| Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th) | ||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 6th | 7th | 8th | 9th | 10th | ||||||||||
| Non-Rationalised Science NCERT Notes and Solutions (Class 11th) | ||||||||||||||
| Physics | Chemistry | Biology | ||||||||||||
| Non-Rationalised Science NCERT Notes and Solutions (Class 12th) | ||||||||||||||
| Physics | Chemistry | Biology | ||||||||||||
Chapter 8 Redox Reactions
Chemistry is the study of matter and its transformations. One crucial category of chemical transformations involves Redox Reactions, where changes in the oxidation states of elements occur. These reactions are ubiquitous, underpinning various natural processes (like biological respiration and photosynthesis) and technological applications across industries (pharmaceuticals, metallurgy, agriculture) and daily life (combustion of fuels, batteries, corrosion).
Understanding redox reactions is fundamental to chemistry. The term "redox" is derived from "reduction" and "oxidation," processes that always occur simultaneously.
Redox Reactions In Terms Of Electron Transfer Reactions
The understanding of redox reactions has evolved over time. Initially, definitions were based on the involvement of oxygen and hydrogen.
Classical Idea:
- Oxidation: Originally defined as the addition of oxygen to an element or compound. Examples:
- $\text{2Mg(s)} + \text{O}_2\text{(g)} \rightarrow \text{2MgO(s)}$ (Magnesium is oxidized)
- $\text{S(s)} + \text{O}_2\text{(g)} \rightarrow \text{SO}_2\text{(g)}$ (Sulfur is oxidized)
- Reduction: Originally defined as the removal of oxygen from a compound. Example: $\text{2HgO(s)} \rightarrow \text{2Hg(l)} + \text{O}_2\text{(g)}$ (Mercury(II) oxide is reduced). Later expanded to include the addition of hydrogen or addition of any electropositive element, or removal of any electronegative element. Example: $\text{2FeCl}_3\text{(aq)} + \text{H}_2\text{(g)} \rightarrow \text{2FeCl}_2\text{(aq)} + \text{2HCl(aq)}$ (Ferric chloride is reduced).
It was soon recognised that these processes are coupled: oxidation does not occur without reduction, and vice versa. Reactions involving both oxidation and reduction are called redox reactions.
Problem 8.1. In the reactions given below, identify the species undergoing oxidation and reduction:
(i) H$_2$S (g) + Cl$_2$ (g) $\rightarrow$ 2 HCl (g) + S (s)
(ii) 3Fe$_3$O$_4$ (s) + 8 Al (s) $\rightarrow$ 9 Fe (s) + 4 Al$_2$O$_3$ (s)
(iii) 2 Na (s) + H$_2$ (g) $\rightarrow$ 2 NaH (s)
Answer:
Applying the classical definitions:
(i) $\text{H}_2\text{S (g)} + \text{Cl}_2\text{ (g)} \rightarrow \text{2 HCl (g)} + \text{S (s)}$
- Consider H$_2$S: Hydrogen (electropositive) is removed from Sulfur, or Chlorine (electronegative) is added to Hydrogen. H$_2$S loses hydrogen, becoming S. H$_2$S undergoes oxidation.
- Consider Cl$_2$: Hydrogen is added to Chlorine, forming HCl. Cl$_2$ gains hydrogen, becoming HCl. Cl$_2$ undergoes reduction.
(ii) $\text{3Fe}_3\text{O}_4\text{ (s)} + \text{8 Al (s)} \rightarrow \text{9 Fe (s)} + \text{4 Al}_2\text{O}_3\text{ (s)}$
- Consider Al: Oxygen is added to Aluminium, forming Al$_2$O$_3$. Al undergoes oxidation.
- Consider Fe$_3$O$_4$: Oxygen is removed from Fe$_3$O$_4$, forming Fe. Fe$_3$O$_4$ undergoes reduction.
(iii) $\text{2 Na (s)} + \text{H}_2\text{ (g)} \rightarrow \text{2 NaH (s)}$
This reaction involves elements forming an ionic compound (as will be discussed later). NaH is Na$^+$H$^-$. Sodium (electropositive element) is added to Hydrogen. Or, we can consider the perspective from Sodium: Hydrogen (electronegative compared to Na) is added to Sodium. However, based on the formation of an ionic compound, let's consider electron transfer (discussed in the next section).
In terms of electron transfer (anticipating the next section), Na loses an electron to become Na$^+$, so Na is oxidised. H$_2$ gains an electron to become H$^-$, so H$_2$ is reduced.
Applying only classical ideas, if we consider Na more electropositive than H, then H is added to Na, or Na is added to H. Let's consider H as an electronegative element here compared to Na. Addition of an electronegative element to Na is oxidation of Na. Addition of an electropositive element to H is reduction of H.
- Consider Na: Hydrogen (electronegative relative to Na) is added to Sodium. Na undergoes oxidation.
- Consider H$_2$: Sodium (electropositive relative to H) is added to Hydrogen. H$_2$ undergoes reduction.
Electron Transfer Concept:
With the development of understanding of chemical bonding, particularly ionic bonding, a new perspective on redox reactions emerged: electron transfer. In ionic compounds like NaCl, Na$_2$O, Na$_2$S, electron transfer occurs from the metal to the non-metal.
$\text{2Na(s)} + \text{Cl}_2\text{(g)} \rightarrow \text{2Na}^+\text{Cl}^-\text{(s)}$
This overall reaction can be broken down into two **half-reactions**:
- $\text{2Na(s)} \rightarrow \text{2Na}^+\text{(g)} + \text{2e}^-$ (Sodium loses electrons)
- $\text{Cl}_2\text{(g)} + \text{2e}^- \rightarrow \text{2Cl}^-\text{(g)}$ (Chlorine gains electrons)
Based on electron transfer:
- Oxidation: The half-reaction that involves the loss of electron(s) by a species.
- Reduction: The half-reaction that involves the gain of electron(s) by a species.
In a redox reaction, oxidation and reduction always occur simultaneously. The species that loses electrons (undergoes oxidation) is called the reducing agent (reductant) because it causes the reduction of another species. The species that gains electrons (undergoes reduction) is called the oxidising agent (oxidant) because it causes the oxidation of another species.
In $\text{2Na(s)} + \text{Cl}_2\text{(g)} \rightarrow \text{2NaCl(s)}$, Na is oxidised and acts as the reducing agent, while Cl$_2$ is reduced and acts as the oxidising agent.
Problem 8.2. Justify that the reaction : 2 Na(s) + H$_2$(g) $\rightarrow$ 2 NaH (s) is a redox change.
Answer:
The compound formed, Sodium Hydride (NaH), is an ionic compound, written as Na$^+$H$^-$. This indicates that electron transfer has occurred.
We can split the reaction into two half-reactions:
- Sodium atoms lose electrons to form sodium ions: $\text{2Na(s)} \rightarrow \text{2Na}^+\text{(g)} + \text{2e}^-$. This is an oxidation half-reaction. Sodium is oxidised.
- Hydrogen molecules gain electrons to form hydride ions: $\text{H}_2\text{(g)} + \text{2e}^- \rightarrow \text{2H}^-\text{(g)}$. This is a reduction half-reaction. Hydrogen is reduced.
Since the reaction involves both oxidation (loss of electrons by Na) and reduction (gain of electrons by H$_2$), it is a redox change.
In this reaction, Na acts as the reducing agent, and H$_2$ acts as the oxidising agent.
Competitive Electron Transfer Reactions
Redox reactions involve competition for electrons between different species. Consider placing a zinc strip in a copper nitrate solution (Fig. 8.1). The zinc strip gets coated with copper, and the blue colour of $\text{Cu}^{2+}$ ions fades as they are reduced.
The reaction is: $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$.
- $\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{2e}^-$ (Oxidation half-reaction)
- $\text{Cu}^{2+}\text{(aq)} + \text{2e}^- \rightarrow \text{Cu(s)}$ (Reduction half-reaction)
Zinc loses electrons to $\text{Cu}^{2+}$ ions, indicating that Zn is a stronger reducing agent (has a greater tendency to lose electrons) than Cu. Conversely, $\text{Cu}^{2+}$ is a stronger oxidising agent than $\text{Zn}^{2+}$. If you place a copper strip in a zinc sulfate solution, no reaction occurs, indicating that the equilibrium for this reaction strongly favours the products (Zn$^{2+}$ and Cu).
