GSEB Class 11 Chemistry Solutions Chapter 8 Redox Reactions

Question 1. Assign oxidation number to the underlined elements in each of the following species.
(a) \( \text{NaH}_2\text{PO}_4 \)
(b) \( \text{NaHSO}_4 \)
(c) \( \text{H}_4\text{P}_2\text{O}_7 \)
(d) \( \text{K}_2\text{MnO}_4 \)
(e) \( \text{CaO}_2 \)
(f) \( \text{NaBH}_4 \)
(g) \( \text{H}_2\text{S}_2\text{O}_7 \)
(h) \( \text{KAl}(\text{SO}_4)_2 \cdot 12\text{H}_2\text{O} \)
Answer:
(a) For \( \text{NaH}_2\text{PO}_4 \): Let the oxidation number of P be \( x \). We know Na is \( +1 \), H is \( +1 \), and O is \( -2 \).
Sum of oxidation numbers: \( 1(+1) + 2(+1) + 1(x) + 4(-2) = 0 \)
\( 1 + 2 + x - 8 = 0 \)
\( x - 5 = 0 \)
\( \implies x = +5 \)
The oxidation number of P in \( \text{NaH}_2\text{PO}_4 \) is \( +5 \).
(b) For \( \text{NaHSO}_4 \): Let the oxidation number of S be \( x \). We know Na is \( +1 \), H is \( +1 \), and O is \( -2 \).
Sum of oxidation numbers: \( 1(+1) + 1(+1) + x + 4(-2) = 0 \)
\( 1 + 1 + x - 8 = 0 \)
\( x - 6 = 0 \)
\( \implies x = +6 \)
The oxidation number of S in \( \text{NaHSO}_4 \) is \( +6 \).
(c) For \( \text{H}_4\text{P}_2\text{O}_7 \): Let the oxidation number of P be \( x \). We know H is \( +1 \) and O is \( -2 \).
Sum of oxidation numbers: \( 4(+1) + 2(x) + 7(-2) = 0 \)
\( 4 + 2x - 14 = 0 \)
\( 2x - 10 = 0 \)
\( \implies 2x = 10 \implies x = +5 \)
The oxidation number of P in \( \text{H}_4\text{P}_2\text{O}_7 \) is \( +5 \).
(d) For \( \text{K}_2\text{MnO}_4 \): Let the oxidation number of Mn be \( x \). We know K is \( +1 \) and O is \( -2 \).
Sum of oxidation numbers: \( 2(+1) + x + 4(-2) = 0 \)
\( 2 + x - 8 = 0 \)
\( x - 6 = 0 \)
\( \implies x = +6 \)
The oxidation number of Mn in \( \text{K}_2\text{MnO}_4 \) is \( +6 \).
(e) For \( \text{CaO}_2 \): Let the oxidation number of oxygen be \( x \). We know Ca is \( +2 \).
Sum of oxidation numbers: \( +2 + 2(x) = 0 \)
\( 2x = -2 \)
\( \implies x = -1 \)
The oxidation number of O in \( \text{CaO}_2 \) is \( -1 \) (it is a peroxide).
(f) For \( \text{NaBH}_4 \): Let the oxidation number of B be \( x \). We know Na is \( +1 \). In metal hydrides, H is \( -1 \).
Sum of oxidation numbers: \( +1 + x + 4(-1) = 0 \)
\( 1 + x - 4 = 0 \)
\( x - 3 = 0 \)
\( \implies x = +3 \)
The oxidation number of B in \( \text{NaBH}_4 \) is \( +3 \).
(g) For \( \text{H}_2\text{S}_2\text{O}_7 \): Let the oxidation number of S be \( x \). We know H is \( +1 \) and O is \( -2 \).
Sum of oxidation numbers: \( 2(+1) + 2(x) + 7(-2) = 0 \)
\( 2 + 2x - 14 = 0 \)
\( 2x - 12 = 0 \)
\( \implies 2x = 12 \implies x = +6 \)
The oxidation number of S in \( \text{H}_2\text{S}_2\text{O}_7 \) is \( +6 \).
(h) For \( \text{KAl}(\text{SO}_4)_2 \cdot 12\text{H}_2\text{O} \): We assume the oxidation number of S is to be determined. We know K is \( +1 \), Al is \( +3 \), O is \( -2 \), and H in water is \( +1 \). Water molecules have an overall oxidation number of 0.
Let the oxidation number of S be \( x \).
Sum of oxidation numbers: \( 1(+1) + 1(+3) + 2(x + 4(-2)) + 12(0) = 0 \)
\( 1 + 3 + 2(x - 8) = 0 \)
\( 4 + 2x - 16 = 0 \)
\( 2x - 12 = 0 \)
\( \implies 2x = 12 \implies x = +6 \)
The oxidation number of S in \( \text{KAl}(\text{SO}_4)_2 \cdot 12\text{H}_2\text{O} \) is \( +6 \).
In simple words: To find the oxidation number, we assign standard values to common elements (like H as +1, O as -2, Na as +1, K as +1, Ca as +2, Al as +3, H in metal hydrides as -1). Then, we set the total sum of all oxidation numbers in the compound or ion to its overall charge (0 for a neutral compound). We solve for the unknown element's oxidation number.

Exam Tip: Remember to consider special cases like peroxides (where O is -1) or metal hydrides (where H is -1). Always ensure the sum of oxidation states equals the molecule's or ion's net charge.

Question 2. What are the oxidation number of the underlined elements in each of the following & how do you rationalise your result?
(a) \( \text{KI}_3 \)
(b) \( \text{H}_2\text{S}_4\text{O}_6 \)
(c) \( \text{Fe}_3\text{O}_4 \)
(d) \( \text{CH}_3\text{CH}_2\text{OH} \)
(e) \( \text{CH}_3\text{COOH} \)
Answer:
(a) For \( \text{KI}_3 \): The compound \( \text{KI}_3 \) is formed from a potassium ion \( (\text{K}^+) \) and a triiodide ion \( (\text{I}_3^-) \). In the \( \text{I}_3^- \) ion, one iodine atom carries a \( -1 \) charge, while the other two iodine atoms have an oxidation number of \( 0 \). The average oxidation number of I in \( \text{KI}_3 \) is \( -1/3 \). However, if we consider the iodine atom directly bonded to K, its oxidation number is \( -1 \), as iodine acts similarly to a halide. Thus, we can rationalise the oxidation number of I in \( \text{KI}_3 \) as \( -1 \), when viewed as \( \text{K}^+\text{I}^- \cdot \text{I}_2 \), where \( \text{I}_2 \) has 0. For the underlined element I in \( \text{KI}_3 \), its oxidation number is \( -1 \).
(b) For \( \text{H}_2\text{S}_4\text{O}_6 \) (tetrathionic acid): The structure of \( \text{H}_2\text{S}_4\text{O}_6 \) is H-O-\(\text{SO}_2\)-S-S-\(\text{SO}_2\)-O-H. In this structure, the two central sulfur atoms are directly bonded to other sulfur atoms and are not bonded to oxygen, so their oxidation number is \( 0 \). The two terminal sulfur atoms (in \( \text{SO}_2 \) groups) are each bonded to two oxygen atoms and one sulfur atom. For these terminal S atoms, their oxidation number is \( +5 \). Therefore, the oxidation number of S in \( \text{H}_2\text{S}_4\text{O}_6 \) varies within the molecule, with some S atoms being \( 0 \) and others being \( +5 \).
(c) For \( \text{Fe}_3\text{O}_4 \):
By the conventional method, let the oxidation number of Fe be \( x \). Since O is \( -2 \), we have:
\( 3x + 4(-2) = 0 \)
\( 3x - 8 = 0 \)
\( \implies x = 8/3 \)
Rationalisation: \( \text{Fe}_3\text{O}_4 \) is a mixed oxide, consisting of \( \text{FeO} \) and \( \text{Fe}_2\text{O}_3 \). In \( \text{FeO} \), Fe has an oxidation number of \( +2 \). In \( \text{Fe}_2\text{O}_3 \), Fe has an oxidation number of \( +3 \). So, iron in \( \text{Fe}_3\text{O}_4 \) actually has oxidation numbers of \( +2 \) and \( +3 \).
(d) For \( \text{CH}_3\text{CH}_2\text{OH} \) (ethanol): We calculate the oxidation number for each carbon atom separately.
For \( \text{C}_2 \) (the carbon in \( \text{CH}_2\text{OH} \)): It is attached to three H-atoms (less electronegative than carbon, so H is \( +1 \)) and one \( -\text{CH}_2\text{OH} \) group (which can be considered to contribute \( -1 \) if thinking about bond polarity from the adjacent carbon). Using the source's calculation: \( 3(+1) + x + 1(-1) = 0 \implies x = -2 \). So, the oxidation number of \( \text{C}_2 \) is \( -2 \).
For \( \text{C}_1 \) (the carbon in \( \text{CH}_3 \)): It is attached to one O-atom by a single bond, two H-atoms, and one \( -\text{CH}_3 \) group. The source's calculation: \( +1 + 2(+1) + x - 1 = 0 \implies x = -2 \). So, the oxidation number of \( \text{C}_1 \) is \( -2 \).
(e) For \( \text{CH}_3\text{COOH} \) (acetic acid): We calculate the oxidation number for each carbon atom.
For \( \text{C}_2 \) (the methyl carbon): It is attached to three H-atoms and one \( -\text{COOH} \) group. If we consider H as \( +1 \) and the \( -\text{COOH} \) group as \( -1 \), then: \( 3(+1) + x + 1(-1) = 0 \implies x = -2 \). So, the oxidation number of \( \text{C}_2 \) is \( -2 \).
For \( \text{C}_1 \) (the carboxyl carbon): It is attached to one O-atom by a double bond \( (-2) \), one OH group \( (-1) \), and one \( -\text{CH}_3 \) group \( (+1) \). The source's calculation: \( +1 + x + (-2) + 1(-1) = 0 \implies x = +2 \). So, the oxidation number of \( \text{C}_1 \) is \( +2 \).
In simple words: For complex molecules, we can find the oxidation number for each atom by considering the electronegativity of the atoms it's bonded to. If an atom is more electronegative, it "takes" electrons, becoming negative. If it's less electronegative, it "gives" electrons, becoming positive. For mixed oxides like \( \text{Fe}_3\text{O}_4 \), the average oxidation state might not be a whole number, showing it's a mix of different oxidation states.

Exam Tip: When calculating oxidation numbers for carbon atoms in organic compounds, remember to consider each carbon atom individually based on its specific bonds to other atoms and their electronegativities.

