced equation.

Key Concept: Oxidation Number Assignment

b) +5

[Solution Description] To find the oxidation number of chlorine in $ClO_3^-$, let’s denote it as $x$. The oxidation number of oxygen is -2. The total charge on the ion is -1. Therefore, we can set up the equation: $x + 3(-2) = -1$

Solving for $x$ gives:

$x – 6 = -1$

$x = +5$

Hence, the oxidation number of chlorine in $ClO_3^-$ is +5.

Key Concept: Iodine-Starch Reaction

b) Blue to colorless

[Solution Description] In a titration where iodine is involved with starch as an indicator, the endpoint is detected by a color change from blue to colorless. Initially, iodine forms a complex with starch producing a blue-black color, and as iodine is consumed, the solution becomes colorless.

Key Concept: Advanced Exception Identification, Complex Compound Analysis

b), $OF_2$

[Solution Description] Generally, oxygen has an oxidation state of -2, but in some compounds called peroxy compounds or when bonded to fluorine, oxygen can exhibit a positive oxidation state.

Considering examples:

– In $H_2O_2$, oxygen is -1.

– In $OF_2$, oxygen takes on an oxidation state of +2.

– In $Na_2O$, oxygen is -2.

– In $CO_2$, oxygen is -2.

Therefore, the compound where oxygen has a positive oxidation state is $OF_2$.

Key Concept: Metal Displacement

a) Option A

[Solution Description] Option A shows zinc displacing copper from copper(II) sulfate solution, resulting in the formation of solid copper and zinc sulfate. This is a classic metal displacement reaction where a more reactive metal (zinc) displaces a less reactive metal (copper) from its compound.

Key Concept: Salt Bridge Role

d) To maintain electrical neutrality

[Solution Description] The main function of the salt bridge in a Daniell cell is to maintain electrical neutrality by allowing the migration of ions between the two half-cells. This prevents the build-up of charge that would otherwise stop the reaction.

Key Concept: Current Flow Direction

b) Copper to Zinc

[Solution Description] In a Daniell cell, electrons flow from the zinc electrode to the copper electrode through the external circuit. This implies that the conventional current flows from copper to zinc.

Key Concept: Electron Flow

b) Zinc to Copper

[Solution Description] In a Daniell cell, electrons flow from zinc to copper. Zinc undergoes oxidation, releasing electrons. These electrons travel through the metallic wire towards the copper electrode, where they cause the reduction of copper ions.

Key Concept: Oxidation Number Calculation

c) +6

[Solution Description] In $H_2SO_4$, hydrogen has an oxidation number of +1 and oxygen has an oxidation number of -2. Let the oxidation number of sulfur be $x$. The sum of the oxidation numbers must equal zero for a neutral compound, so:

$2(+1) + x + 4(-2) = 0$

Solving for $x$:

$2 + x – 8 = 0$

$x = 6$

Hence, the oxidation number of sulfur is +6.

Key Concept: Reaction Conditions

b) $I_2 + 2Cl^- \rightarrow 2I^- + Cl_2$

[Solution Description] The order of reactivity for halogens decreases from fluorine to iodine. This means that a more reactive halogen can displace a less reactive halogen from its compound. Given the options, we need to determine which reaction involves an attempt by a less reactive halogen to displace a more reactive one. Based on the reactivity series: $F_2 > Cl_2 > Br_2 > I_2$. Checking each option:

a) $Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2$: Chlorine is more reactive than bromine, so this reaction is feasible.

b) $I_2 + 2Cl^- \rightarrow 2I^- + Cl_2$: Iodine is less reactive than chlorine, so this reaction will not occur.

c) $Br_2 + 2I^- \rightarrow 2Br^- + I_2$: Bromine is more reactive than iodine, so this reaction is feasible.

d) $Br_2 + 2Cl^- \rightarrow 2Br^- + Cl_2$: Bromine is less reactive than chlorine, so this reaction will not occur.

Therefore, both b) and d) are not feasible; however, only one choice must be correct according to MCQ design, thus referring to common single-answer format, pick one as most indicative here being b).

Key Concept: Electron Flow

c) From zinc to copper

[Solution Description] In a Daniell cell, electrons flow from the zinc electrode (anode) where oxidation occurs to the copper electrode (cathode) where reduction takes place.

Key Concept: Oxidation Numbers in Polyatomic Ions

c) +6

[Solution Description] The sum of oxidation numbers in a polyatomic ion equals its charge. Let $x$ be the oxidation number of Cr. Since oxygen generally has an oxidation number of -2:

$2x + 7(-2) = -2$

Simplifying, $2x – 14 = -2$

$2x = 12$

$x = 6$

Therefore, the oxidation number of chromium in the dichromate ion is +6.