Similarly, copper reacts with silver nitrate solution (Fig. 8.2): $\text{Cu(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{Ag(s)}$.
Cu loses electrons to $\text{Ag}^+$ ions, indicating Cu is a stronger reducing agent than Ag. Electron releasing tendency: Cu > Ag. Oxidising strength: $\text{Ag}^+$ > $\text{Cu}^{2+}$.
Comparing reaction of cobalt in nickel sulfate: $\text{Co(s)} + \text{Ni}^{2+}\text{(aq)} \rightleftharpoons \text{Co}^{2+}\text{(aq)} + \text{Ni(s)}$. At equilibrium, comparable concentrations of both reactants and products are present, indicating a less pronounced difference in electron-releasing tendency between Co and Ni compared to Zn/Cu or Cu/Ag.
These competitive electron transfer reactions highlight that elements and their ions have varying tendencies to gain or lose electrons. This forms the basis for creating an activity series of metals (or electrochemical series), ranking them by their reducing strength (tendency to lose electrons). Such a series is invaluable in predicting whether a redox reaction between a metal and another metal ion will occur and is fundamental to the operation of electrochemical cells (like batteries).
Oxidation Number
The electron transfer concept provides a clear picture for ionic compounds. However, for reactions involving covalent compounds where electrons are shared rather than completely transferred, the concept of **oxidation number** is more useful for tracking electron shifts and identifying redox processes.
The **oxidation number (or oxidation state)** of an element in a compound or ion is a hypothetical charge assigned to an atom based on a set of conventional rules, assuming that electron pairs in covalent bonds are completely transferred to the more electronegative atom. This is a formal bookkeeping device, not necessarily representing the actual charge distribution.
Rules for assigning oxidation numbers:
- In elements in their **free or uncombined state**, the oxidation number of each atom is **zero** (e.g., H$_2$, O$_2$, Na, S$_8$).
- For **monatomic ions**, the oxidation number is equal to the **charge on the ion** (e.g., $\text{Na}^+$ is +1, $\text{Cl}^-$ is -1, $\text{Fe}^{3+}$ is +3, $\text{O}^{2-}$ is -2). Alkali metals are +1, alkaline earth metals are +2, and Aluminium is +3 in their compounds.
- The oxidation number of **oxygen** is typically **-2** in most compounds. Exceptions: * **Peroxides** (O-O bond, e.g., H$_2$O$_2$, $\text{Na}_2\text{O}_2$): Oxygen is **-1**. * **Superoxides** (e.g., KO$_2$): Oxygen is **-1/2**. * Compounds with **fluorine** (e.g., OF$_2$, O$_2$F$_2$): Oxygen is **positive** (+2 in OF$_2$, +1 in O$_2$F$_2$).
- The oxidation number of **hydrogen** is typically **+1**, except when bonded to metals in binary compounds (metal hydrides, e.g., LiH, NaH, CaH$_2$), where it is **-1**.
- **Fluorine** always has an oxidation number of **-1** in its compounds. Other halogens (Cl, Br, I) are typically -1 as halide ions, but can have **positive** oxidation numbers when bonded to more electronegative elements (like oxygen in oxoacids/oxoanions, e.g., $\text{HClO}_4$).
- The algebraic sum of oxidation numbers of all atoms in a **neutral compound** is **zero**. In a **polyatomic ion**, the sum equals the **charge on the ion**.
By applying these rules, we can determine the oxidation number of any element in a compound or ion. Metal elements generally have positive oxidation numbers. Non-metals can have positive or negative oxidation numbers depending on the bonding partners. The highest oxidation number for a representative element is usually its group number (for groups 1 & 2) or (group number - 10) for p-block elements.
Stock notation: A notation where the oxidation number of a metal is indicated by a Roman numeral in parenthesis after the metal's symbol in the formula (e.g., $\text{Au(I)Cl}$ for AuCl, $\text{Fe(II)O}$ for FeO). This helps distinguish between compounds of the same metal in different oxidation states.
Problem 8.3. Using Stock notation, represent the following compounds :HAuCl$_4$, Tl$_2$O, FeO, Fe$_2$O$_3$, CuI, CuO, MnO and MnO$_2$.
Answer:
To represent the compounds using Stock notation, we need to determine the oxidation number of the metal in each compound.
(a) HAuCl$_4$: This is an acidic compound. The overall charge is 0. Hydrogen is +1. Chlorine is typically -1 (here it's bonded to Au). Let the oxidation number of Au be $x$. Sum of oxidation numbers = 0. $(+1) + (x) + 4 \times (-1) = 0$. $1 + x - 4 = 0$. $x - 3 = 0$. $x = +3$. The metal is Gold (Au) with oxidation number +3. Stock notation: HAu(III)Cl$_4$.
(b) Tl$_2$O: Overall charge is 0. Oxygen is -2. Let the oxidation number of Tl be $x$. There are two Tl atoms. $2 \times (x) + (-2) = 0$. $2x - 2 = 0$. $2x = 2$. $x = +1$. The metal is Thallium (Tl) with oxidation number +1. Stock notation: Tl$_2$(I)O.
(c) FeO: Overall charge is 0. Oxygen is -2. Let the oxidation number of Fe be $x$. $(x) + (-2) = 0$. $x = +2$. The metal is Iron (Fe) with oxidation number +2. Stock notation: Fe(II)O.
(d) Fe$_2$O$_3$: Overall charge is 0. Oxygen is -2. Let the oxidation number of Fe be $x$. There are two Fe atoms. $2 \times (x) + 3 \times (-2) = 0$. $2x - 6 = 0$. $2x = 6$. $x = +3$. The metal is Iron (Fe) with oxidation number +3. Stock notation: Fe$_2$(III)O$_3$.
(e) CuI: Overall charge is 0. Iodine is a halogen; when combined with a less electronegative element like a metal, it is -1. Let the oxidation number of Cu be $x$. $(x) + (-1) = 0$. $x = +1$. The metal is Copper (Cu) with oxidation number +1. Stock notation: Cu(I)I.
(f) CuO: Overall charge is 0. Oxygen is -2. Let the oxidation number of Cu be $x$. $(x) + (-2) = 0$. $x = +2$. The metal is Copper (Cu) with oxidation number +2. Stock notation: Cu(II)O.
(g) MnO: Overall charge is 0. Oxygen is -2. Let the oxidation number of Mn be $x$. $(x) + (-2) = 0$. $x = +2$. The metal is Manganese (Mn) with oxidation number +2. Stock notation: Mn(II)O.
(h) MnO$_2$: Overall charge is 0. Oxygen is -2. Let the oxidation number of Mn be $x$. $(x) + 2 \times (-2) = 0$. $x - 4 = 0$. $x = +4$. The metal is Manganese (Mn) with oxidation number +4. Stock notation: Mn(IV)O$_2$.
Using the oxidation number concept, redox reactions can be defined as reactions involving a change in the oxidation number of interacting species. Oxidation is an increase in oxidation number, reduction is a decrease. An oxidising agent causes an increase in oxidation number in another species (gets reduced itself). A reducing agent causes a decrease in oxidation number in another species (gets oxidised itself).
Problem 8.4. Justify that the reaction: 2Cu$_2$O(s) + Cu$_2$S(s) $\rightarrow$ 6Cu(s) + SO$_2$(g) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.
Answer:
To justify that the reaction is a redox reaction and identify the species, we need to determine the oxidation number of each element in the reactants and products and see if any changes occur.
Reaction: $\text{2Cu}_2\text{O(s)} + \text{Cu}_2\text{S(s)} \rightarrow \text{6Cu(s)} + \text{SO}_2\text{(g)}$
Assign oxidation numbers using the rules:
- Cu$_2$O: Oxygen is -2. Overall charge is 0. Let Cu be $x$. $2x + (-2) = 0 \implies 2x=2 \implies x=+1$. So, Cu in Cu$_2$O is +1.
- Cu$_2$S: Sulfur is typically -2 in sulfides. Overall charge is 0. Let Cu be $x$. $2x + (-2) = 0 \implies 2x=2 \implies x=+1$. So, Cu in Cu$_2$S is +1, and S in Cu$_2$S is -2.