Question 3. Justify that the following reactions are redox reactions :
(a) \( \text{CuO}(\text{S}) + \text{H}_2(\text{g}) \rightarrow \text{Cu}(\text{S}) + \text{H}_2\text{O}(\text{g}) \)
(b) \( \text{Fe}_2\text{O}_3(\text{S}) + 3\text{CO}(\text{g}) \rightarrow 2\text{Fe}(\text{S}) + 3\text{CO}_2(\text{g}) \)
(c) \( 4\text{BCl}_3(\text{g}) + 3\text{LiAlH}_4(\text{S}) \rightarrow 2\text{B}_2\text{H}_6(\text{g}) + 3\text{LiCl}(\text{s}) + \text{AlCl}_3 \)
(d) \( 2\text{K}(\text{S}) + \text{F}_2(\text{g}) \rightarrow 2\text{K}^+\text{F}^-(\text{S}) \)
(e) \( 4\text{NH}_3(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 4\text{NO}(\text{g}) + 6\text{H}_2\text{O}(\text{g}) \)
Answer:
(a) For \( \text{CuO}(\text{S}) + \text{H}_2(\text{g}) \rightarrow \text{Cu}(\text{S}) + \text{H}_2\text{O}(\text{g}) \):
In this reaction, copper in \( \text{CuO} \) goes from an oxidation number of \( +2 \) to \( 0 \) in \( \text{Cu(S)} \), which shows a decrease in oxidation number (reduction). Hydrogen in \( \text{H}_2(\text{g}) \) goes from \( 0 \) to \( +1 \) in \( \text{H}_2\text{O}(\text{g}) \), which shows an increase in oxidation number (oxidation). Since both oxidation and reduction occur, this is a redox reaction.
\( \text{Cu} \overset{+2}{\text{O}} (\text{s}) + \text{H}_2 \overset{0}{(\text{g})} \rightarrow \text{Cu} \overset{0}{(\text{s})} + \text{H}_2 \overset{+1}{\text{O}} (\text{g}) \)
(b) For \( \text{Fe}_2\text{O}_3(\text{S}) + 3\text{CO}(\text{g}) \rightarrow 2\text{Fe}(\text{S}) + 3\text{CO}_2(\text{g}) \):
Iron in \( \text{Fe}_2\text{O}_3 \) goes from an oxidation number of \( +3 \) to \( 0 \) in \( \text{Fe(S)} \), which is a reduction. Carbon in \( \text{CO}(\text{g}) \) goes from \( +2 \) to \( +4 \) in \( \text{CO}_2(\text{g}) \), which is an oxidation. As both oxidation and reduction take place, this is a redox reaction.
\( \text{Fe}_2 \overset{+3}{\text{O}}_3 (\text{s}) + 3\text{C} \overset{+2}{\text{O}} (\text{g}) \rightarrow 2\text{Fe} \overset{0}{(\text{s})} + 3\text{C} \overset{+4}{\text{O}}_2 (\text{g}) \)
(c) For \( 4\text{BCl}_3(\text{g}) + 3\text{LiAlH}_4(\text{S}) \rightarrow 2\text{B}_2\text{H}_6(\text{g}) + 3\text{LiCl}(\text{s}) + \text{AlCl}_3 \):
Boron in \( \text{BCl}_3 \) goes from an oxidation number of \( +3 \) to \( -3 \) in \( \text{B}_2\text{H}_6 \), showing a reduction. Hydrogen in \( \text{LiAlH}_4 \) goes from \( -1 \) to \( +1 \) in \( \text{B}_2\text{H}_6 \), showing an oxidation. Because of these changes in oxidation numbers, this reaction is a redox reaction.
(d) For \( 2\text{K}(\text{S}) + \text{F}_2(\text{g}) \rightarrow 2\text{K}^+\text{F}^-(\text{S}) \):
Potassium goes from an oxidation number of \( 0 \) in \( \text{K}(\text{S}) \) to \( +1 \) in \( \text{K}^+\text{F}^-(\text{S}) \), which is an oxidation. Fluorine in \( \text{F}_2(\text{g}) \) goes from \( 0 \) to \( -1 \) in \( \text{K}^+\text{F}^-(\text{S}) \), which is a reduction. Since both processes occur, this is a redox reaction.
\( 2\text{K} \overset{0}{(\text{s})} + \text{F}_2 \overset{0}{(\text{g})} \rightarrow 2\text{K} \overset{+1}{\text{F}} \overset{-1}{(\text{s})} \)
(e) For \( 4\text{NH}_3(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 4\text{NO}(\text{g}) + 6\text{H}_2\text{O}(\text{g}) \):
Nitrogen in \( \text{NH}_3 \) goes from an oxidation number of \( -3 \) to \( +2 \) in \( \text{NO} \), showing an oxidation. Oxygen in \( \text{O}_2 \) goes from \( 0 \) to \( -2 \) in \( \text{H}_2\text{O} \) (and also in \( \text{NO} \)), showing a reduction. As these changes happen, this is a redox reaction.
\( 4\text{N} \overset{-3}{\text{H}}_3 (\text{g}) + 5\text{O}_2 \overset{0}{(\text{g})} \rightarrow 4\text{N} \overset{+2}{\text{O}} (\text{g}) + 6\text{H}_2 \overset{-2}{\text{O}} (\text{g}) \)
In simple words: A redox reaction always involves one element losing electrons (oxidation, its oxidation number goes up) and another element gaining electrons (reduction, its oxidation number goes down). If you see these changes, it's a redox reaction.

Exam Tip: To quickly identify if a reaction is redox, check the oxidation numbers of all elements in the reactants and products. If any element's oxidation number changes, it's a redox reaction.

Question 4. Fluorine reacts with ice and result in the change. \( \text{H}_2\text{O}(\text{s}) + \text{F}_2(\text{g}) \rightarrow \text{HF}(\text{g}) + \text{HOF}(\text{g}) \). Justify this reaction is a redox reaction.
Answer:
In this reaction, the oxidation number of fluorine changes. Elemental \( \text{F}_2 \) has an oxidation number of \( 0 \). In \( \text{HF} \), fluorine has an oxidation number of \( -1 \), meaning it has been reduced. In \( \text{HOF} \), fluorine has an oxidation number of \( +1 \), meaning it has been oxidized.
Since \( \text{F}_2 \) is both reduced to \( \text{F}^- \) in \( \text{HF} \) and oxidized to \( \text{F}^+ \) in \( \text{HOF} \), this reaction is a disproportionation reaction, which is a specific type of redox reaction where a single element undergoes both oxidation and reduction. \( \text{HOF} \) is known to be an unstable compound and decomposes to \( \text{H}_2\text{O} \) and \( \text{OF}_2 \). However, fluorine itself does not typically undergo disproportionation in its stable state of \( -1 \), as it can usually only be reduced, not oxidized.
In simple words: This reaction is a redox reaction because fluorine changes its "electron count" in two different ways. Some fluorine atoms gain electrons (get reduced) to become \( \text{HF} \), and other fluorine atoms lose electrons (get oxidized) to become \( \text{HOF} \). This type of reaction, where one element does both, is called disproportionation.

Exam Tip: Remember that a disproportionation reaction is a special kind of redox reaction where the same element is both oxidized and reduced simultaneously.

Question 5. Determine the oxidation number of sulphur, chromium, and nitrogen in \( \text{H}_2\text{SO}_5 \), \( \text{CrO}_5 \) and \( \text{NO}_3^- \). Suggest structures of these compounds. Account for any fallacies.
Answer:
(i) Oxidation number of S in \( \text{H}_2\text{SO}_5 \):
By the conventional method, let the oxidation number of S be \( x \). H is \( +1 \) and O is \( -2 \).
\( 2(+1) + x + 5(-2) = 0 \)
\( 2 + x - 10 = 0 \)
\( \implies x = +8 \) (wrong)
Fallacy: This result is incorrect because the maximum oxidation number for sulfur cannot be greater than \( +6 \), as sulfur only possesses six valence electrons. The actual oxidation number of S in \( \text{H}_2\text{SO}_5 \) (Caro's acid) is \( +6 \).
Rationalisation based on structure: The structure of \( \text{H}_2\text{SO}_5 \) involves a peroxide linkage (O-O bond).
H-O-\(\text{SO}_2\)-O-O-H
In this structure, two oxygen atoms are part of a peroxide linkage, so their oxidation number is \( -1 \). The remaining three oxygen atoms have an oxidation number of \( -2 \).
Thus, the calculation becomes: \( 2(+1) + x + 3(-2) + 2(-1) = 0 \)
\( 2 + x - 6 - 2 = 0 \)
\( \implies x = +6 \)
The oxidation number of S in \( \text{H}_2\text{SO}_5 \) is \( +6 \).
(ii) Oxidation number of Cr in \( \text{CrO}_5 \):
By the conventional method, let the oxidation number of Cr be \( x \). O is \( -2 \).
\( x + 5(-2) = 0 \)
\( \implies x = +10 \) (wrong)
Fallacy: This result is incorrect as the maximum oxidation number for chromium cannot be greater than \( +6 \). Chromium has five 3d electrons and one 4s electron, totaling six valence electrons.
Rationalisation based on structure: In \( \text{CrO}_5 \) (chromium pentoxide), the structure is a butterfly shape with two peroxide bonds. Out of the five oxygen atoms, four oxygen atoms are part of two peroxide bonds (each O is \( -1 \)), and one oxygen atom is double-bonded to chromium (O is \( -2 \)).
Thus, the calculation becomes: \( x + 4(-1) + 1(-2) = 0 \)
\( x - 4 - 2 = 0 \)
\( \implies x = +6 \)
The oxidation number of Cr in \( \text{CrO}_5 \) is \( +6 \).
(iii) Oxidation number of N in \( \text{NO}_3^- \):
By the conventional method, let the oxidation number of N be \( x \). O is \( -2 \), and the ion has a \( -1 \) charge.
\( x + 3(-2) = -1 \)
\( x - 6 = -1 \)
\( \implies x = +5 \)
The oxidation number of N in \( \text{NO}_3^- \) is \( +5 \). There is no fallacy here, as \( +5 \) is a stable oxidation state for nitrogen.
Structure of \( \text{NO}_3^- \) ion: The nitrate ion has a central nitrogen atom bonded to three oxygen atoms. One oxygen is double-bonded, and the other two are single-bonded, with the negative charge delocalized across the oxygen atoms. A simplified representation shows N at the center with three O atoms surrounding it.
In simple words: When a simple calculation gives an oxidation number higher than an element's maximum valence electrons, it signals a "fallacy" or a hidden structural feature. For \( \text{H}_2\text{SO}_5 \) and \( \text{CrO}_5 \), this means there are peroxide (O-O) bonds, where oxygen has a \( -1 \) oxidation state instead of \( -2 \). For \( \text{NO}_3^- \), the calculation is correct, and nitrogen's oxidation state is \( +5 \).