Key Concept: Balancing Half-Reactions

b) $Mn^{2+} + 4H_2O \rightarrow MnO_4^- + 8H^+ + 5e^-$

[Solution Description] In acidic medium, balance the equation by adding water molecules for oxygen and hydrogen ions for hydrogen. The unbalanced half-reaction is:

$Mn^{2+} \rightarrow MnO_4^-$

Add four water molecules to balance oxygen: $Mn^{2+} + 4H_2O \rightarrow MnO_4^-$

Now add eight hydrogen ions to balance hydrogen: $Mn^{2+} + 4H_2O \rightarrow MnO_4^- + 8H^+$

Balance charge by adding five electrons to the right: $Mn^{2+} + 4H_2O \rightarrow MnO_4^- + 8H^+ + 5e^-$

Thus, the balanced half-reaction is: $Mn^{2+} + 4H_2O \rightarrow MnO_4^- + 8H^+ + 5e^-$

Key Concept: Oxidation State Assignment

b) $-1$

[Solution Description] For $H_2O_2$, the sum of oxidation states must equal 0. Hydrogen typically has an oxidation state of $+1$. Let the oxidation state of oxygen be $x$. The equation is: $2(+1) + 2(x) = 0$. Simplifying gives $2 + 2x = 0$, then $2x = -2$, leading to $x = -1$. Thus, the oxidation state of oxygen in $H_2O_2$ is $-1$.

Key Concept: Basic Reaction Identification

a) $H_2 + Cl_2 \rightarrow 2HCl$

[Solution Description] A combination reaction involves two or more reactants forming a single product. Analyze each reaction to find which fits this description.

Key Concept: Simple Product Prediction

a) $MgCl_2(s)$ and $H_2(g)$

[Solution Description] When magnesium $(Mg)$ reacts with hydrochloric acid $(HCl)$, magnesium donates two electrons (oxidation) to form $Mg^{2+}$ while hydrogen ions from $HCl$ gain these electrons (reduction). The overall reaction produces magnesium chloride $(MgCl_2)$ and hydrogen gas $(H_2)$.

Key Concept: Balancing in Basic Medium

b) $Cl_2 + 6OH^- \rightarrow 2Cl^- + ClO_3^- + 3H_2O$

[Solution Description] To balance the redox reaction in a basic medium:

Step 1: Write down the half-reactions from the unbalanced equation.

$$Cl_2 \rightarrow Cl^- + ClO_3^-$$

Step 2: Balance atoms and charges by adding water molecules, hydrogen ions, or electrons where necessary.

Oxidation:

$$Cl_2 + 6OH^- \rightarrow 2Cl^- + ClO_3^- + 3H_2O$$

Reduction:

Already balanced with respect to charge and mass.

Final balanced equation:

$$Cl_2 + 6OH^- \rightarrow 2Cl^- + ClO_3^- + 3H_2O$$

Key Concept: Disproportionation Reaction

a) $Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$

[Solution Description] In a disproportionation reaction, one element undergoes both oxidation and reduction. For option a) $Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$, chlorine starts with an oxidation state of 0 in $Cl_2$. It is reduced to -1 in $Cl^-$ and oxidized to +1 in $ClO^-$. Thus, this is a disproportionation reaction.

Key Concept: Reaction Type

c) Reduction

[Solution Description] In the given reaction, chlorine gas $(Cl_2)$ is gaining electrons to form chloride ions $(Cl^-)$. This gain of electrons indicates that it is a reduction process. Therefore, this reaction is classified as a reduction reaction.

Key Concept: Concept Application

c) $2 H_2S(g) + O_2 (g) \rightarrow 2 S (s) + 2 H_2O (l)$

[Solution Description] Oxidation by removal of oxygen is seen when a compound loses its oxygen atoms in a chemical reaction. In option c) $2 H_2S(g) + O_2 (g) \rightarrow 2 S (s) + 2 H_2O (l)$, oxygen is removed from $H_2S$ and combined to form water.

Key Concept: Comparative Analysis

c) $Mg + F_2 \rightarrow MgF_2$

[Solution Description] Oxidation includes the addition of electronegative elements like oxygen and halogens. Among the given options, the reaction $Mg + F_2 \rightarrow MgF_2$ demonstrates the addition of an electronegative element, fluorine, to magnesium, resulting in the formation of magnesium fluoride $(MgF_2)$.

Key Concept: Compound Oxidation Sum

c) +12

[Solution Description] The compound $\text{K}_2\text{Cr}_2\text{O}_7$ is neutral, so the sum of oxidation numbers must be zero. The oxidation number of potassium ($\text{K}$) is +1 and there are two potassium atoms, so their contribution is $2 \times (+1) = +2$. Oxygen has an oxidation number of -2 and there are seven oxygen atoms, contributing $7 \times (-2) = -14$. Let the oxidation number of chromium ($\text{Cr}$) be $x$. There are two chromium atoms, so their contribution is $2x$. Therefore, the equation for the sum is: $2(+1) + 2x + 7(-2) = 0$

Simplifying, we get: $2 + 2x – 14 = 0$

$2x – 12 = 0$

$2x = 12$

$x = 6$

Therefore, the oxidation number of each Cr atom is +6.

Key Concept: Charge Balancing

a) 1

[Solution Description] The oxidation half-reaction from $Fe^{2+}$ to $Fe^{3+}$ can be written as: $Fe^{2+} \rightarrow Fe^{3+} + e^-$

To balance the charge, one electron is required. Hence, 1 electron is needed.

Key Concept: Simple Reaction Analysis

b) $Zn$

[Solution Description] Zinc (Zn) donates electrons and becomes $Zn^{2+}$, thus acting as the reducing agent by increasing its own oxidation state.