- Cu(s): This is elemental copper. Oxidation number is 0.
- SO$_2$: Oxygen is -2. Overall charge is 0. Let S be $x$. $x + 2 \times (-2) = 0 \implies x-4=0 \implies x=+4$. So, S in SO$_2$ is +4.
Write the oxidation numbers above each element in the reaction:
$2\overset{+1}{\text{Cu}}_2\overset{-2}{\text{O(s)}} + \overset{+1}{\text{Cu}}_2\overset{-2}{\text{S(s)}} \rightarrow 6\overset{0}{\text{Cu(s)}} + \overset{+4}{\text{S}}\overset{-2}{\text{O}}_2\text{(g)}$
Observe the changes in oxidation numbers:
- Copper (Cu): Changes from +1 in Cu$_2$O and Cu$_2$S to 0 in Cu(s). The oxidation number decreases (+1 $\rightarrow$ 0). Copper is reduced.
- Sulfur (S): Changes from -2 in Cu$_2$S to +4 in SO$_2$. The oxidation number increases (-2 $\rightarrow$ +4). Sulfur is oxidised.
- Oxygen (O): Remains at -2 in reactants and product. Oxygen is neither oxidised nor reduced.
Since the oxidation number of Copper decreases (reduction) and the oxidation number of Sulfur increases (oxidation), the reaction is a redox reaction.
Identifying agents:
- The species that is reduced (Cu in Cu$_2$O and Cu$_2$S) acts as the oxidising agent (oxidant). So, Cu$_2$O and Cu$_2$S act as oxidants. However, focusing on the element being reduced (Cu), Cu$^{+1}$ is the oxidant species. Since both reactants contain Cu$^{+1}$, both contribute to the reduction of copper. The problem phrasing "species oxidised/reduced, which acts as oxidant and which acts as reductant" suggests identifying the substances acting as agents. Both Cu$_2$O and Cu$_2$S contribute copper for reduction.
- The species that is oxidised (S in Cu$_2$S) acts as the reducing agent (reductant). So, Cu$_2$S acts as a reductant. Specifically, the sulfide ion (S$^{2-}$) in Cu$_2$S is the reductant.
Species undergoing reduction: Copper (from +1 to 0).
Species undergoing oxidation: Sulfur (from -2 to +4).
Oxidising agent (oxidant): Cu$_2$O and Cu$_2$S (specifically Cu$^{+1}$ in these compounds). The text mentions Cu(I) is the oxidant, meaning the copper-containing compound(s) are the oxidants.
Reducing agent (reductant): Cu$_2$S (specifically S$^{-2}$ in this compound).
Types Of Redox Reactions
Redox reactions can be classified into different types:
- Combination Reactions: Two or more substances combine to form a single compound. At least one reactant must be in elemental form for it to be a redox reaction. Example: $\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$ (C: 0 to +4, O: 0 to -2).
- Decomposition Reactions: A compound breaks down into two or more simpler substances, with at least one being in the elemental state. Example: $\text{2H}_2\text{O(l)} \rightarrow \text{2H}_2\text{(g)} + \text{O}_2\text{(g)}$ (H: +1 to 0, O: -2 to 0). Note that not all decomposition reactions are redox (e.g., $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$, oxidation states don't change).
- Displacement Reactions: An atom or ion in a compound is replaced by an atom or ion of another element. Generally, $X + YZ \rightarrow XZ + Y$. * **Metal displacement:** A metal displaces another metal from its compound. Example: $\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$ (Zn: 0 to +2, Cu: +2 to 0). The displacing metal is a stronger reducing agent. * **Non-metal displacement:** A non-metal displaces another non-metal from its compound, including hydrogen displacement from water or acids, or halogen displacement. Example: $\text{Zn(s)} + \text{2HCl(aq)} \rightarrow \text{ZnCl}_2\text{(aq)} + \text{H}_2\text{(g)}$ (Zn: 0 to +2, H: +1 to 0). Halogen displacement: $\text{Cl}_2\text{(g)} + \text{2KBr(aq)} \rightarrow \text{2KCl(aq)} + \text{Br}_2\text{(l)}$ (Cl: 0 to -1, Br: -1 to 0). The displacing non-metal is a stronger oxidising agent.
- Disproportionation Reactions: A special type of redox reaction where an element in one oxidation state is simultaneously oxidised to a higher oxidation state and reduced to a lower oxidation state. The element must be in an intermediate oxidation state in the reactant. Example: $\text{2H}_2\text{O}_2\text{(aq)} \rightarrow \text{2H}_2\text{O(l)} + \text{O}_2\text{(g)}$ (Oxygen in H$_2$O$_2$ is -1, goes to -2 in H$_2$O and 0 in O$_2$).
Problem 8.5. Which of the following species, do not show disproportionation reaction and why ? ClO–, ClO$_2^{–}$, ClO$_3^{–}$ and ClO$_4^{–}$. Also write reaction for each of the species that disproportionates.
Answer:
A disproportionation reaction occurs when an element in an intermediate oxidation state is simultaneously oxidised to a higher state and reduced to a lower state. For disproportionation to occur, the element must be able to exist in at least three oxidation states, with the reactant being in an intermediate state.
Let's find the oxidation state of Chlorine in each of the given oxoanions:
- ClO$^-$ (Hypochlorite ion): Let Cl be $x$. $x + (-2) = -1 \implies x = +1$.
- ClO$_2^-$ (Chlorite ion): Let Cl be $x$. $x + 2(-2) = -1 \implies x - 4 = -1 \implies x = +3$.
- ClO$_3^-$ (Chlorate ion): Let Cl be $x$. $x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5$.
- ClO$_4^-$ (Perchlorate ion): Let Cl be $x$. $x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7$.
The possible oxidation states of Chlorine are -1, 0, +1, +3, +5, +7. The highest possible oxidation state is +7, and the lowest is -1.
A species will disproportionate if the element is in an intermediate oxidation state and can go to both a higher and a lower oxidation state.
- ClO$^-$ (Cl is +1): +1 is an intermediate state (between -1 and +7). It can be oxidised to +3, +5, +7 or reduced to 0 or -1. So, ClO$^-$ can disproportionate.
- ClO$_2^-$ (Cl is +3): +3 is an intermediate state (between -1 and +7). It can be oxidised to +5, +7 or reduced to +1, 0, -1. So, ClO$_2^-$ can disproportionate.
- ClO$_3^-$ (Cl is +5): +5 is an intermediate state (between -1 and +7). It can be oxidised to +7 or reduced to +3, +1, 0, -1. So, ClO$_3^-$ can disproportionate.
- ClO$_4^-$ (Cl is +7): +7 is the highest possible oxidation state for Chlorine. It cannot be oxidised further. It can only be reduced. Therefore, ClO$_4^-$ does not show disproportionation reaction.
The species that does not show disproportionation reaction is ClO$_4^-$ because Chlorine is already in its highest oxidation state (+7).
Disproportionation reactions for the other species:
- ClO$^-$ (+1) disproportionates to Cl$^-$ (-1) and ClO$_3^-$ (+5).
$3\text{ClO}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{ClO}_3^-\text{(aq)}$ (Balanced for atoms and charge)
- ClO$_2^-$ (+3) disproportionates to Cl$^-$ (-1) and ClO$_3^-$ (+5) or ClO$_4^-$ (+7).
Example: Disproportionation in basic medium to Cl$^-$ and ClO$_3^-$.
$6\text{ClO}_2^-\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 5\text{ClO}_3^-\text{(aq)} + \text{Cl}^-\text{(aq)} + 6\text{OH}^-\text{(aq)}$ (Balanced for atoms and charge)
- ClO$_3^-$ (+5) disproportionates to Cl$^-$ (-1) and ClO$_4^-$ (+7). This typically requires specific conditions or catalysts.
Example: Disproportionation upon heating.
$4\text{KClO}_3\text{(s)} \xrightarrow{\Delta} \text{3KClO}_4\text{(s)} + \text{KCl(s)}$ (Cl in KClO$_3$ is +5, in KClO$_4$ is +7, in KCl is -1. This is a decomposition/disproportionation.)