Exam Tip: Be cautious when calculating oxidation numbers that exceed the element's group number (e.g., +8 for sulfur, +10 for chromium). This often indicates the presence of peroxide bonds or unusual structures not accounted for by standard rules.

Question 6. Write formulas for the following compounds :
(a) Mercury (II) chloride
(b) Nickel (IV) sulphate
(c) Tin (IV) oxide
(d) Thallium (I) Sulphate
(e) Iron (III) Sulphate
(f) Chromium (III) Oxide
Answer:
(a) Mercury (II) chloride: \( \text{HgCl}_2 \)
(b) Nickel (IV) sulphate: \( \text{Ni}(\text{SO}_4)_2 \)
(c) Tin (IV) oxide: \( \text{SnO}_2 \)
(d) Thallium (I) Sulphate: \( \text{Tl}_2\text{SO}_4 \)
(e) Iron (III) Sulphate: \( \text{Fe}_2(\text{SO}_4)_3 \)
(f) Chromium (III) Oxide: \( \text{Cr}_2\text{O}_3 \)
In simple words: To write a chemical formula, you need to know the charge of each ion. The Roman numeral tells you the positive ion's charge. Then, you combine them so the total positive charge equals the total negative charge, making the compound neutral.

Exam Tip: Always balance the total positive and negative charges to zero when forming a neutral chemical compound formula.

Question 7. Suggest a list of the substance where. Carbon can exhibit oxidation state from \( -4 \) to \( +4 \) and nitrogen from \( -3 \) to \( +5 \)
Answer:
Substances containing Carbon showing oxidation states from \( -4 \) to \( +4 \):

Substances containing N showing oxidation states from \( -3 \) to \( +5 \):

Exam Tip: Understanding the common oxidation states of elements helps predict their chemical behavior and the types of compounds they can form. For elements like carbon and nitrogen, which form many compounds, a diverse range of oxidation states is expected.

Question 8. While sulphur dioxide & hydrogen peroxide can act as oxidizing as well as reducing agent both in their reactions, ozone, and nitric acid acts only as oxidants. Why?
Answer:
Sulphur dioxide \( (\text{SO}_2) \) and hydrogen peroxide \( (\text{H}_2\text{O}_2) \) can both increase and decrease their oxidation numbers. For instance, in \( \text{SO}_2 \), sulphur has an oxidation number of \( +4 \). It can be oxidized to \( +6 \) (e.g., in \( \text{SO}_3 \) or \( \text{H}_2\text{SO}_4 \)) or reduced to \( 0 \) (elemental S) or \( -2 \) (e.g., in \( \text{H}_2\text{S} \)). Similarly, in \( \text{H}_2\text{O}_2 \), oxygen has an oxidation number of \( -1 \). It can be oxidized to \( 0 \) (in \( \text{O}_2 \)) or reduced to \( -2 \) (in \( \text{H}_2\text{O} \)). Because they can do both, they act as both reducing and oxidizing agents.
However, ozone \( (\text{O}_3) \) and nitric acid \( (\text{HNO}_3) \) function only as oxidants. In \( \text{O}_3 \), oxygen has an average oxidation number of \( 0 \), but it can only decrease its oxidation state (e.g., to \( -1 \) or \( -2 \)). It cannot easily increase its oxidation state. In \( \text{HNO}_3 \), nitrogen is already in its maximum possible oxidation state of \( +5 \). Therefore, it can only decrease its oxidation number (e.g., to \( +4 \) in \( \text{NO}_2 \), \( +2 \) in \( \text{NO} \), or \( -3 \) in \( \text{NH}_3 \)). Since it cannot be oxidized further, \( \text{HNO}_3 \) acts only as an oxidizing agent.
In simple words: \( \text{SO}_2 \) and \( \text{H}_2\text{O}_2 \) are flexible; they can gain or lose electrons depending on the reaction, so they can be both oxidizers and reducers. But \( \text{O}_3 \) and \( \text{HNO}_3 \) are already at their "highest" or easily reducible state for part of their structure (oxygen in ozone, nitrogen in nitric acid), so they can only take electrons from other substances, acting solely as oxidizers.

Exam Tip: An element can act as both an oxidizing and reducing agent if it can exist in intermediate oxidation states, allowing it to either gain or lose electrons. If an element is already at its maximum (or minimum) oxidation state, it can only undergo reduction (or oxidation, respectively).

Question 9. Consider the reaction:
(a) \( 6\text{CO}_2(\text{g}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(\text{aq}) + 6\text{O}_2(\text{g}) \)
(b) \( \text{O}_3(\text{g}) + \text{H}_2\text{O}_2(\text{l}) \rightarrow \text{H}_2\text{O}(\text{l}) + 2\text{O}_2(\text{g}) \)
Why it is more appropriate to write these reactions as :
(a) \( 6\text{CO}_2(\text{g}) + 12\text{H}_2\text{O}(\text{l}) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) + 6\text{O}_2(\text{g}) \)
(b) \( \text{O}_3(\text{g}) + \text{H}_2\text{O}_2(\text{l}) \rightarrow \text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g}) + \text{O}_2(\text{g}) \)
Also suggest a technique to investigate the path of the above (a) & (b) reactions.
Answer:
(a) The reaction \( 6\text{CO}_2(\text{g}) + 12\text{H}_2\text{O}(\text{l}) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) + 6\text{O}_2(\text{g}) \) is more appropriate for photosynthesis. In this equation, \( 12 \) molecules of water are shown as reactants, and \( 6 \) molecules of water are produced. This clarifies the role of water as both a reactant and a product, particularly in tracing the origin of oxygen gas. Atmospheric \( \text{CO}_2 \) is reduced to \( \text{C}_6\text{H}_{12}\text{O}_6 \), and six carbon atoms are reduced. Similarly, out of \( 12 \) water molecules, the oxidation number of oxygen in \( 6\text{H}_2\text{O} \) increases from \( -2 \) to \( 0 \) in \( 6\text{O}_2 \) molecules, while in the remaining \( 6\text{H}_2\text{O} \) molecules, its oxidation number remains the same.
(b) The reaction \( \text{O}_3(\text{g}) + \text{H}_2\text{O}_2(\text{l}) \rightarrow \text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g}) + \text{O}_2(\text{g}) \) is more appropriate because it highlights that \( \text{O}_3 \) is reduced to \( \text{O}_2 \) by the reducing action of \( \text{H}_2\text{O}_2 \). The two \( \text{O}_2(\text{g}) \) products originate from different sources, which can be distinguished using isotopic labeling. The first \( \text{O}_2 \) comes from the ozone, while the second \( \text{O}_2 \) comes from the hydrogen peroxide. This way of writing helps to understand the mechanism and atom transfer.
To investigate the path of the above reactions, a common technique is **isotopic labeling** (using tracer atoms). For instance, in photosynthesis (reaction (a)), if \( \text{H}_2\text{O} \) with heavy oxygen \( (^{18}\text{O}) \) is used, the \( \text{O}_2 \) produced will also contain \( ^{18}\text{O} \), proving that oxygen gas comes from water and not \( \text{CO}_2 \). Similarly, in reaction (b), using \( \text{H}_2^{18}\text{O}_2 \) would reveal that the \( ^{18}\text{O} \) atoms appear in one of the \( \text{O}_2 \) molecules, confirming its origin from hydrogen peroxide.
In simple words: The longer equations show more clearly where all the atoms go during the reaction. For example, in photosynthesis, the extra water on the reactant side helps us see that the oxygen gas we breathe comes from water, not carbon dioxide. For the ozone reaction, writing two separate \( \text{O}_2 \) molecules helps explain that they come from different starting chemicals. To figure out exactly which atoms go where, scientists use special "labeled" atoms (like heavier oxygen) to trace their journey.

Exam Tip: For complex reactions like photosynthesis, balanced equations that explicitly show all intermediates or recycled molecules often provide more mechanistic insight. Isotopic labeling is a powerful tool to track atom movements and confirm reaction pathways.

Question 10. The compound \( \text{AgF}_2 \) is unstable compound. However if formed, the compound acts as a here strong oxidizing agent. Why?
Answer:
The compound \( \text{AgF}_2 \) is unstable because silver is in the \( +2 \) oxidation state, which is an unusually high and less common oxidation state for silver. Silver typically prefers the \( +1 \) oxidation state. Due to this instability, \( \text{AgF}_2 \) readily decomposes to silver in a lower oxidation state and elemental fluorine gas:
\( \text{AgF}_2 \rightarrow \text{Ag} + \text{F}_2 \)
In this decomposition, silver reduces its oxidation state from \( +2 \) to \( 0 \). To do this, it must gain electrons, which means it causes other substances to lose electrons (i.e., it oxidizes them). Therefore, \( \text{AgF}_2 \) acts as a very powerful oxidizing agent, especially because it tends to release \( \text{F}_2(\text{g}) \), which is itself a strong oxidizing agent.
In simple words: \( \text{AgF}_2 \) is not stable because silver wants to be in a simpler form. Since it contains silver in an unusually high \( +2 \) state, it likes to "grab" electrons from other chemicals to change back into a more stable silver form. This means it makes other chemicals lose electrons, which is what a strong oxidizing agent does. It's like it's holding too much energy and wants to get rid of it by oxidizing something else.

Exam Tip: Compounds with elements in unusually high or unstable oxidation states are generally strong oxidizing agents because they have a strong tendency to gain electrons and return to a more stable oxidation state.