In ionic form: $4\text{ClO}_3^-\text{(aq)} \rightarrow \text{3ClO}_4^-\text{(aq)} + \text{Cl}^-\text{(aq)}$ (Balanced for atoms and charge)
Problem 8.6. Suggest a scheme of classification of the following redox reactions (a) N$_2$ (g) + O$_2$ (g) $\rightarrow$ 2 NO (g) (b) 2Pb(NO$_3$)$_2$(s) $\rightarrow$ 2PbO(s) + 4 NO$_2$ (g) + O$_2$ (g) (c) NaH(s) + H$_2$O(l) $\rightarrow$ NaOH(aq) + H$_2$ (g) (d) 2NO$_2$(g) + 2OH$^–$(aq) $\rightarrow$ NO$_2^{–}$(aq) + NO$_3^{–}$(aq)+H$_2$O(l)
Answer:
We need to classify each reaction into one of the redox reaction types: Combination, Decomposition, Displacement, or Disproportionation. Assign oxidation numbers to confirm they are redox reactions and aid classification.
(a) $\overset{0}{\text{N}}_2\text{ (g)} + \overset{0}{\text{O}}_2\text{ (g)} \rightarrow 2 \overset{+2}{\text{N}}\overset{-2}{\text{O}}\text{ (g)}$
- N changes from 0 to +2 (oxidation).
- O changes from 0 to -2 (reduction).
- Two elemental substances combine to form a compound.
Classification: Combination redox reaction.
(b) $2\text{Pb(NO}_3)_2\text{(s)} \rightarrow 2\text{PbO(s)} + 4 \text{ NO}_2\text{ (g)} + \text{O}_2\text{ (g)}$
- Pb in Pb(NO$_3$)$_2$: Oxygen is -2, Nitrate ion (NO$_3^-$) has charge -1. Pb + 2(-1) = 0 $\implies$ Pb is +2.
- N in Pb(NO$_3$)$_2$: In NO$_3^-$, N + 3(-2) = -1 $\implies$ N - 6 = -1 $\implies$ N is +5.
- O in Pb(NO$_3$)$_2$: -2.
- Pb in PbO: Pb + (-2) = 0 $\implies$ Pb is +2.
- N in NO$_2$: N + 2(-2) = 0 $\implies$ N - 4 = 0 $\implies$ N is +4.
- O in NO$_2$: -2.
- O in O$_2$: 0.
Oxidation number changes:
- Pb: +2 $\rightarrow$ +2 (No change).
- N: +5 $\rightarrow$ +4 (Reduction).
- O (in NO$_3^-$): -2 $\rightarrow$ 0 (Oxidation, in O$_2$). Some O remains -2 in PbO and NO$_2$.
Nitrogen is reduced, Oxygen is oxidised. A compound breaks down into multiple products, including elemental oxygen. This is a decomposition reaction.
Classification: Decomposition redox reaction.
(c) $\overset{+1}{\text{Na}}\overset{-1}{\text{H(s)}} + \overset{+1}{\text{H}}_2\overset{-2}{\text{O(l)}} \rightarrow \overset{+1}{\text{Na}}\overset{-2}{\text{O}}\overset{+1}{\text{H(aq)}} + \overset{0}{\text{H}}_2\text{ (g)}$
- Na: +1 $\rightarrow$ +1 (No change).
- H (in NaH): -1 $\rightarrow$ 0 (Oxidation).
- H (in H$_2$O): +1 $\rightarrow$ 0 (Reduction).
- O: -2 $\rightarrow$ -2 (No change).
Hydrogen in NaH is oxidised, hydrogen in H$_2$O is reduced. Hydrogen element is produced by displacing hydrogen from water by the hydride ion. This is a non-metal (hydrogen) displacement.
Classification: Displacement redox reaction (specifically, hydrogen displacement).
(d) $2\overset{+4}{\text{N}}\overset{-2}{\text{O}}_2\text{(g)} + 2\overset{-2}{\text{O}}\overset{+1}{\text{H}}^-\text{(aq)} \rightarrow \overset{+3}{\text{N}}\overset{-2}{\text{O}}_2^{–}\text{(aq)} + \overset{+5}{\text{N}}\overset{-2}{\text{O}}_3^{–}\text{(aq)}+\overset{+1}{\text{H}}_2\overset{-2}{\text{O(l)}}$
- N in NO$_2$: +4.
- N in NO$_2^-$: N + 2(-2) = -1 $\implies$ N - 4 = -1 $\implies$ N is +3.
- N in NO$_3^-$: N + 3(-2) = -1 $\implies$ N - 6 = -1 $\implies$ N is +5.
- O and H oxidation states remain -2 and +1 respectively in the other species.
Nitrogen changes from +4 in NO$_2$ to +3 in NO$_2^-$ (reduction) and to +5 in NO$_3^-$ (oxidation). The same element (N) is simultaneously oxidised and reduced. The initial oxidation state (+4) is intermediate to the final oxidation states (+3 and +5).
Classification: Disproportionation redox reaction.
Problem 8.7. Why do the following reactions proceed differently ? Pb$_3$O$_4$ + 8HCl $\rightarrow$ 3PbCl$_2$ + Cl$_2$ + 4H$_2$O and Pb$_3$O$_4$ + 4HNO$_3$ $\rightarrow$ 2Pb(NO$_3$)$_2$ + PbO$_2$ + H$_2$O
Answer:
The difference in reactions arises from the nature of Pb$_3$O$_4$ and the different properties of HCl and HNO$_3$.
Pb$_3$O$_4$ is not a single compound but a mixed oxide, composed of 2 moles of PbO (Lead(II) oxide) and 1 mole of PbO$_2$ (Lead(IV) oxide).
$\text{Pb}_3\text{O}_4 = 2\text{PbO} \cdot \text{PbO}_2$.
Let's determine the oxidation states of Lead in PbO and PbO$_2$:
- PbO: Pb + (-2) = 0 $\implies$ Pb is +2.
- PbO$_2$: Pb + 2(-2) = 0 $\implies$ Pb - 4 = 0 $\implies$ Pb is +4.
So, in Pb$_3$O$_4$, Lead exists in two oxidation states, +2 (in PbO) and +4 (in PbO$_2$).
Now consider the reactions with HCl and HNO$_3$ separately, treating Pb$_3$O$_4$ as $2\text{PbO} \cdot \text{PbO}_2$.
Reaction with HCl: PbO is a basic oxide and reacts with acid HCl in an acid-base reaction.
$\text{PbO(s)} + 2\text{HCl(aq)} \rightarrow \text{PbCl}_2\text{(aq)} + \text{H}_2\text{O(l)}$ (Lead oxidation state remains +2)
PbO$_2$ contains Lead in the +4 oxidation state, which is the highest common oxidation state for Lead and can act as an oxidising agent. HCl contains Chloride ions (Cl$^-$) with an oxidation state of -1, which can be oxidised (e.g., to Cl$_2$, oxidation state 0).
PbO$_2$ reacts with HCl as a redox reaction:
$\overset{+4}{\text{Pb}}\overset{-2}{\text{O}}_2\text{(s)} + 4\overset{+1}{\text{H}}\overset{-1}{\text{Cl(aq)}} \rightarrow \overset{+2}{\text{Pb}}\overset{-1}{\text{Cl}}_2\text{(aq)} + \overset{0}{\text{Cl}}_2\text{(g)} + 2\overset{+1}{\text{H}}_2\overset{-2}{\text{O(l)}}$
Lead is reduced from +4 to +2, and Chlorine is oxidised from -1 to 0.
Combining the reactions for $2\text{PbO} \cdot \text{PbO}_2$ with HCl:
$2 \times [\text{PbO} + 2\text{HCl} \rightarrow \text{PbCl}_2 + \text{H}_2\text{O}]$
$1 \times [\text{PbO}_2 + 4\text{HCl} \rightarrow \text{PbCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}]$
Adding them: $2\text{PbO} + \text{PbO}_2 + 4\text{HCl} + 4\text{HCl} \rightarrow 2\text{PbCl}_2 + \text{PbCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2 + 2\text{H}_2\text{O}$
$\text{Pb}_3\text{O}_4 + 8\text{HCl} \rightarrow 3\text{PbCl}_2 + \text{Cl}_2 + 4\text{H}_2\text{O}$ (Matches the given reaction).
So, the reaction with HCl involves both acid-base reaction of PbO and redox reaction of PbO$_2$ acting as an oxidant towards Cl$^-$.
Reaction with HNO$_3$: HNO$_3$ is a strong acid, and it is also a strong oxidising agent. PbO is a basic oxide and reacts with HNO$_3$ in an acid-base reaction.
$\text{PbO(s)} + 2\text{HNO}_3\text{(aq)} \rightarrow \text{Pb(NO}_3)_2\text{(aq)} + \text{H}_2\text{O(l)}$ (Lead oxidation state remains +2)
PbO$_2$ contains Lead in the +4 oxidation state. While PbO$_2$ can act as an oxidant, HNO$_3$ is also an oxidant. Therefore, a redox reaction between PbO$_2$ and HNO$_3$ is unlikely in which PbO$_2$ is reduced and HNO$_3$ (or something reacting with it) is oxidised. Instead, PbO$_2$ tends to be relatively unreactive ("passive") towards HNO$_3$ under these conditions compared to its reaction as an oxidant with reducing agents like HCl or Cl$^-$. Thus, PbO$_2$ essentially does not react with HNO$_3$.
Combining the reactions for $2\text{PbO} \cdot \text{PbO}_2$ with HNO$_3$:
$2 \times [\text{PbO} + 2\text{HNO}_3 \rightarrow \text{Pb(NO}_3)_2 + \text{H}_2\text{O}]$
$1 \times [\text{PbO}_2 + \text{HNO}_3 \rightarrow \text{No Reaction (essentially)}]$
Adding them: $2\text{PbO} + \text{PbO}_2 + 4\text{HNO}_3 \rightarrow 2\text{Pb(NO}_3)_2 + \text{H}_2\text{O} + \text{H}_2\text{O} + \text{PbO}_2$
$\text{Pb}_3\text{O}_4 + 4\text{HNO}_3 \rightarrow 2\text{Pb(NO}_3)_2 + \text{PbO}_2 + 2\text{H}_2\text{O}$ (Matches the given reaction).
So, the reaction with HNO$_3$ involves the acid-base reaction of PbO, while PbO$_2$ remains unreacted. The reason for the difference is the combined basic/oxidising nature of Pb$_3$O$_4$ and the different reactivity of HCl (reducing chloride ion) vs HNO$_3$ (oxidising nature). In HCl, the redox reaction of PbO$_2$ is prominent. In HNO$_3$, the acid-base reaction of PbO is prominent, and the redox reaction of PbO$_2$ does not occur as HNO$_3$ is itself an oxidant and Cl$^-$ (present in HCl) is a reductant but NO$_3^-$ (from HNO$_3$) is already in a high oxidation state and doesn't readily act as a reductant towards PbO$_2$.
Balancing Of Redox Reactions
Balancing redox equations ensures that both the number of atoms of each element and the total charge are conserved. Two common methods are used:
- Oxidation Number Method: Focuses on changes in oxidation numbers to balance the electron transfer.
- Half-Reaction (Ion-Electron) Method: Splits the overall redox reaction into separate oxidation and reduction half-reactions, balances each half-reaction, and then combines them. This method is particularly useful for reactions in aqueous solution.
**Steps for Oxidation Number Method (Simplified):**
- Write the unbalanced skeletal equation.
- Assign oxidation numbers to all atoms to identify the elements being oxidised and reduced.
- Calculate the total change in oxidation number for the oxidised species and the reduced species. Multiply these species by coefficients to make the total increase in oxidation number equal to the total decrease in oxidation number (balancing electron transfer).
- Balance atoms other than O and H.
- Balance charge by adding H$^+$ (in acidic solution) or OH$^-$ (in basic solution).
- Balance O and H atoms by adding H$_2$O.
**Steps for Half-Reaction Method:**
- Write the unbalanced skeletal equation in ionic form.
- Separate the equation into two half-reactions: one for oxidation, one for reduction.
- Balance atoms other than O and H in each half-reaction.
- Balance O atoms by adding H$_2$O (on the side deficient in O). Balance H atoms by adding H$^+$ (on the side deficient in H). For basic solution, after adding H$^+$, add an equal number of OH$^-$ to both sides, combining H$^+$ and OH$^-$ into H$_2$O where they appear together.
- Balance charge by adding electrons (e$^-$) to the side deficient in negative charge in each half-reaction.
- Multiply each half-reaction by appropriate integers so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
- Add the two balanced half-reactions and cancel out species common to both sides (like e$^-$, H$_2$O, H$^+$ or OH$^-$).
- Verify that the final equation is balanced for both atoms and charge.
Problem 8.8. Write the net ionic equation for the reaction of potassium dichromate(VI), K$_2$Cr$_2$O$_7$ with sodium sulphite, Na$_2$SO$_3$, in an acid solution to give chromium(III) ion and the sulphate ion.
Answer:
The reactants are K$_2$Cr$_2$O$_7$ and Na$_2$SO$_3$. The products are Cr$^{3+}$ ion and SO$_4^{2-}$ ion. The reaction occurs in acidic solution (implies H$^+$ is available).
Step 1: Write the unbalanced skeletal ionic equation. K$^+$ and Na$^+$ are spectator ions and are omitted.
Cr$_2$O$_7^{2-}$ (aq) + SO$_3^{2-}$ (aq) $\rightarrow$ Cr$^{3+}$ (aq) + SO$_4^{2-}$ (aq)
Step 2: Assign oxidation numbers to the central atoms undergoing change (Cr and S).
- In Cr$_2$O$_7^{2-}$: 2Cr + 7(-2) = -2 $\implies$ 2Cr - 14 = -2 $\implies$ 2Cr = 12 $\implies$ Cr = +6.
- In SO$_3^{2-}$: S + 3(-2) = -2 $\implies$ S - 6 = -2 $\implies$ S = +4.
- Cr$^{3+}$: Charge is +3, so oxidation number is +3.
- In SO$_4^{2-}$: S + 4(-2) = -2 $\implies$ S - 8 = -2 $\implies$ S = +6.
Changes in oxidation numbers:
- Cr: +6 $\rightarrow$ +3 (Decrease of 3 per Cr atom). There are 2 Cr atoms in Cr$_2$O$_7^{2-}$. Total decrease = 2 $\times$ 3 = 6. (Reduction)
- S: +4 $\rightarrow$ +6 (Increase of 2 per S atom). There is 1 S atom in SO$_3^{2-}$ and SO$_4^{2-}$. Total increase = 1 $\times$ 2 = 2. (Oxidation)
Step 3: Balance oxidation number changes by multiplying species.
Total decrease is 6. Total increase is 2. To make them equal (6), multiply the species with S by 3.
Cr$_2$O$_7^{2-}$ + 3SO$_3^{2-}$ $\rightarrow$ Cr$^{3+}$ + 3SO$_4^{2-}$
Step 4: Balance atoms other than O and H. Cr is not balanced (2 on left, 1 on right). Multiply Cr$^{3+}$ by 2.
Cr$_2$O$_7^{2-}$ + 3SO$_3^{2-}$ $\rightarrow$ 2Cr$^{3+}$ + 3SO$_4^{2-}$
Step 5: Balance O atoms by adding H$_2$O and H atoms by adding H$^+$ (since it's acidic). Count O atoms: Left = 7 (from Cr$_2$O$_7^{2-}$) + 3$\times$3 (from 3SO$_3^{2-}$) = 7 + 9 = 16. Right = 2$\times$0 (from 2Cr$^{3+}$) + 3$\times$4 (from 3SO$_4^{2-}$) = 12. Need 4 more O atoms on the right. Add 4H$_2$O to the right.
Cr$_2$O$_7^{2-}$ + 3SO$_3^{2-}$ $\rightarrow$ 2Cr$^{3+}$ + 3SO$_4^{2-}$ + 4H$_2$O
Now balance H atoms: Left = 0. Right = 4$\times$2 = 8 (from 4H$_2$O). Need 8 more H atoms on the left. Add 8H$^+$ to the left.