Question 11. Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
Answer:
The statement explains that the final oxidation state of the product often depends on which reactant (oxidizing or reducing agent) is present in excess. If the reducing agent is in excess, it can reduce the other reactant more completely, leading to a lower oxidation state in the product. Conversely, if the oxidizing agent is in excess, it can oxidize the other reactant more extensively, resulting in a higher oxidation state.
Here are three illustrations:
(i) **Reducing agent in excess (formation of lower oxidation state):**
When iron(III) oxide \( (\text{Fe}_3\text{O}_4) \) reacts with excess hydrogen gas \( (\text{H}_2) \), it is reduced to elemental iron \( (\text{Fe}) \). In \( \text{Fe}_3\text{O}_4 \), iron has an average oxidation state of \( +8/3 \). In elemental \( \text{Fe} \), its oxidation state is \( 0 \), which is a lower oxidation state.
\( \text{Fe}_3\text{O}_4(\text{s}) + 4\text{H}_2(\text{g}) \text{ (excess)} \rightarrow 3\text{Fe}(\text{s}) + 4\text{H}_2\text{O}(\text{g}) \)
(ii) **Oxidizing agent in excess (formation of higher oxidation state):**
When iron(II) chloride \( (\text{FeCl}_2) \) reacts with excess chlorine gas \( (\text{Cl}_2) \), it is oxidized to iron(III) chloride \( (\text{FeCl}_3) \). In \( \text{FeCl}_2 \), iron has an oxidation state of \( +2 \). In \( \text{FeCl}_3 \), its oxidation state is \( +3 \), which is a higher oxidation state.
\( 2\text{FeCl}_2 + \text{Cl}_2 \text{ (excess)} \rightarrow 2\text{FeCl}_3 \)
(iii) **Oxidizing agent in excess (another example):**
When hydrogen sulfide \( (\text{H}_2\text{S}) \) reacts with excess bromine \( (\text{Br}_2) \), sulfur is oxidized to its elemental form. In \( \text{H}_2\text{S} \), sulfur has an oxidation state of \( -2 \). In elemental \( \text{S} \), its oxidation state is \( 0 \), which is a higher oxidation state.
\( \text{H}_2\text{S} + \text{Br}_2 \text{ (excess)} \rightarrow 2\text{HBr} + \text{S} \)
In simple words: The amount of reactant you have matters. If you have a lot of the chemical that adds electrons (reducing agent), it will push the other chemical to gain as many electrons as possible, resulting in a lower oxidation state. If you have a lot of the chemical that takes electrons (oxidizing agent), it will make the other chemical lose as many electrons as possible, resulting in a higher oxidation state.

Exam Tip: In redox reactions, controlling the stoichiometry (relative amounts of reactants) is crucial for directing the reaction towards a specific product with a desired oxidation state, especially when multiple oxidation states are possible.

Question 12. How do you count for me following observations?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer:
(a) Toluene \( (\text{C}_6\text{H}_5\text{CH}_3) \) is a non-polar organic liquid. It does not dissolve well in aqueous solutions, whether they are acidic or alkaline. Since an effective reaction requires the reactants to mix well, alcoholic potassium permanganate \( (\text{KMnO}_4) \) is preferred. Toluene is soluble in alcohol, allowing for a homogeneous reaction mixture and efficient oxidation.
The simplified oxidation reaction is:
\( \text{C}_6\text{H}_5\text{CH}_3 + 3[\text{O}] \xrightarrow{\text{alc. KMnO}_4} \text{C}_6\text{H}_5\text{COOH} + \text{H}_2\text{O} \)
(b) When concentrated sulfuric acid \( (\text{H}_2\text{SO}_4) \) is added to a chloride-containing mixture, a color-less, pungent-smelling gas, hydrogen chloride \( (\text{HCl}) \), is produced. This is primarily an acid-base reaction where \( \text{H}_2\text{SO}_4 \) acts as a stronger acid, displacing the weaker acid \( \text{HCl} \). \( \text{HCl} \) is not easily oxidized by \( \text{H}_2\text{SO}_4 \):
\( \text{MCl} + \text{H}_2\text{SO}_4 \rightarrow \text{MHSO}_4 + \text{HCl} \)
However, when concentrated \( \text{H}_2\text{SO}_4 \) is added to a bromide-containing mixture, a red vapor of bromine \( (\text{Br}_2) \) is obtained. This is because bromide ions \( (\text{Br}^-) \) are more easily oxidized than chloride ions \( (\text{Cl}^-) \). Concentrated sulfuric acid is a strong enough oxidizing agent to oxidize \( \text{HBr} \) (formed initially) to elemental bromine, which appears as a red vapor:
\( \text{MBr} + \text{H}_2\text{SO}_4 \rightarrow \text{MHSO}_4 + \text{HBr} \)
\( 2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \)
In simple words: (a) We use alcoholic \( \text{KMnO}_4 \) to make benzoic acid from toluene because toluene doesn't mix well with water-based solutions. Alcohol helps everything dissolve, so the reaction works better. (b) When you mix concentrated sulfuric acid with a chloride, you get \( \text{HCl} \) gas, which is just an acid reacting. But with a bromide, you get red bromine vapor because bromide is easier to oxidize than chloride, so the sulfuric acid acts as an oxidizer and turns it into bromine.

Exam Tip: In organic reactions, solvent choice is critical for solubility and reaction efficiency. For halide displacement reactions, remember that the reducing strength of halides increases down the group, making \( \text{Br}^- \) and \( \text{I}^- \) susceptible to oxidation by concentrated \( \text{H}_2\text{SO}_4 \), unlike \( \text{Cl}^- \).

Question 13. Identify the substance oxidised reduced, oxidising agent an reducing agent for each of the following reactions :
(a) \( 2\text{AgBr}(\text{s}) + \text{C}_6\text{H}_6\text{O}_2(\text{aq}) \rightarrow 2\text{Ag}(\text{s}) + 2\text{HBr}(\text{aq}) + \text{C}_6\text{H}_4\text{O}_2(\text{aq}) \)
(b) \( \text{HCHO}(\text{l}) + 2[\text{Ag}(\text{NH}_3)_2]^+(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow 2\text{Ag}(\text{s}) + \text{HCOO}^-(\text{aq}) + 4\text{NH}_3(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \)
(c) \( \text{HCHO}(\text{l}) + 2\text{Cu}^{2+}(\text{aq}) + 5\text{OH}^-(\text{aq}) \rightarrow \text{Cu}_2\text{O}(\text{s}) + \text{HCOO}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \)
(d) \( \text{N}_2\text{H}_4(\text{l}) + 2\text{H}_2\text{O}_2(\text{l}) \rightarrow \text{N}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l}) \)
(e) \( \text{Pb}(\text{s}) + \text{PbO}_2(\text{s}) + 2\text{H}_2\text{SO}_4(\text{aq}) \rightarrow 2\text{PbSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \)
Answer:
(a) For \( 2\text{AgBr}(\text{s}) + \text{C}_6\text{H}_6\text{O}_2(\text{aq}) \rightarrow 2\text{Ag}(\text{s}) + 2\text{HBr}(\text{aq}) + \text{C}_6\text{H}_4\text{O}_2(\text{aq}) \):
Substance reduced: \( \text{AgBr} \) (Ag goes from \( +1 \) to \( 0 \))
Substance oxidized: \( \text{C}_6\text{H}_6\text{O}_2 \) (Quinol, carbon atoms change oxidation states)
Oxidizing agent: \( \text{AgBr} \)
Reducing agent: \( \text{C}_6\text{H}_6\text{O}_2 \)
(b) For \( \text{HCHO}(\text{l}) + 2[\text{Ag}(\text{NH}_3)_2]^+(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow 2\text{Ag}(\text{s}) + \text{HCOO}^-(\text{aq}) + 4\text{NH}_3(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \):
Substance reduced: \( [\text{Ag}(\text{NH}_3)_2]^+ \) (Ag goes from \( +1 \) to \( 0 \))
Substance oxidized: \( \text{HCHO} \) (Carbon goes from \( 0 \) to \( +2 \))
Oxidizing agent: \( [\text{Ag}(\text{NH}_3)_2]^+ \)
Reducing agent: \( \text{HCHO} \)
(c) For \( \text{HCHO}(\text{l}) + 2\text{Cu}^{2+}(\text{aq}) + 5\text{OH}^-(\text{aq}) \rightarrow \text{Cu}_2\text{O}(\text{s}) + \text{HCOO}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \):
Substance reduced: \( \text{Cu}^{2+} \) (Cu goes from \( +2 \) to \( +1 \) in \( \text{Cu}_2\text{O} \))
Substance oxidized: \( \text{HCHO} \) (Carbon goes from \( 0 \) to \( +2 \))
Oxidizing agent: \( \text{Cu}^{2+} \)
Reducing agent: \( \text{HCHO} \)
(d) For \( \text{N}_2\text{H}_4(\text{l}) + 2\text{H}_2\text{O}_2(\text{l}) \rightarrow \text{N}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l}) \):
Substance reduced: \( \text{H}_2\text{O}_2 \) (Oxygen goes from \( -1 \) to \( -2 \))
Substance oxidized: \( \text{N}_2\text{H}_4 \) (Nitrogen goes from \( -2 \) to \( 0 \))
Oxidizing agent: \( \text{H}_2\text{O}_2 \)
Reducing agent: \( \text{N}_2\text{H}_4 \)
(e) For \( \text{Pb}(\text{s}) + \text{PbO}_2(\text{s}) + 2\text{H}_2\text{SO}_4(\text{aq}) \rightarrow 2\text{PbSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \):
Substance oxidized: \( \text{Pb}(\text{s}) \) (Pb goes from \( 0 \) to \( +2 \))
Substance reduced: \( \text{PbO}_2(\text{s}) \) (Pb goes from \( +4 \) to \( +2 \))
Oxidizing agent: \( \text{PbO}_2 \)
Reducing agent: \( \text{Pb} \)
In simple words: An oxidizing agent makes another substance lose electrons (gets oxidized) and itself gains electrons (gets reduced). A reducing agent makes another substance gain electrons (gets reduced) and itself loses electrons (gets oxidized). By looking at which element's oxidation number changes, you can tell what role each substance plays.

Exam Tip: To identify agents, first assign oxidation states to all atoms. Then, identify which elements are oxidized (increase in oxidation state) and reduced (decrease in oxidation state). The substance containing the element that is oxidized is the reducing agent, and vice versa.