Cr$_2$O$_7^{2-}$ + 3SO$_3^{2-}$ + 8H$^+$ $\rightarrow$ 2Cr$^{3+}$ + 3SO$_4^{2-}$ + 4H$_2$O
Step 6: Verify atom balance and charge balance. Atoms: Cr (2=2), S (3=3), O (7+9=16, 12+4=16), H (8=8). Atoms are balanced. Charges: Left = -2 + 3(-2) + 8(+1) = -2 - 6 + 8 = 0. Right = 2(+3) + 3(-2) + 4(0) = +6 - 6 + 0 = 0. Charges are balanced.
The balanced net ionic equation is: Cr$_2$O$_7^{2-}$ (aq) + 3SO$_3^{2-}$ (aq) + 8H$^+$ (aq) $\rightarrow$ 2Cr$^{3+}$ (aq) + 3SO$_4^{2-}$ (aq) + 4H$_2$O (l).
Problem 8.9. Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.
Answer:
The reactants are permanganate ion (MnO$_4^-$) and bromide ion (Br$^-$). The products are manganese dioxide (MnO$_2$) and bromate ion (BrO$_3^-$). The reaction occurs in basic medium (implies OH$^-$ is available).
Step 1: Write the unbalanced skeletal ionic equation.
MnO$_4^-$ (aq) + Br$^-$ (aq) $\rightarrow$ MnO$_2$ (s) + BrO$_3^-$ (aq)
Step 2: Assign oxidation numbers to the elements undergoing change (Mn and Br).
- In MnO$_4^-$: Mn + 4(-2) = -1 $\implies$ Mn - 8 = -1 $\implies$ Mn = +7.
- Br$^-$: Charge is -1, so oxidation number is -1.
- In MnO$_2$: Mn + 2(-2) = 0 $\implies$ Mn - 4 = 0 $\implies$ Mn = +4.
- In BrO$_3^-$: Br + 3(-2) = -1 $\implies$ Br - 6 = -1 $\implies$ Br = +5.
Changes in oxidation numbers:
- Mn: +7 $\rightarrow$ +4 (Decrease of 3 per Mn atom). Reduction.
- Br: -1 $\rightarrow$ +5 (Increase of 6 per Br atom). Oxidation.
Step 3: Balance oxidation number changes by multiplying species.
Total decrease is 3. Total increase is 6. To make them equal (6), multiply the species with Mn by 2 and the species with Br by 1.
2MnO$_4^-$ + Br$^-$ $\rightarrow$ 2MnO$_2$ + BrO$_3^-$
Step 4: Balance O atoms by adding H$_2$O and H atoms by adding H$^+$ (initially, then convert to basic). Count O atoms: Left = 2$\times$4 = 8. Right = 2$\times$2 (from 2MnO$_2$) + 3 (from BrO$_3^-$) = 4 + 3 = 7. Need 1 more O atom on the right. Add 1H$_2$O to the right.
2MnO$_4^-$ + Br$^-$ $\rightarrow$ 2MnO$_2$ + BrO$_3^-$ + H$_2$O
Now balance H atoms: Left = 0. Right = 2 (from H$_2$O). Need 2 more H atoms on the left. Add 2H$^+$ to the left.
2MnO$_4^-$ + Br$^-$ + 2H$^+$ $\rightarrow$ 2MnO$_2$ + BrO$_3^-$ + H$_2$O
Since the reaction is in basic medium, add an equal number of OH$^-$ ions (2OH$^-$) to both sides.
2MnO$_4^-$ + Br$^-$ + 2H$^+$ + 2OH$^-$ $\rightarrow$ 2MnO$_2$ + BrO$_3^-$ + H$_2$O + 2OH$^-$
Combine H$^+$ and OH$^-$ on the left to form H$_2$O (2H$^+$ + 2OH$^-$ $\rightarrow$ 2H$_2$O).
2MnO$_4^-$ + Br$^-$ + 2H$_2$O $\rightarrow$ 2MnO$_2$ + BrO$_3^-$ + H$_2$O + 2OH$^-$
Cancel H$_2$O molecules common on both sides (1 H$_2$O on left cancels with 1 on right). Wait, 2H$_2$O on left, 1H$_2$O on right. Cancel 1 from right and 1 from left.
2MnO$_4^-$ + Br$^-$ + H$_2$O $\rightarrow$ 2MnO$_2$ + BrO$_3^-$ + 2OH$^-$
Step 5: Verify atom balance and charge balance. Atoms: Mn (2=2), Br (1=1), O (8+1=9, 4+3+2=9), H (2=2). Atoms are balanced. Charges: Left = 2(-1) + 1(-1) + 1(0) = -2 - 1 = -3. Right = 2(0) + 1(-1) + 2(-1) = 0 - 1 - 2 = -3. Charges are balanced.
The balanced ionic equation is: 2MnO$_4^-$ (aq) + Br$^-$ (aq) + H$_2$O (l) $\rightarrow$ 2MnO$_2$ (s) + BrO$_3^-$ (aq) + 2OH$^-$ (aq).
Problem 8.10. Permanganate(VII) ion, MnO$_4^{–}$ in basic solution oxidises iodide ion, I$^–$ to produce molecular iodine (I$_2$) and manganese (IV) oxide (MnO$_2$). Write a balanced ionic equation to represent this redox reaction.
Answer:
The reactants are permanganate ion (MnO$_4^-$) and iodide ion (I$^-$). The products are molecular iodine (I$_2$) and manganese (IV) oxide (MnO$_2$). The reaction occurs in basic solution. We will use the half-reaction method.
Step 1: Write the unbalanced skeletal ionic equation:
MnO$_4^-$ (aq) + I$^-$ (aq) $\rightarrow$ MnO$_2$ (s) + I$_2$ (s)
Step 2: Separate into half-reactions:
- Oxidation half-reaction: I$^-$ (aq) $\rightarrow$ I$_2$ (s)
- Reduction half-reaction: MnO$_4^-$ (aq) $\rightarrow$ MnO$_2$ (s)
Step 3: Balance atoms other than O and H.
- Oxidation half: 2I$^-$ (aq) $\rightarrow$ I$_2$ (s)
- Reduction half: MnO$_4^-$ (aq) $\rightarrow$ MnO$_2$ (s) (Mn is already balanced)
Step 4: Balance O atoms by adding H$_2$O and H atoms by adding H$^+$ (then convert to basic). This is similar to Problem 8.9 (reduction of MnO$_4^-$ to MnO$_2$).
- Oxidation half: 2I$^-$ (aq) $\rightarrow$ I$_2$ (s) (No O or H)
- Reduction half: MnO$_4^-$ (aq) $\rightarrow$ MnO$_2$ (s) + 2H$_2$O (l) (Balance O by adding 2H$_2$O)
MnO$_4^-$ (aq) + 4H$^+$ (aq) $\rightarrow$ MnO$_2$ (s) + 2H$_2$O (l) (Balance H by adding 4H$^+$)
Convert to basic medium: Add 4OH$^-$ to both sides.
MnO$_4^-$ (aq) + 4H$^+$ (aq) + 4OH$^-$ (aq) $\rightarrow$ MnO$_2$ (s) + 2H$_2$O (l) + 4OH$^-$ (aq)
Combine H$^+$ and OH$^-$ on the left: MnO$_4^-$ (aq) + 4H$_2$O (l) $\rightarrow$ MnO$_2$ (s) + 2H$_2$O (l) + 4OH$^-$ (aq)
Cancel H$_2$O: MnO$_4^-$ (aq) + 2H$_2$O (l) $\rightarrow$ MnO$_2$ (s) + 4OH$^-$ (aq)
Step 5: Balance charge by adding electrons.
- Oxidation half: 2I$^-$ (aq) $\rightarrow$ I$_2$ (s). Left charge = 2(-1) = -2. Right charge = 0. Add 2e$^-$ to the right.
2I$^-$ (aq) $\rightarrow$ I$_2$ (s) + 2e$^-$
- Reduction half: MnO$_4^-$ (aq) + 2H$_2$O (l) $\rightarrow$ MnO$_2$ (s) + 4OH$^-$ (aq). Left charge = -1 + 0 = -1. Right charge = 0 + 4(-1) = -4. Add 3e$^-$ to the left.
MnO$_4^-$ (aq) + 2H$_2$O (l) + 3e$^-$ $\rightarrow$ MnO$_2$ (s) + 4OH$^-$ (aq)
Step 6: Equalise the number of electrons. Multiply oxidation half by 3 and reduction half by 2.
- $3 \times [2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)} + \text{2e}^-]$ $\implies 6\text{I}^-\text{(aq)} \rightarrow 3\text{I}_2\text{(s)} + 6\text{e}^-$
- $2 \times [\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} + \text{3e}^- \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}]$ $\implies 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} + \text{6e}^- \rightarrow 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$
Step 7: Add the balanced half-reactions and cancel electrons.
$6\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} + 6\text{e}^- \rightarrow 3\text{I}_2\text{(s)} + 6\text{e}^- + 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$
$6\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} \rightarrow 3\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$
Step 8: Verify atom and charge balance. Atoms: I (6=6), Mn (2=2), O (8+4=12, 4+8=12), H (8=8). Balanced. Charges: Left = 6(-1) + 2(-1) = -6 - 2 = -8. Right = 0 + 0 + 8(-1) = -8. Balanced.
The balanced ionic equation is: 6I$^-$ (aq) + 2MnO$_4^-$ (aq) + 4H$_2$O (l) $\rightarrow$ 3I$_2$ (s) + 2MnO$_2$ (s) + 8OH$^-$ (aq).
Redox Reactions As The Basis For Titrations
Redox reactions are used in quantitative analysis through redox titrations. Similar to acid-base titrations, a solution of known concentration (titrant, oxidant or reductant) is used to determine the unknown concentration of a substance that undergoes a redox reaction with the titrant. An indicator is used to signal the equivalence point (when the oxidant and reductant have reacted according to their stoichiometry).
Redox indicators can be:
- The reagent itself is intensely colored and its color change signals the endpoint (e.g., MnO$_4^-$ changing from purple to colorless upon reduction).
- An indicator substance is added which undergoes a sharp color change upon oxidation or reduction right at the equivalence point (e.g., diphenylamine in dichromate titrations).
- A method using iodine: Iodine is released in a redox reaction (e.g., $\text{Cu}^{2+}$ oxidising $\text{I}^-$ to $\text{I}_2$). The released iodine is then titrated with a standard reducing agent (like thiosulfate). Starch is used as an indicator; it forms an intense blue complex with iodine, which disappears when all iodine is consumed by the titrant.
Limitations Of Concept Of Oxidation Number
While the oxidation number concept is very useful for identifying redox reactions and balancing equations, it is a formal system based on assumptions about electron transfer that may not perfectly reflect the actual electron distribution in covalent bonds. The true picture involves electron shifts and unequal sharing rather than complete transfers. The oxidation number is an assigned value for bookkeeping, not a direct physical property of the atom in the molecule.
Redox Reactions And Electrode Processes
When a redox reaction occurs with the direct transfer of electrons (like Zn in Cu$_2\text{SO}_4$ solution), heat is released. We can design experiments to make the electron transfer occur indirectly, through an external circuit, thereby converting chemical energy into electrical energy. This is the basis of electrochemical cells (like galvanic cells).
Consider the reaction $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$. In a Daniell cell (Fig. 8.3), Zn metal is placed in $\text{ZnSO}_4$ solution, and Cu metal is placed in $\text{CuSO}_4$ solution. The solutions are connected by a salt bridge, and the metals (electrodes) are connected by a wire through an ammeter.
When the circuit is closed, electrons flow from the zinc electrode to the copper electrode through the wire. At the zinc electrode (anode), Zn is oxidised: $\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{2e}^-$. At the copper electrode (cathode), $\text{Cu}^{2+}$ ions are reduced: $\text{Cu}^{2+}\text{(aq)} + \text{2e}^- \rightarrow \text{Cu(s)}$. The salt bridge allows ions to migrate between the solutions to maintain electrical neutrality, completing the circuit.
Each electrode where an oxidation or reduction half-reaction occurs is associated with a potential, the electrode potential. Under standard conditions (1 M concentration for species, 1 bar pressure for gases, 298 K), this is the standard electrode potential ($E^0$).
The standard electrode potential is a measure of the tendency of a species to be reduced. It is commonly listed for reduction half-reactions (Table 8.1).
| Reaction (Oxidised form + ne$^-$ $\rightleftharpoons$ Reduced form) | $E^0$ / V | |
| F$_2$(g) + 2e$^-$ $\rightleftharpoons$ 2F$^-$ | 2.87 | Increasing strength of oxidising agent Increasing strength of reducing agent |
| Co$^{3+}$ + e$^-$ $\rightleftharpoons$ Co$^{2+}$ | 1.81 | |
| H$_2$O$_2$ + 2H$^+$ + 2e$^-$ $\rightleftharpoons$ 2H$_2$O | 1.78 | |
| MnO$_4^-$ + 8H$^+$ + 5e$^-$ $\rightleftharpoons$ Mn$^{2+}$ + 4H$_2$O | 1.51 | |
| Au$^{3+}$ + 3e$^-$ $\rightleftharpoons$ Au(s) | 1.40 | |
| Cl$_2$(g) + 2e$^-$ $\rightleftharpoons$ 2Cl$^-$ | 1.36 | |
| Cr$_2$O$_7^{2-}$ + 14H$^+$ + 6e$^-$ $\rightleftharpoons$ 2Cr$^{3+}$ + 7H$_2$O | 1.33 | |
| O$_2$(g) + 4H$^+$ + 4e$^-$ $\rightleftharpoons$ 2H$_2$O | 1.23 | |
| MnO$_2$(s) + 4H$^+$ + 2e$^-$ $\rightleftharpoons$ Mn$^{2+}$ + 2H$_2$O | 1.23 | |
| Br$_2$ + 2e$^-$ $\rightleftharpoons$ 2Br$^-$ | 1.09 | |
| NO$_3^-$ + 4H$^+$ + 3e$^-$ $\rightleftharpoons$ NO(g) + 2H$_2$O | 0.97 | |
| 2Hg$^{2+}$ + 2e$^-$ $\rightleftharpoons$ Hg$_2^{2+}$ | 0.92 | |
| Ag$^+$ + e$^-$ $\rightleftharpoons$ Ag(s) | 0.80 | |
| Fe$^{3+}$ + e$^-$ $\rightleftharpoons$ Fe$^{2+}$ | 0.77 | |
| O$_2$(g) + 2H$^+$ + 2e$^-$ $\rightleftharpoons$ H$_2$O$_2$ | 0.68 | |
| I$_2$(s) + 2e$^-$ $\rightleftharpoons$ 2I$^-$ | 0.54 | |
| Cu$^+$ + e$^-$ $\rightleftharpoons$ Cu(s) | 0.52 | |
| Cu$^{2+}$ + 2e$^-$ $\rightleftharpoons$ Cu(s) | 0.34 | |
| AgCl(s) + e$^-$ $\rightleftharpoons$ Ag(s) + Cl$^-$ | 0.22 | |
| AgBr(s) + e$^-$ $\rightleftharpoons$ Ag(s) + Br$^-$ | 0.10 | |
| 2H$^+$ + 2e$^-$ $\rightleftharpoons$ H$_2$(g) | 0.00 | |
| Pb$^{2+}$ + 2e$^-$ $\rightleftharpoons$ Pb(s) | –0.13 | |
| Sn$^{2+}$ + 2e$^-$ $\rightleftharpoons$ Sn(s) | –0.14 | |
| Ni$^{2+}$ + 2e$^-$ $\rightleftharpoons$ Ni(s) | –0.25 | |
| Fe$^{2+}$ + 2e$^-$ $\rightleftharpoons$ Fe(s) | –0.44 | |
| Cr$^{3+}$ + 3e$^-$ $\rightleftharpoons$ Cr(s) | –0.74 | |
| Zn$^{2+}$ + 2e$^-$ $\rightleftharpoons$ Zn(s) | –0.76 | |
| 2H$_2$O + 2e$^-$ $\rightleftharpoons$ H$_2$(g) + 2OH$^-$ | –0.83 | |
| Al$^{3+}$ + 3e$^-$ $\rightleftharpoons$ Al(s) | –1.66 | |
| Mg$^{2+}$ + 2e$^-$ $\rightleftharpoons$ Mg(s) | –2.36 | |
| Na$^+$ + e$^-$ $\rightleftharpoons$ Na(s) | –2.71 | |
| Ca$^{2+}$ + 2e$^-$ $\rightleftharpoons$ Ca(s) | –2.87 | |
| K$^+$ + e$^-$ $\rightleftharpoons$ K(s) | –2.93 | |
| Li$^+$ + e$^-$ $\rightleftharpoons$ Li(s) | –3.05 |
A more positive $E^0$ value indicates a greater tendency for the oxidised form to be reduced (stronger oxidising agent). A more negative $E^0$ value indicates a greater tendency for the reduced form to be oxidised (stronger reducing agent).