Question 14. Consider the reactions : \( 2\text{S}_2\text{O}_3^{2-}(\text{aq}) + \text{I}_2(\text{s}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \) and \( \text{S}_2\text{O}_3^{2-}(\text{aq}) + 2\text{Br}_2(\text{l}) + 5\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + 4\text{Br}^-(\text{aq}) + 10\text{H}^+(\text{aq}) \). Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
The reductant, thiosulphate \( (\text{S}_2\text{O}_3^{2-}) \), reacts differently with iodine \( (\text{I}_2) \) and bromine \( (\text{Br}_2) \) due to the difference in their oxidizing strengths. Bromine \( (\text{Br}_2) \) is a stronger oxidizing agent than iodine \( (\text{I}_2) \).
In the first reaction, with the weaker oxidizing agent \( \text{I}_2 \), thiosulphate is oxidized to tetrathionate \( (\text{S}_4\text{O}_6^{2-}) \). In \( \text{S}_2\text{O}_3^{2-} \), the average oxidation state of sulfur is \( +2 \), while in \( \text{S}_4\text{O}_6^{2-} \), it is \( +2.5 \). This is a partial oxidation.
In the second reaction, with the stronger oxidizing agent \( \text{Br}_2 \), thiosulphate is oxidized further to sulfate \( (\text{SO}_4^{2-}) \). In \( \text{SO}_4^{2-} \), sulfur has an oxidation state of \( +6 \). This indicates a more complete oxidation due to the stronger oxidizing power of bromine.
Thus, the stronger oxidizing agent \( \text{Br}_2 \) forces sulfur in thiosulphate to reach a higher oxidation state \( (+6) \), whereas the weaker oxidizing agent \( \text{I}_2 \) only oxidizes it to an intermediate oxidation state \( (+2.5) \).
In simple words: Thiosulphate reacts in different ways with iodine and bromine because bromine is a stronger "electron-taker" (oxidizing agent) than iodine. A strong oxidizer like bromine can take more electrons from thiosulphate, pushing sulfur to a higher oxidation state. A weaker oxidizer like iodine can only take some electrons, leading to a lower oxidation state for sulfur.

Exam Tip: The outcome of a redox reaction, specifically the final oxidation state of the product, is often determined by the relative strengths of the oxidizing and reducing agents involved.

Question 15. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
**Fluorine as the best oxidant:**
Among the halogens \( (\text{F}_2, \text{Cl}_2, \text{Br}_2, \text{I}_2) \), fluorine \( (\text{F}_2) \) is the strongest oxidizing agent. This is indicated by its high standard electrode potential, \( \text{E}^{\circ}_{\text{red}} = +2.87 \text{ V} \). A higher positive reduction potential means a greater tendency to be reduced and thus a stronger oxidizing ability. Fluorine can displace other halide ions \( (\text{Cl}^-, \text{Br}^-, \text{I}^-) \) from their salt solutions, which demonstrates its superior oxidizing power:
\( \text{F}_2 + 2\text{e}^- \rightarrow 2\text{F}^- \)
\( \text{F}_2(\text{g}) + 2\text{NaCl}(\text{aq}) \rightarrow 2\text{NaF}(\text{aq}) + \text{Cl}_2(\text{g}) \)
\( \text{F}_2(\text{g}) + 2\text{NaBr}(\text{aq}) \rightarrow 2\text{NaF}(\text{aq}) + \text{Br}_2(\text{l}) \)
\( \text{F}_2(\text{g}) + 2\text{NaI}(\text{aq}) \rightarrow 2\text{NaF}(\text{aq}) + \text{I}_2(\text{s}) \)
Conversely, a reaction like \( \text{Cl}_2 + 2\text{NaF} \) does not occur, confirming that fluorine is the most readily reduced halogen.
**Hydroiodic acid (HI) as the best reductant:**
The reducing power of hydrohalic acids \( (\text{HX}) \) depends on how easily they decompose to form hydrogen gas \( (\text{H}_2) \) and elemental halogen \( (\text{X}_2) \). This ease of decomposition is inversely related to the bond dissociation energy of the \( \text{H-X} \) bond. A weaker bond means easier decomposition and stronger reducing power.
The bond dissociation energies of hydrohalic acids increase in the order:
\( \text{HI} < \text{HBr} < \text{HCl} < \text{HF} \)
Consequently, the reducing power of these acids increases in the reverse direction:
\( \text{HF} < \text{HCl} < \text{HBr} < \text{HI} \)
Therefore, hydroiodic acid \( (\text{HI}) \) is the strongest reducing agent because its H-I bond is the weakest and most easily broken, allowing \( \text{I}^- \) to be readily oxidized to \( \text{I}_2 \). For example:
\( 2\text{HI} \rightarrow \text{H}_2 + \text{I}_2 \)
An illustration is the reaction of iodide with concentrated sulfuric acid, where iodide is oxidized to iodine:
\( 2\text{NaI} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{SO}_2 + \text{I}_2 + 2\text{H}_2\text{O} \)
In simple words: Fluorine is the strongest oxidizing agent because it loves to gain electrons more than any other halogen, making it capable of "stealing" electrons from other halides. On the other hand, hydroiodic acid is the strongest reducing agent because its bond is the weakest, allowing it to easily lose its electron (from \( \text{I}^- \)) and reduce other substances.

Exam Tip: Remember that strong oxidizing agents have high reduction potentials and easily gain electrons, while strong reducing agents have low oxidation potentials and easily lose electrons. For hydrohalic acids, reducing power increases with decreasing bond strength.

Question 16. Why does the following reaction occur? \( \text{XeO}_4^{6-}(\text{aq}) + 2\text{F}^-(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow \text{XeO}_3(\text{g}) + \text{F}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l}) \). What conclusion about the compound \( \text{Na}_4\text{XeO}_6 \) (of which \( \text{XeO}_4^{6-} \) is a part) can be drawn from the reaction.
Answer:
This reaction occurs because \( \text{Na}_4\text{XeO}_6 \) (containing the \( \text{XeO}_4^{6-} \) ion) is an incredibly powerful oxidizing agent. In the \( \text{XeO}_4^{6-} \) ion, xenon is in its highest possible oxidation state of \( +8 \). Fluoride ions \( (\text{F}^-) \), on the other hand, are very difficult to oxidize because fluorine is the most electronegative element and prefers to exist in its \( -1 \) oxidation state.
In the given reaction:
\( \text{XeO}_4^{6-} (\text{aq}) + 2\text{F}^- (\text{aq}) + 6\text{H}^+ (\text{aq}) \rightarrow \text{XeO}_3 (\text{g}) + \text{F}_2 (\text{g}) + 3\text{H}_2\text{O} (\text{l}) \)
- Xenon is reduced from \( +8 \) in \( \text{XeO}_4^{6-} \) to \( +6 \) in \( \text{XeO}_3 \).
- Fluorine is oxidized from \( -1 \) in \( \text{F}^- \) to \( 0 \) in \( \text{F}_2 \).
The conclusion about \( \text{Na}_4\text{XeO}_6 \) is that it is a very strong oxidizing agent. It is strong enough to oxidize fluoride ions \( (\text{F}^-) \) to elemental fluorine \( (\text{F}_2) \), which is a remarkable feat as fluoride is one of the most stable ions and is rarely oxidized.
In simple words: This reaction happens because the xenon compound \( (\text{Na}_4\text{XeO}_6) \) is an incredibly strong "electron-taker" (oxidizing agent). It's so powerful that it can even take electrons from fluoride ions \( (\text{F}^-) \), which usually hold onto their electrons very tightly. This makes the fluoride turn into fluorine gas. So, the main conclusion is that \( \text{Na}_4\text{XeO}_6 \) is one of the strongest oxidizers known.

Exam Tip: The ability of a compound to oxidize a very stable species (like \( \text{F}^- \) to \( \text{F}_2 \)) is a strong indicator of its extreme oxidizing power. Elements in their highest oxidation states are often powerful oxidizing agents.

Question 17. Consider the reactions;
(a) \( \text{H}_3\text{PO}_2(\text{aq}) + 4\text{AgNO}_3(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_3\text{PO}_4(\text{aq}) + 4\text{Ag}(\text{s}) + 4\text{HNO}_3(\text{aq}) \)
(b) \( \text{H}_3\text{PO}_2(\text{aq}) + 2\text{CuSO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_3\text{PO}_4(\text{aq}) + 2\text{Cu}(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \)
(c) \( \text{C}_6\text{H}_5\text{CHO}(\text{l}) + 2[\text{Ag}(\text{NH}_3)_2]^+(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{C}_6\text{H}_5\text{COO}^-(\text{aq}) + 2\text{Ag}(\text{s}) + 4\text{NH}_3(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \)
(d) \( \text{C}_6\text{H}_5\text{CHO}(\text{l}) + 2\text{Cu}^{2+}(\text{aq}) + 5\text{OH}^-(\text{aq}) \rightarrow \text{No change observed} \).
What inference do you draw about the behaviour of \( \text{Ag}^+ \) and \( \text{Cu}^{2+} \) from these reactions?
Answer:
From the given reactions, we can make the following inferences about the behavior of \( \text{Ag}^+ \) and \( \text{Cu}^{2+} \):
(a) In this reaction, \( \text{Ag}^+ \) ions from \( \text{AgNO}_3 \) are reduced to elemental silver \( (\text{Ag}) \). This means \( \text{Ag}^+ \) acts as an oxidizing agent, gaining electrons from \( \text{H}_3\text{PO}_2 \), which is simultaneously oxidized to \( \text{H}_3\text{PO}_4 \).
(b) Similarly, in this reaction, \( \text{Cu}^{2+} \) ions from \( \text{CuSO}_4 \) are reduced to elemental copper \( (\text{Cu}) \). This also shows that \( \text{Cu}^{2+} \) acts as an oxidizing agent, gaining electrons from \( \text{H}_3\text{PO}_2 \), which gets oxidized to \( \text{H}_3\text{PO}_4 \).
(c) In this reaction, benzaldehyde \( (\text{C}_6\text{H}_5\text{CHO}) \) reduces the silver complex ion \( [\text{Ag}(\text{NH}_3)_2]^+ \) to elemental silver \( (\text{Ag}) \). This indicates that the silver complex also functions as an oxidizing agent.
(d) However, when benzaldehyde \( (\text{C}_6\text{H}_5\text{CHO}) \) is mixed with \( \text{Cu}^{2+} \) ions, no change is observed. This means that benzaldehyde is unable to reduce \( \text{Cu}^{2+} \) ions.
**Inference:** Comparing reactions (c) and (d), we can conclude that \( \text{Ag}^+ \) (or its complex \( [\text{Ag}(\text{NH}_3)_2]^+) \) is a stronger oxidizing agent than \( \text{Cu}^{2+} \). Benzaldehyde is capable of reducing \( \text{Ag}^+ \) but not \( \text{Cu}^{2+} \). This means that \( \text{Cu}^{2+} \) ions are less reactive as oxidizing agents compared to \( \text{Ag}^+ \) ions.
In simple words: These reactions show how good \( \text{Ag}^+ \) and \( \text{Cu}^{2+} \) are at taking electrons from other chemicals (being oxidizing agents). Both \( \text{Ag}^+ \) and \( \text{Cu}^{2+} \) can take electrons from \( \text{H}_3\text{PO}_2 \). But a chemical called benzaldehyde can only take electrons from \( \text{Ag}^+ \) and not from \( \text{Cu}^{2+} \). This tells us that \( \text{Ag}^+ \) is a "better" or stronger electron-taker than \( \text{Cu}^{2+} \).

Exam Tip: The outcome of a reaction, particularly the observation of "no change," provides valuable information about the relative strengths of oxidizing and reducing agents. If one reductant can reduce A but not B, then A is a stronger oxidizing agent than B.

Question 18. Balance the following redox reactions by ion – electron method :
(a) \( MnO^{-} _{4}(aq) + I^{-}(aq) \rightarrow MnO_{2}(s) + I_{2}(s) \) (in basic medium)
(b) \( MnO^{-} _{4}(aq) + SO_{2}(g) \rightarrow Mn^{2+}(aq) + HSO^{-} _{4}(aq) \) (in acidic solution)
(c) \( H_{2}O_{2}(aq) + Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + H_{2}O(l) \) (in acidic solution)
(d) \( Cr_{2}O^{2-} _{7} + SO_{2}(g) \rightarrow Cr^{3+}(aq) + SO^{2-} _{4}(aq) \) (in acidic solution)
Answer:
(a) \( MnO^{-} _{4}(aq) + I^{-}(aq) \rightarrow MnO_{2}(s) + I_{2}(s) \) in basic medium.
Oxidation half reaction is:
(i) \( I^{-}(aq) \rightarrow I_{2}(s) \)
(ii) \( 2I^{-}(aq) \rightarrow I_{2}(s) \)
(iii) \( 2I^{-}(aq) \rightarrow I_{2}(s) + 2e^{-} \) ... (1)
Reduction half reaction is:
(i) \( MnO^{-} _{4} \rightarrow MnO_{2} + 2H_{2}O \) (To balance O atom).
(ii) \( MnO^{-} _{4} + 4H_{2}O \rightarrow MnO_{2}(s) + 2H_{2}O + 4OH^{-} \) (To balance H atom)
(iii) \( MnO^{-} _{4}(aq) + 2H_{2}O \rightarrow MnO_{2} + 4OH(aq) \) (Remove duplication)
(iv) \( MnO^{-} _{4}(aq) + 2H_{2}O + 3e^{-} \rightarrow MnO_{2} + 4OH^{-}(aq) \) ... (2)
To equalize the number of electrons in both half reactions, multiply equation (i) by (3) and equation no. (2) by 2, then add the two half reactions to cancel out electrons.
\( 6I^{-}(aq) \rightarrow 3I_{2}(s) + 6e^{-} \)
\( 6I^{-}(aq) + 2MnO^{-} _{4}(aq) + 4H_{2}O(aq) + 6e^{-} \rightarrow 2MnO_{2} + 8OH^{-} + 3I_{2}(s) \)
To balance charge and atoms on both sides.
\( 2(-1) + 6(-1) = 8(-1) \)
\( -8 = -8 \)
(b) \( MnO^{-} _{4}(aq) + SO_{2}(g) \rightarrow Mn^{2+}(aq) + HSO^{-} _{4}(aq) \) (in acidic solution)
Oxidation half reaction is:
(i) \( SO_{2}(g) \rightarrow HSO^{-} _{4}(aq) \)
There is an increase in oxidation number from +4 in \( SO_{2} \) to +6 in \( HSO^{-} _{4} \).
(ii) \( SO_{2}(g) + 2H_{2}O(l) \rightarrow HSO^{-} _{4}(aq) \) (to balance O atom)
\( SO_{2}(g) + 2H_{2}O(l) \rightarrow HSO^{-} _{4} + 3H^{+}(aq) \) (to balance H atom)
\( SO_{2}(g) + 2H_{2}O(l) \rightarrow HSO^{-} _{4} + 3H^{+} + 2e^{-} \) ... (1) (to balance charge)
Reduction half - reaction is:
(i) \( MnO^{-} _{4}(aq) \rightarrow Mn^{2+}(aq) \)
There has been a decrease in the oxidation number of Mn from +7 to +2.
(ii) \( MnO^{-} _{4}(aq) \rightarrow Mn^{2+} + 4H_{2}O(l) \) (To balance oxygen atom)
\( MnO^{-} _{4} + 8H^{+}(aq) \rightarrow Mn^{2+}(aq) + 4H_{2}O(l) \)
\( MnO^{-} _{4} + 8H^{+}(aq) + 5e^{-} \rightarrow Mn^{2+}(aq) + 4H_{2}O(l) \) (To balance charge) ... (2)
Multiply equation (1) by (5) and equation number (2) by 2 to equalize the number of electrons and add the two half reactions, cancelling electrons.
\( 5SO_{2}(g) + 10H_{2}O(l) \rightarrow 5HSO^{-} _{4} + 15H^{+} + 10e^{-} \)
\( 2MnO^{-} _{4}(aq) + 16H^{+}(aq) + 10e^{-} \rightarrow 2Mn^{2+} + 8H_{2}O \)
\( 2MnO^{-} _{4}(aq) + 5SO_{2}(g) + 2H_{2}O + H^{+} \rightarrow 2Mn^{2+} + 5HSO^{-} _{4} \)
To balance atoms and charges
\( 2(-1) + 1 = 2(+2) + (-1) \)
\( -1 = -1 \)
(c) \( H_{2}O_{2}(aq) + Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + H_{2}O(l) \) (In acidic medium)
(i) The oxidation half reaction is
\( Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) \)
(ii) \( Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^{-} \) ... (1)
The reduction half reaction is
\( H_{2}O_{2}(aq) \rightarrow H_{2}O(l) \)
(Oxidation number of O decreases from -1 in \( H_{2}O_{2} \) to -2 in \( H_{2}O \)) (To balance oxygen atoms)
(iii) \( H_{2}O_{2}(aq) + 2H^{+}(aq) \rightarrow 2H_{2}O(l) \) (To balance H atoms in acidic medium)
(iv) \( H_{2}O_{2}(aq) + 2H^{+} + 2e^{-} \rightarrow 2H_{2}O(l) \) ... (2) (To balance charge)
(v) Multiply equation number (1) by 2 to balance the number of electrons in both the half reaction and add them.
\( 2Fe^{2+}(aq) + H_{2}O_{2} + 2H^{+}(aq) \rightarrow 2Fe^{3+}(aq) + 2H_{2}O(l) \)
To balance atoms & charge
\( 2(+2) + 2(+1) = 2(+3) \)
\( +4+2 = +6 \)
\( +6 = +6 \)
(d) \( Cr_{2}O^{2-} _{7}(aq) + SO_{2}(aq) \rightarrow Cr^{3+}(aq) + SO^{2-} _{4}(aq) \) (in acidic solution)
Divide the equation into two parts.
Oxidation half reaction -
(i) \( SO_{2}(aq) \rightarrow SO^{2-} _{4}(aq) \)
[Oxidation number of S increases from +4 in \( SO_{2}(aq) \) to +6 in \( SO^{2-} _{4}(aq) \)]
(ii) \( SO_{2}(aq) + 2H_{2}O(l) \rightarrow SO^{2-} _{4}(aq) \) (To balance O atom)
(iii) \( SO_{2}(aq) + 2H_{2}O(l) \rightarrow SO^{2-} _{4} + 4H^{+}(aq) \) (To balance H atom, add \( H^{+} \) to the deficient side as the medium is acidic)
(iv) \( SO_{2}(aq) + 2H_{2}O(l) \rightarrow SO^{2-} _{4} + 4H^{+}(aq) + 2e^{-} \) ... (1) (To balance charge)
Reduction half reaction -
(i) \( Cr_{2}O^{2-} _{7}(aq) \rightarrow Cr^{3+}(aq) \)
(oxidation number of Cr decreases from +6 to +3)
(ii) \( Cr_{2}O^{2-} _{7}(aq) \rightarrow 2Cr^{3+}(aq) \) (to balance all atoms except H & O)
(iii) \( Cr_{2}O^{2-} _{7}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_{2}O(l) \) (to balance O atom)
(iv) \( Cr_{2}O^{2-} _{7}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_{2}O(l) \) (to balance H atoms in acidic medium)
(v) \( Cr_{2}O^{2-} _{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O(l) \) ... (2) (to balance charge)
(vi) Multiply equation number (1) by 3 to balance electrons in both half reactions & add them to cancel electrons.
\( Cr_{2}O^{2-} _{7}(aq) + 14H_{3}^{+}(aq) + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O(l) \)
\( 3SO_{2}(aq) + 6H_{2}O(l) \rightarrow 3SO^{2-} _{4}(aq) + 12H^{+}(aq) + 6e^{-} \)
\( Cr_{2}O^{2-} _{7}(aq) + 3SO_{2}(aq) + 2H^{+}(aq) \rightarrow 2Cr^{3+} + 3SO^{2-} _{4} + H_{2}O(l) \)
\( -2+2 = +6-6 \)
\( 0 = 0 \)
In simple words: To balance these redox reactions, we separate them into oxidation and reduction half-reactions. Then, we balance atoms like oxygen and hydrogen by adding water and H+ (or OH- in basic medium) and finally balance charges by adding electrons. We adjust coefficients to make electron counts equal and combine the half-reactions.

Exam Tip: Remember to identify the acidic or basic medium first, as it dictates how you balance oxygen and hydrogen atoms using \( H_{2}O \), \( H^{+} \), or \( OH^{-} \).

Question 19. Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) \( P_{4}(s) + OH^{-}(aq) \rightarrow PH_{3}(g) + HPO^{2-} _{2}(aq) \)
(b) \( N_{2}H_{4}(l) + ClO^{-}_{3}(aq) \rightarrow NO(g) + Cl^{-}(g) \)
(c) \( Cl_{2}O_{7}(g) + H_{2}O_{2}(aq) \rightarrow ClO^{-}_{2}(aq) + O_{2}(g) + H^{+} \)
Answer:
Balance the equation by ion electron method.
(a) \( P_{4}(s) + OH^{-}(aq) \rightarrow PH_{3}(g) + H_{2}PO^{-}_{2}(aq) \) (basic medium)
Oxidation half reaction increase in oxidation number
\( P \rightarrow H_{2}PO^{-}_{2} \)
\( P + 2OH^{-} \rightarrow H_{2}PO^{-}_{2} \)
\( P + 2OH^{-} \rightarrow H_{2}PO^{-}_{2} + e^{-} \) ... (1)
Reduction half reaction decrease in oxidation number
\( P \rightarrow PH_{3} \)
\( P + 3H_{2}O \rightarrow PH_{3} + 3OH^{-} \)
\( P + 3H_{2}O + 3e^{-} \rightarrow PH_{3} + 3OH^{-} \) ... (2)
Multiply equation (1) by 3 to equalize the number of electrons in both half reaction.
Add the two half reactions
\( 3P + 6OH^{-} \rightarrow 3H_{2}PO^{-}_{2} + 3e^{-} \)
\( P + 3H_{2}O + 3e^{-} \rightarrow PH_{3} + 3OH^{-} \)
\( 4P + 3H_{2}O + 3OH^{-} \rightarrow 3H_{2}PO^{-}_{2} + PH_{3}(g) \)
or \( P_{4} + 3H_{2}O(l) + 3OH^{-}(aq) \rightarrow 3H_{2}PO_{2} + PH_{3}(g) \)
Balance the atoms and charge on both sides
\( 3(-1) = +3(-1) \)
\( -3 = -3 \)
It is a disproportionation reaction. Here phosphorus undergoes oxidation as well as reduction. Therefore P is both an oxidizing and reducing agent.
(b) \( N_{2}H_{4}(l) + ClO^{-}_{3}(aq) \rightarrow NO(g) + Cl^{-}(g) \)
Oxidation half reaction increase oxidation number by 4 per atom of N
\( N_{2}H_{4}(l) \rightarrow NO(g) \)
Decrease by 6
\( N 18 \times 3 \)
\( Cl \uparrow 6 \times 4 \)
\( 3N_{2}H_{4}(l) + 4ClO^{-}_{3} (aq) \rightarrow 6NO + 4Cl^{-} \)
Now balance H and O atom
\( 3N_{2}H_{4}(l) + 4ClO^{-}_{3}(aq) \rightarrow 6NO(g) + 4Cl^{-} + 6H_{2}O \)
\( 4(-1) = 4(-1) \)
\( -4 = -4 \)
Here \( ClO^{-}_{3} \) ions act as an oxidant and \( N_{2}H_{4} \) acts as a reductant.
(c) \( Cl_{2}O_{7}(g) + H_{2}O_{2}(aq) \rightarrow ClO^{-}_{2}(aq) + O_{2}(g) \)
Oxidation half reaction
\( H_{2}O_{2}(aq) \rightarrow O_{2}(g) \) (basic medium)
\( 2OH^{-} + H_{2}O_{2} \rightarrow O_{2} + 2H_{2}O \)
\( H_{2}O_{2} + 2OH^{-} \rightarrow O_{2} + 2H_{2}O + 2e^{-} \) ... (i)
Reduction half reaction
\( Cl_{2}O_{7}(g) \rightarrow ClO^{-}_{2} \)
\( Cl_{2}O_{7}(g) \rightarrow 2ClO^{-}_{2} \)
\( Cl_{2}O_{7}(g) \rightarrow 2ClO^{-}_{2} + 3H_{2}O \)
\( Cl_{2}O_{7}(g) + 6H_{2}O(l) \rightarrow 2ClO^{-}_{2} + 3H_{2}O + 6OH^{-} \)
\( Cl_{2}O_{7}(g) + 3H_{2}O(l) \rightarrow 2ClO^{-}_{2} + 6OH^{-} \)
\( Cl_{2}O_{7}(g) + 3H_{2}O + 8e^{-} \rightarrow 2ClO^{-}_{2} + 7OH^{-} \)
To equalize electrons in both half reactions multiply equation number (1) by 4. Add both the half reactions to cancel out electrons.
\( 4H_{2}O_{2}(aq) + 8OH^{-}(aq) \rightarrow 4O_{2} + 8H_{2}O(l) + 8e^{-} \)
\( Cl_{2}O_{7} + 3H_{2}O(l) + 8e^{-} \rightarrow ClO_{2}^{-} + 6OH^{-} \)
\( Cl_{2}O_{7}(g) + 4H_{2}O (aq) + 2OH^{-} (aq) \rightarrow 2ClO^{-}_{2}(aq) + 5H_{2}O + 4O_{2} \)
To balance atoms and charges
\( 2(-1) = 2(-1) \)
\( -2 = -2 \)
\( Cl_{2}O_{7} \) is an oxidizing agent and \( H_{2}O_{2} \) acts as a reducing agent.
In simple words: Balancing redox reactions involves breaking them into oxidation and reduction parts. We balance the atoms first, then charges with electrons. For basic solutions, add water to balance oxygen and hydroxide ions to balance hydrogen. For acidic solutions, use water and hydrogen ions.

Exam Tip: For disproportionation reactions, the same element gets both oxidized and reduced. Be sure to show both half-reactions where the element’s oxidation state changes in opposite directions.

Question 20. What sorts of informations can you draw from the following reaction?
\( (CN)_{2}(g) + 2OH^{-}(aq) \rightarrow CN^{-}(aq) + CNO^{-}(aq) + H_{2}O(l) \)
Answer:
\( (CN)_{2}(g) + 2OH^{-}(aq) \rightarrow CN^{-}(aq) + CNO^{-}(aq) + H_{2}O(l) \)
(i) In the above reaction, \( (CN)_{2} \) is known as a pseudohalogen, while \( CN^{-} \) and \( CNO^{-} \) ions are called pseudohalides ions.
(ii) Ions that contain two or more electronegative atoms, with at least one being nitrogen, and possess properties similar to halides ions, are named pseudohalides ions.
(iii) Just as halide ions (like \( Cl^{-} \)) have corresponding diatomic molecules (like \( Cl_{2} \)), pseudohalides ions (like \( CN^{-} \), \( CNO^{-} \)) also have corresponding dimeric molecules (like \( (CN)_{2} \), \( (CNO)_{2} \)). These are termed pseudohalogens and display properties similar to halogens. \( (CN)_{2} \) is cyanogen and resembles \( Cl_{2} \).
(iv) They undergo reactions similar to halogens, as \( Cl_{2}(g) \) disproportionates to give \( Cl^{-} \) ions and \( ClO^{-} \) ions in basic medium. So does \( (CN)_{2} \).
In simple words: This reaction shows that \( (CN)_{2} \) behaves like a halogen, being both oxidized and reduced. It forms \( CN^{-} \) and \( CNO^{-} \) ions, which are like halide ions. This means \( (CN)_{2} \) is a "pseudohalogen" because it acts like real halogens.

Exam Tip: When asked to draw conclusions from a reaction, break down each product and reactant to identify its role and classification, especially for unusual compounds like pseudohalogens.

Question 21. The \( Mn^{3+} \) ion is unstable in solution and undergoes disproportionation to give \( Mn^{2+} \), \( MnO_{2} \), and \( H^{+} \) ion. Write a balanced ionic equation for the reaction.
Answer:
The skeletal equation is :
\( Mn^{3+} \rightarrow Mn^{2+} + MnO_{2} + H^{+}(aq) \)
Oxidation half reaction
\( Mn^{3+}(aq) \rightarrow MnO_{2}(s) \)
\( Mn^{3+}(aq) + 2H_{2}O(l) \rightarrow MnO_{2}(s) \) (to balance O atom)
\( Mn^{3+}(aq) + 2H_{2}O(l) \rightarrow MnO_{2} + 4H^{+} \) (to balance \( H^{+} \) atom)
\( Mn^{3+} + 2H_{2}O(l) \rightarrow MnO_{2} + 4H^{+} + e^{-} \) ... (1) (to balance charge)
Reduction half reaction
\( Mn^{3+}(aq) \rightarrow Mn^{2+}(aq) \)
\( Mn^{3+}(aq) + e^{-} \rightarrow Mn^{2+}(aq) \) ... (2)
Add both the half reaction (1) & (2)
\( 2Mn^{3+} + 2H_{2}O \rightarrow MnO_{2} + Mn^{2+} + 4H^{+} \)
It is the net balanced ionic equation
\( 2(+3) = +2 + 4(+1) \)
\( +6 = +6 \)
In simple words: The \( Mn^{3+} \) ion breaks down in solution. Some of it turns into \( Mn^{2+} \) (reduction), and some turns into \( MnO_{2} \) (oxidation). The balanced equation shows how two \( Mn^{3+} \) ions become one \( Mn^{2+} \) and one \( MnO_{2} \) with some \( H^{+} \) and water.

Exam Tip: For disproportionation reactions, ensure the element in question appears in three different oxidation states: its initial state, a higher state (oxidized product), and a lower state (reduced product).

Question 22. Consider the elements : Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.
Answer:
(a) F exhibits only a negative oxidation state.
(b) Cs exhibits only a positive oxidation state (as \( Cs^{+} \)).
(c) I exhibits both negative and positive oxidation states, like \( I^{-} \) and \( I^{+} \).
(d) Ne shows neither a positive nor a negative oxidation state.
In simple words: Fluorine only takes electrons, so it's always negative. Cesium always gives electrons away, so it's always positive. Iodine can do both, sometimes gaining and sometimes losing electrons. Neon is a noble gas, so it doesn't usually gain or lose electrons, meaning it has no oxidation state.

Exam Tip: Remember that highly electronegative elements (like F) will almost always have negative oxidation states, while highly electropositive elements (like Cs) will always have positive oxidation states.

Question 23. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Answer:
\( Cl_{2}(g) + SO_{2}(aq) + 2H_{2}O(l) \rightarrow 2HCl(aq) + H_{2}SO_{4}(aq) \)
Here \( Cl_{2}(g) \) undergoes reduction to form HCl, whereas \( SO_{2} \) undergoes oxidation to form \( H_{2}SO_{4} \).
In simple words: When too much chlorine is in water, sulfur dioxide is added to fix it. The chlorine changes into HCl, and the sulfur dioxide changes into sulfuric acid. This reaction uses up the extra chlorine, making the water safe.

Exam Tip: When balancing redox reactions in water, ensure all atoms (especially O and H) and charges are balanced by adding \( H_{2}O \) and \( H^{+} \) (or \( OH^{-} \) for basic solutions).

Question 24. Refer to the periodic table given in your book and now answer the following questions:
(a) Identify the non-metals that can show disproportionation reaction.
(b) Identify the metals that can show disproportionation, reaction.
Answer:
(a) Oxygen, chlorine, bromine, phosphorus, etc.
(b) Copper, manganese.
In simple words: Non-metals like oxygen and chlorine can undergo disproportionation, meaning they can both get oxidized and reduced in the same reaction. Similarly, metals such as copper and manganese can also do this, showing different oxidation states in one process.

Exam Tip: Disproportionation reactions often occur with elements that have at least three common oxidation states, where the intermediate state can both increase and decrease.

Question 25. In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam.
What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?
Answer:
\( 4NH_{3}(g) + 5O_{2}(g) \rightarrow 4NO(g) + 6H_{2}O(g) \)
\( (4 \times 14 + 3 \times 4 \times 1) = 68.0 \) gm
\( 5 \times 32 = 160 \) gm
\( 4(14 + 16) = 120 \) gm
Amount of \( NH_{3} \) given = 10.00 gm
Amount of \( O_{2} \) given = 20 gm.
Let's assume \( NH_{3} \) is the limiting reactant. If it were, it would need \( (160/68) \times 10 = 23.53 \) g of \( O_{2} \). We only have 20 g of \( O_{2} \), so \( O_{2} \) is the limiting reactant.
Mass of \( NH_{3} \) required = \( \frac{68}{160} \times 20 = 8.5 \) gm
This means 20 gm of \( O_{2} \) will need 8.5 gm of \( NH_{3} \).
Therefore, oxygen must be the limiting reactant, and \( NH_{3} \) is present in excess (10 gm).
Now, if the entire mass of \( O_{2} \) reacts, how much NO is formed?
Mass of NO produced from 20 gm of \( O_{2} \)
\( \frac{120}{160} \times 20 = 15 \) gm of NO.
Maximum weight of NO produced = 15 gm.
In simple words: We have 10 grams of ammonia and 20 grams of oxygen. We first figure out which ingredient runs out first, which is oxygen. Based on the amount of oxygen, the most nitric oxide we can make is 15 grams.

Exam Tip: Always identify the limiting reactant first in stoichiometry problems; this determines the maximum amount of product that can be formed.

Question 26. Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:
(a) \( Fe^{3+}(aq) \) and \( I^{-}(aq) \)
(b) \( Ag^{3+}(aq) \) and \( Cu(s) \)
(c) \( Fe^{3+}(aq) \) and \( Cu(s) \)
(d) \( Ag(s) \) and \( Fe^{3+}(aq) \)
(e) \( Br^{-}(aq) \) and \( Fe^{2+}(aq) \).
Answer:
(a) \( Fe^{3+}(aq) \) and \( I^{-}(aq) \)
Oxidation half reaction is \( 2I^{-} \rightarrow I_{2} + 2e^{-} \)
\( E^{\circ} = -0.54 \) V.
Reduction half reaction is \( [Fe^{3+}(aq) + e^{-} \rightarrow Fe^{2+}] \times 2 \)
\( E^{\circ} = +0.77 \) V.
\( 2I^{-} + 2Fe^{3+} \rightarrow I_{2}(g) + 2Fe^{2+} \)
E.M.F. = \( -0.54 + 0.77 = +0.23 \)V
Since the EMF is positive, the reaction is feasible.
(b) Here Cu(s) loses electrons and \( Ag^{+}(aq) \) gains electrons. Thus, oxidation half cell reaction is
\( Cu(s) \rightarrow Cu^{2+} + 2e^{-} ] E^{\circ} = -0.34 \)V
Reduction \( Ag^{+} + e^{-} \rightarrow Ag(s) \times 2 E^{\circ} = -0.80 \) V
Overall reaction is :
\( Cu(s) + 2Ag^{+} \rightarrow Cu^{2+}(aq) + 2Ag(s) \)
\( E^{\circ}_{cell} = -0.46 \)V
Since \( E^{\circ}_{cell} \) is negative, the reaction is not feasible.
(c) Oxidation \( Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{-} \), \( E^{\circ} = -0.34 \)V
Reduction \( Fe^{3+} + e^{-} \rightarrow Fe^{2+} \), \( E^{\circ} = +0.77 \)V
Probable cell
\( Cu(s) | Cu^{2+} :: Fe^{3+} | Fe^{2+} \)
\( E^{\circ}_{cell} = E^{\circ}_{red}(R.H.S) - E^{\circ}_{ox}L.H.S. \)
\( = [+0.77 - 0.34] V = 0.43 \)V.
Since \( E^{\circ}_{cell} \) is positive, the reaction is feasible.
(d) \( Ag(s) \) and \( Fe^{3+}(aq) \)
Oxidation \( Ag(s) \rightarrow Ag^{+} + e^{-} \), \( E^{\circ} = -0.80 \)V
Reduction \( Fe^{3+} + e^{-} \rightarrow Fe^{2+} \), \( E^{\circ} = + 0.77 \)V
\( E^{\circ}_{cell} = -0.03 \)V
Since \( E^{\circ}_{cell} \) is negative, the reaction is not feasible.
(e) \( Br_{2} \) and \( Fe^{2+} \)
Here \( Fe^{2+} \) lose electrons and \( Br_{2} \) gains them.
Oxidation \( Fe^{2+} \rightarrow Fe^{3+} + e^{-} ] \times 2 [E^{\circ} = 0.77 \)V
Reduction \( Br_{2} + 2e^{-} \rightarrow 2Br^{-} ] (E^{\circ} = + 1.08 \)V)
\( E^{\circ}_{cell} = +0.31 \)V
\( 2Fe^{2+} + Br_{2} \rightarrow 2Fe^{3+} + 2Br^{-} \)
Since \( E^{\circ}_{cell} \) is positive, the reaction is feasible.
In simple words: We check if a reaction can happen by calculating its overall voltage, or EMF. If the EMF is a positive number, the reaction is likely to occur on its own. If it's a negative number, the reaction won't happen naturally without energy input.

Exam Tip: A positive standard cell potential (\( E^{\circ}_{cell} \)) indicates a spontaneous (feasible) reaction under standard conditions, while a negative \( E^{\circ}_{cell} \) indicates a non-spontaneous reaction.

Question 27. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of \( AgNO_{3} \) with silver electrodes
(ii) An aqueous solution of \( AgNO_{3} \) with platinum electrodes
(iii) A dilute solution of \( H_{2}SO_{4} \) with platinum electrodes
(iv) An aqueous solution of \( CuCl_{2} \) with platinum electrodes.
Answer:
(i) \( AgNO_{3} \rightleftharpoons Ag^{+} + NO^{-}_{3} \)
At cathode: \( Ag^{+} + e^{-} \rightarrow Ag \) (silver is deposited)
At anode: \( NO^{-}_{3} \) ions move, but the silver electrode dissolves instead. An equivalent amount of silver from the anode is consumed, so the \( Ag^{+} \) concentration in the \( AgNO_{3} \) solution remains unchanged.
(ii) At cathode: Ag will be liberated.
At anode: \( O_{2} \) gas will be liberated.
(iii) At anode: \( O_{2} \) gas will be liberated
At cathode: \( H_{2} \) gas will be liberated.
(iv) At cathode: Cu will be liberated.
At anode: \( Cl_{2} \) gas will be liberated.
In simple words: During electrolysis, positive ions go to the negative electrode (cathode) and negative ions go to the positive electrode (anode). What gets released depends on the type of electrodes and the ions present. For example, silver metal deposits on the cathode if there are silver ions, and oxygen or hydrogen gas forms if water is electrolyzed.

Exam Tip: The products of electrolysis depend on the nature of the electrolyte (molten or aqueous) and the electrodes (inert or active). Always consider the relative reduction/oxidation potentials.

Question 28. Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Answer:
Any metal will displace another metal from its salt solution if it has a higher reduction potential value. This depends on the values of their reduction potential.
The order of displacement is thus :
Mg = - 2.37
Al = - 1.66
Zn = - 0.76
Fe = + 0.44
Cu = + 0.34
Thus, the decreasing order is :
Mg < Al < Zn < Fe < Cu
In simple words: The metals can displace each other based on how reactive they are, which we can tell from their reduction potential numbers. A metal with a lower reduction potential can push out metals with higher reduction potentials from their salt solutions. So, Magnesium is the most reactive, then Aluminum, Zinc, Iron, and Copper is the least reactive in this group.

Exam Tip: Metals with more negative (or less positive) standard reduction potentials are stronger reducing agents and can displace metals with less negative (or more positive) potentials from their salt solutions.

Question 29. Given the standard electrode potentials,
(i) \( K^{+}/K = - 2.93 \)V
(ii) \( Mg^{2+}/Mg = - 2.37 \)V
(iii) \( Ag^{+}/Ag = 0.80 \)V
(iv) \( Hg^{2+}/Hg = 0.79 \)V
(v) \( Cr^{3+}/Cr = - 0.74 \)V
arrange these metals in their increasing order of reducing power.
Answer:
(i) \( K^{+}/K = - 2.93 \)V
(ii) \( Mg^{2+}/Mg = - 2.37 \)V
(iii) \( Cr^{3+}/Cr = - 0.74 \)V
(iv) \( Hg^{2+}/Hg = 0.79 \)V
(v) \( Ag^{+}/Ag = 0.80 \)V
The lower the reduction potential, the greater the tendency to get oxidized, and thus the greater the reducing power.
Hence, the increasing order of reducing power is:
Ag < Hg < Cr < Mg < K
In simple words: The smaller the reduction potential number (more negative), the better the metal is at giving away electrons, making it a stronger reducing agent. So, Potassium is the best reducing agent, followed by Magnesium, Chromium, Mercury, and finally Silver is the weakest.

Exam Tip: Remember that a more negative (or less positive) standard reduction potential means a greater tendency to lose electrons, indicating stronger reducing power.

Question 30. Depict the galvanic cell in which the reaction
\( Zn(s) + 2Ag^{+}(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s) \) takes place,
Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.
Answer:
The galvanic cell will be
\( Zn(s) | Zn^{2+}(aq) || Ag^{+}(aq) | Ag(s) \)
(i) The Zn electrode will be negatively charged.
(ii) Current will flow from the silver electrode to the zinc electrode. Electrons flow from Zn to Ag outside, and ions carry current in the cell.
(iii) Reaction at each electrode will be:
At anode: oxidation half reaction
\( Zn(s) \rightarrow Zn^{2+} + 2e^{-} \)
At cathode: reduction half reaction
\( 2Ag^{+}(aq) + 2e^{-} \rightarrow 2Ag(s) \)
In simple words: This setup is like a battery. The zinc metal is the negative part, where electrons are released. These electrons travel through a wire to the silver metal, which is the positive part. Inside the cell, charged particles (ions) move to keep the electricity flowing.

Exam Tip: In a galvanic cell, oxidation always occurs at the anode (negative electrode), and reduction always occurs at the cathode (positive electrode). Electrons flow from anode to cathode externally.