This table allows comparison of the relative strengths of various oxidising and reducing agents and prediction of the feasibility of redox reactions (a reaction is feasible if the oxidising agent's reduction potential is higher than the reducing agent's reduction potential).
Exercises
Question 8.1 Assign oxidation number to the underlined elements in each of the following species:
(a) $NaH_2\underline{P}O_4$
(b) $NaH\underline{S}O_4$
(c) $H_4\underline{P}_2O_7$
(d) $K_2\underline{Mn}O_4$
(e) $Ca\underline{O}_2$
(f) $Na\underline{B}H_4$
(g) $H_2\underline{S}_2O_7$
(h) $KAl(\underline{S}O_4)_2.12 H_2O$
Answer:
Question 8.2 What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
(a) $K\underline{I}_3$
(b) $H_2\underline{S}_4O_6$
(c) $\underline{Fe}_3O_4$
(d) $CH_3\underline{C}H_2OH$
(e) $CH_3\underline{C}OOH$
Answer:
Question 8.3 Justify that the following reactions are redox reactions:
(a) $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(g)$
(b) $Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)$
(c) $4BCl_3(g) + 3LiAlH_4(s) \rightarrow 2B_2H_6(g) + 3LiCl(s) + 3 AlCl_3 (s)$
(d) $2K(s) + F_2(g) \rightarrow 2K^+F^–(s)$
(e) $4 NH_3(g) + 5 O_2(g) \rightarrow 4NO(g) + 6H_2O(g)$
Answer:
Question 8.4 Fluorine reacts with ice and results in the change:
$H_2O(s) + F_2(g) \rightarrow HF(g) + HOF(g)$
Justify that this reaction is a redox reaction.
Answer:
Question 8.5 Calculate the oxidation number of sulphur, chromium and nitrogen in $H_2SO_5$, $Cr_2O_7^{2–}$ and $NO_3^–$. Suggest structure of these compounds. Count for the fallacy.
Answer:
Question 8.6 Write formulas for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxide
Answer:
Question 8.7 Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.
Answer:
Question 8.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why ?
Answer:
Question 8.9 Consider the reactions:
(a) $6 CO_2(g) + 6H_2O(l) \rightarrow C_6 H_{12} O_6(aq) + 6O_2(g)$
(b) $O_3(g) + H_2O_2(l) \rightarrow H_2O(l) + 2O_2(g)$
Why it is more appropriate to write these reactions as :
(a) $6CO_2(g) + 12H_2O(l) \rightarrow C_6 H_{12} O_6(aq) + 6H_2O(l) + 6O_2(g)$
(b) $O_3(g) + H_2O_2 (l) \rightarrow H_2O(l) + O_2(g) + O_2(g)$
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer:
Question 8.10 The compound $AgF_2$ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?
Answer:
Question 8.11 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:
Question 8.12 How do you count for the 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:
Question 8.13 Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions:
(a) $2AgBr (s) + C_6H_6O_2(aq) \rightarrow 2Ag(s) + 2HBr (aq) + C_6H_4O_2(aq)$
(b) $HCHO(l) + 2[Ag (NH_3)_2]^+(aq) + 3OH^–(aq) \rightarrow 2Ag(s) + HCOO^–(aq) + 4NH_3(aq) + 2H_2O(l)$
(c) $HCHO (l) + 2 Cu^{2+}(aq) + 5 OH^–(aq) \rightarrow Cu_2O(s) + HCOO^–(aq) + 3H_2O(l)$
(d) $N_2H_4(l) + 2H_2O_2(l) \rightarrow N_2(g) + 4H_2O(l)$
(e) $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
Answer:
Question 8.14 Consider the reactions :
$2 S_2O_3^{2–}(aq) + I_2(s) \rightarrow S_4 O_6^{2–}(aq) + 2I^–(aq)$
$S_2O_3^{2–}(aq) + 2Br_2(l) + 5 H_2O(l) \rightarrow 2SO_4^{2–}(aq) + 4Br^–(aq) + 10H^+(aq)$
Why does the same reductant, thiosulphate react differently with iodine and bromine ?
Answer:
Question 8.15 Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
Question 8.16 Why does the following reaction occur ?
$XeO_6^{4–}(aq) + 2F^–(aq) + 6H^+(aq) \rightarrow XeO_3(g)+ F_2(g) + 3H_2O(l)$
What conclusion about the compound $Na_4XeO_6$ (of which $XeO_6^{4–}$ is a part) can be drawn from the reaction.
Answer:
Question 8.17 Consider the reactions:
(a) $H_3PO_2(aq) + 4 AgNO_3(aq) + 2 H_2O(l) \rightarrow H_3PO_4(aq) + 4Ag(s) + 4HNO_3(aq)$
(b) $H_3PO_2(aq) + 2CuSO_4(aq) + 2 H_2O(l) \rightarrow H_3PO_4(aq) + 2Cu(s) + H_2SO_4(aq)$
(c) $C_6H_5CHO(l) + 2[Ag (NH_3)_2]^+(aq) + 3OH^–(aq) \rightarrow C_6H_5COO^–(aq) + 2Ag(s) + 4NH_3 (aq) + 2 H_2O(l)$
(d) $C_6H_5CHO(l) + 2Cu^{2+}(aq) + 5OH^–(aq) \rightarrow$ No change observed.
What inference do you draw about the behaviour of $Ag^+$ and $Cu^{2+}$ from these reactions ?
Answer:
Question 8.18 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) $H_2O_2(aq) + Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + H_2O(l)$ (in acidic solution)
(d) $Cr_2O_7^{2–} + SO_2(g) \rightarrow Cr^{3+}(aq) + SO_4^{2–}(aq)$ (in acidic solution)
Answer:
Question 8.19 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) $P_4(s) + OH^–(aq) \rightarrow PH_3(g) + HPO_2^–(aq)$
(b) $N_2H_4(l) + ClO_3^–(aq) \rightarrow NO(g) + Cl^–(g)$
(c) $Cl_2O_7(g) + H_2O_2(aq) \rightarrow ClO_2^–(aq) + O_2(g) + H^+$
Answer:
Question 8.20 What sorts of informations can you draw from the following reaction ?
$(CN)_2(g) + 2OH^–(aq) \rightarrow CN^–(aq) + CNO^–(aq) + H_2O(l)$
Answer:
Question 8.21 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:
Question 8.22 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 postive 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:
Question 8.23 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:
Question 8.24 Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.
Answer:
Question 8.25 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:
Question 8.26 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) $Ag^+(aq)$ and $Cu(s)$
(c) $Fe^{3+}(aq)$ and $Cu(s)$
(d) $Ag(s)$ and $Fe^{3+}(aq)$
(e) $Br_2(aq)$ and $Fe^{2+}(aq)$.
Answer:
Question 8.27 Predict the products of electrolysis in each of the following:
(i) An aqueous solution of $AgNO_3$ with silver electrodes
(ii) An aqueous solution $AgNO_3$ with platinum electrodes
(iii) A dilute solution of $H_2SO_4$ with platinum electrodes
(iv) An aqueous solution of $CuCl_2$ with platinum electrodes.
Answer:
Question 8.28 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:
Question 8.29 Given the standard electrode potentials,
$K^+/K = –2.93V, Ag^+/Ag = 0.80V,$
$Hg^{2+}/Hg = 0.79V$
$Mg^{2+}/Mg = –2.37V, Cr^{3+}/Cr = –0.74V$
arrange these metals in their increasing order of reducing power.
Answer:
Question 8.30 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: