Redox Reactions

The chapter builds redox chemistry in a deliberate ladder. It opens with the classical, oxygen-and-hydrogen definition of oxidation and reduction, then reinterprets the same reactions as electron transfer, introducing oxidising and reducing agents. From there it develops oxidation number as a universal bookkeeping tool with six precise rules, including the tricky cases of peroxides, superoxides and fluorides of oxygen, and shows how oxidation number reproduces the electron-transfer definitions of oxidation and reduction even for covalent compounds where no full charge is actually transferred. The chapter then sorts redox reactions into combination, decomposition, displacement and disproportionation reactions, teaches the oxidation number method and the half reaction (ion-electron) method for balancing equations in both acidic and basic medium, and closes by connecting redox behaviour to electrode processes, standard electrode potential, the electrochemical series and the Daniell cell, the direct bridge to electrochemistry in Class 12.

The classical idea of oxidation and reduction

Chemists first defined oxidation narrowly, as the addition of oxygen to a substance, seen in reactions such as $2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$ and $\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g)$. Reaction $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$ can equally be read as removal of hydrogen from methane, so the definition of oxidation broadened to include loss of hydrogen. Extending the pattern further, reactions such as $\text{Mg}(s) + \text{Cl}_2(g) \rightarrow \text{MgCl}_2(s)$ show oxidation by addition of any electronegative element, not just oxygen, and by extension, removal of an electropositive element also counts as oxidation. Reduction, mirror image of oxidation, was first defined as removal of oxygen, seen in $2\text{HgO}(s) \rightarrow 2\text{Hg}(l) + \text{O}_2(g)$. The definition broadened the same way, to cover removal of any electronegative element, as in $2\text{FeCl}_3(aq) + \text{H}_2(g) \rightarrow 2\text{FeCl}_2(aq) + 2\text{HCl}(aq)$, and addition of hydrogen or an electropositive element, as in $\text{CH}_2 = \text{CH}_2(g) + \text{H}_2(g) \rightarrow \text{H}_3\text{C} - \text{CH}_3(g)$. Because oxidation and reduction turn out to occur together in every one of these reactions, for instance in $2\text{HgCl}_2(aq) + \text{SnCl}_2(aq) \rightarrow \text{Hg}_2\text{Cl}_2(s) + \text{SnCl}_4(aq)$ where mercury is reduced while tin is simultaneously oxidised, chemists coined the composite word redox for the whole class. The classical picture is intuitive but limited, it cannot easily classify a reaction like $2\text{Na}(s) + \text{H}_2(g) \rightarrow 2\text{NaH}(s)$ without appealing to electronegativity, which is exactly the gap the electron-transfer picture closes.

Redox reactions as electron transfer

Ionic compounds such as sodium chloride, sodium oxide and sodium sulphide are more honestly written as $\text{Na}^+\text{Cl}^-$, $(\text{Na}^+)_2\text{O}^{2-}$ and $(\text{Na}^+)_2\text{S}^{2-}$, which suggests splitting each formation reaction into two separate steps, one where sodium loses electrons and one where the other element gains them. For the formation of sodium chloride these half reactions are

where: each equation is called a half reaction because it explicitly shows electrons appearing on one side; adding the two half reactions together, with the electrons cancelling exactly, regenerates the overall equation $2\text{Na}(s) + \text{Cl}_2(g) \rightarrow 2\text{NaCl}(s)$. This gives the modern operational definitions used throughout the chapter: oxidation is loss of electrons by a species, reduction is gain of electrons by a species, an oxidising agent (oxidant) is the acceptor of electrons, and a reducing agent (reductant) is the donor of electrons. Sodium, which donates electrons, is therefore the reducing agent in reaction (7.1), while chlorine, which accepts them, is the oxidising agent. Placing a strip of metallic zinc into aqueous copper nitrate makes the strip turn reddish with deposited copper while the blue $\text{Cu}^{2+}$ colour fades, the visible signature of

Copper metal placed in silver nitrate solution behaves the same way, $\text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)$, with equilibrium strongly favouring products in both cases, but copper metal placed in zinc sulphate solution shows no reaction at all, no detectable $\text{Cu}^{2+}$ ever forms. This one-way competition for electrons, zinc gives electrons to copper, copper gives electrons to silver, but the reverse does not happen spontaneously, is the seed of the electrochemical series developed later in the chapter: it establishes an order of electron-releasing tendency $\text{Zn} > \text{Cu} > \text{Ag}$.

Oxidation number: a universal bookkeeping tool

Electron-transfer language works cleanly for ionic compounds, but for a covalent molecule like water, formed in $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$, no electron is truly and fully transferred from hydrogen to oxygen, the bonding electrons are merely shared unevenly. Chemists handle this by an accounting device, the oxidation number, defined by assuming that the shared electron pair in every bond belongs entirely to the more electronegative atom. Six rules fix the oxidation number of any atom in any species.

Rule 1. Every atom of an element in its free, uncombined state has an oxidation number of zero; this covers $\text{H}_2$, $\text{O}_2$, $\text{Cl}_2$, $\text{O}_3$, $\text{P}_4$, $\text{S}_8$ and metals like $\text{Na}$ or $\text{Al}$.

Rule 2. For a monatomic ion, the oxidation number equals the charge on the ion; $\text{Na}^+$ is $+1$, $\text{Mg}^{2+}$ is $+2$, $\text{Cl}^-$ is $-1$. In their compounds, alkali metals are always $+1$, alkaline earth metals always $+2$, and aluminium always $+3$.

Rule 3. Oxygen is usually $-2$, with two exceptions: in peroxides such as $\text{H}_2\text{O}_2$ and $\text{Na}_2\text{O}_2$ it is $-1$, in superoxides such as $\text{KO}_2$ it is $-\tfrac{1}{2}$, and when bonded to the more electronegative fluorine, as in $\text{OF}_2$ and $\text{O}_2\text{F}_2$, it is forced positive, $+2$ and $+1$ respectively.

Rule 4. Hydrogen is usually $+1$, except in ionic metal hydrides such as $\text{NaH}$ and $\text{CaH}_2$, where it is $-1$ because the metal is less electronegative.

Rule 5. Fluorine, the most electronegative element, is always $-1$ in every compound. The other halogens are $-1$ when present as simple halide ions, but take positive oxidation numbers when bonded to oxygen, as in oxoacids and oxoanions.

Rule 6. The algebraic sum of the oxidation numbers of all atoms in a neutral compound must be zero, and in a polyatomic ion the sum must equal the ion's overall charge; this is what lets us solve for an unknown oxidation number by simple algebra. The oxidation number of a metal is often written using Stock notation, a Roman numeral in parentheses after the element symbol, so aurous and auric chloride become $\text{Au(I)Cl}$ and $\text{Au(III)Cl}_3$, and stannous and stannic chloride become $\text{Sn(II)Cl}_2$ and $\text{Sn(IV)Cl}_4$. With oxidation number in hand, oxidation and reduction get sharper, general definitions that subsume the classical and electron-transfer pictures: oxidation is an increase in the oxidation number of an element, reduction is a decrease, an oxidising agent is a reagent that increases another element's oxidation number (while its own falls), and a reducing agent lowers another element's oxidation number (while its own rises). The highest oxidation number a representative element can show equals its group number for groups 1 and 2, and the group number minus ten for groups 13 to 17, so across period 3 it climbs in single steps from $+1$ in $\text{NaCl}$ up to $+7$ in $\text{HClO}_4$, passing through $+2$ ($\text{MgSO}_4$), $+3$ ($\text{AlF}_3$), $+4$ ($\text{SiCl}_4$), $+5$ ($\text{P}_4\text{O}_{10}$) and $+6$ ($\text{SF}_6$) on the way.

Fractional oxidation numbers and their real meaning

Rule 6 sometimes returns a fraction, the oxidation number of carbon in carbon suboxide $\text{C}_3\text{O}_2$ works out to $+\tfrac{4}{3}$, of bromine in $\text{Br}_3\text{O}_8$ to $+\tfrac{16}{3}$, and of sulphur in the tetrathionate ion $\text{S}_4\text{O}_6^{2-}$ to $+2.5$. Electrons are never shared in fractions, so a fractional value is always an average across chemically distinct atoms of the same element. The structure of $\text{C}_3\text{O}_2$ (O=C=C=C=O) shows two terminal carbons at $+2$ flanking a central carbon at $0$, averaging to $+\tfrac{4}{3}$; the structure of $\text{S}_4\text{O}_6^{2-}$ shows two terminal sulphurs at $+5$ flanking two central sulphurs at $0$, averaging to $2.5$. Whenever a fractional oxidation number appears, the honest statement is that the element occupies more than one whole-number oxidation state within that single species, and only the structure reveals which atoms are which; $\text{Fe}_3\text{O}_4$, $\text{Mn}_3\text{O}_4$ and $\text{Pb}_3\text{O}_4$ are further familiar examples, each a stoichiometric mixture of two whole-number oxidation states of the metal.

Four families of redox reactions

Redox reactions sort neatly into four types. A combination reaction, $A + B \rightarrow C$, is redox only if at least one of $A$ and $B$ is an element in the free state, as in $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$; note that combining $\text{CaO}$ with $\text{CO}_2$ to give $\text{CaCO}_3$ is not redox, because neither reactant is elemental and no oxidation number changes. A decomposition reaction breaks one compound into two or more products, at least one of which must be elemental, as in $2\text{KClO}_3(s) \rightarrow 2\text{KCl}(s) + 3\text{O}_2(g)$; decomposing $\text{CaCO}_3$ into $\text{CaO}$ and $\text{CO}_2$ is, by contrast, not redox, since no oxidation number changes there either. A displacement reaction, $X + YZ \rightarrow XZ + Y$, replaces one element in a compound with another, and splits into metal displacement, such as $\text{CuSO}_4(aq) + \text{Zn}(s) \rightarrow \text{Cu}(s) + \text{ZnSO}_4(aq)$, widely used in metallurgy, and non-metal displacement, chiefly hydrogen displacement from water or acid, such as $2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)$ and $\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$, and the rarer halogen displacement such as $\text{Cl}_2(g) + 2\text{KBr}(aq) \rightarrow 2\text{KCl}(aq) + \text{Br}_2(l)$, which underlies the classic layer test for bromide and iodide. A disproportionation reaction is the special case in which one element, present in an intermediate oxidation state, is simultaneously oxidised and reduced, forming both a higher and a lower oxidation state of itself; this always requires an element capable of at least three oxidation states. The decomposition of hydrogen peroxide is the textbook example, $2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)$, in which oxygen in $\text{H}_2\text{O}_2$ at $-1$ splits into oxygen at $-2$ in water and oxygen at $0$ in $\text{O}_2$. Chlorine gas in cold alkali similarly disproportionates, $\text{Cl}_2(g) + 2\text{OH}^-(aq) \rightarrow \text{Cl}^-(aq) + \text{ClO}^-(aq) + \text{H}_2\text{O}(l)$, the reaction behind household bleach; fluorine is the one halogen that never disproportionates, because being the most electronegative element of all, it cannot be pushed into any positive oxidation state.

Balancing redox equations

Two systematic routes turn an unbalanced skeletal redox equation into a fully balanced one, and both rest on the same conservation law, electrons lost must equal electrons gained. The oxidation number method tracks the total rise and fall in oxidation number across the whole molecule or ion; the half reaction (ion-electron) method splits the equation into an oxidation half and a reduction half, balances each independently for atoms and charge, and adds them so the electrons cancel. Both methods are worked in full, step by step, in the derivation blocks below, using the chapter's own examples, dichromate oxidising sulphite in acid, and permanganate oxidising iodide in base.

Redox reactions as the basis of titrations

Because redox reactions proceed to a sharp stoichiometric end point, they underlie a family of volumetric titrations parallel to acid-base titrations, tracked with a redox-sensitive indicator instead of a pH indicator. Permanganate titrations need no added indicator: $\text{MnO}_4^-$ is so intensely purple that the first permanent faint pink, appearing once the last trace of reductant such as $\text{Fe}^{2+}$ or $\text{C}_2\text{O}_4^{2-}$ is consumed, is itself the end point, a self-indicating titration. Dichromate titrations need an external indicator, diphenylamine, oxidised to an intense blue only after the equivalence point, since $\text{Cr}_2\text{O}_7^{2-}$ is not coloured enough to serve alone. Iodometric titrations use a third trick: an oxidant such as $\text{Cu}^{2+}$ first liberates iodine from iodide, $2\text{Cu}^{2+}(aq) + 4\text{I}^-(aq) \rightarrow \text{Cu}_2\text{I}_2(s) + \text{I}_2(aq)$, and the liberated iodine, which turns starch blue, is titrated against standard thiosulphate, $\text{I}_2(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow 2\text{I}^-(aq) + \text{S}_4\text{O}_6^{2-}(aq)$, the disappearance of the blue marking the end point.

Redox reactions and electrode processes

When zinc metal sits directly in copper sulphate solution, electrons transfer at the metal surface and the reaction's energy is released only as heat. Separating the two half reactions physically, zinc in zinc sulphate solution in one beaker, copper in copper sulphate solution in another, joined by a wire and a salt bridge, forces that same electron transfer through the external wire instead, and that flow of electrons is an electric current; this arrangement is the Daniell cell, shown above. The oxidised and reduced forms of one species together form a redox couple, written oxidised form before reduced form and separated by a slash, so the two couples at work in the Daniell cell are $\text{Zn}^{2+}/\text{Zn}$ and $\text{Cu}^{2+}/\text{Cu}$. Each electrode develops an electrode potential that measures its couple's tendency to hold onto or release electrons. When every species is at unit concentration (or, for a gas, one atmosphere) at $298\ \text{K}$, this is the standard electrode potential, $E^{\circ}$. By convention the standard hydrogen electrode, $2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)$, is fixed at $E^{\circ} = 0.00\ \text{V}$, and every other couple's $E^{\circ}$ is measured against it. A negative $E^{\circ}$ marks a couple that is a stronger reducing agent than $\text{H}^+/\text{H}_2$; a positive $E^{\circ}$ marks a weaker one. Arranging couples by $E^{\circ}$ gives the electrochemical series, from the most positive (strongest oxidant, $\text{F}_2/\text{F}^-$ at $+2.87\ \text{V}$) to the most negative (strongest reductant, $\text{Li}^+/\text{Li}$ at $-3.05\ \text{V}$); it explains, quantitatively, why zinc displaces copper and copper displaces silver, developed with a full worked trend in the third derivation below.

Limitations of the oxidation number concept

The oxidation number is a bookkeeping convention, not a claim about literal, physically measured charge on an atom; the true electron distribution in a covalent bond like H–Cl is only partially shifted, never a full unit charge. The modern, more physical description reframes oxidation as a decrease in electron density around an atom and reduction as an increase, agreeing with oxidation number everywhere without the artificial assumption of complete electron transfer in covalent bonds. Despite this caveat, the oxidation number method remains the fastest, most reliable practical tool for spotting and balancing redox reactions, the one used throughout this chapter and in Class 12 electrochemistry.

Board level - assigning oxidation numbers to classify an oxidant and a reductant

Show that $2\text{Cu}_2\text{O}(s) + \text{Cu}_2\text{S}(s) \rightarrow 6\text{Cu}(s) + \text{SO}_2(g)$ is a redox reaction and identify which species is the oxidant and which is the reductant (extending NCERT Problem 7.4).

Step 1. Assign oxidation numbers using the six rules. In $\text{Cu}_2\text{O}$, oxygen is $-2$, so copper is $+1$; in $\text{Cu}_2\text{S}$, copper stays $+1$ and sulphur is $-2$. In elemental copper on the right, the oxidation number is $0$ by rule 1. In $\text{SO}_2$, oxygen is $-2$, so sulphur is $+4$.

Step 2. Track every change. Copper falls from $+1$ to $0$, a decrease, so copper is reduced. Sulphur rises from $-2$ to $+4$, a six-unit increase, so sulphur is oxidised. One element reduced while another is oxidised in the same equation confirms the reaction is redox.

Step 3. Name the oxidant and reductant. Copper(I) in $\text{Cu}_2\text{O}$ pushes sulphur to a higher oxidation number, so $\text{Cu}_2\text{O}$ is the oxidising agent; sulphur in $\text{Cu}_2\text{S}$ pulls copper down to the zero state, so $\text{Cu}_2\text{S}$ is the reducing agent. Answer: the reaction is redox, with copper reduced from +1 to 0 and sulphur oxidised from −2 to +4; Cu2O is the oxidising agent and Cu2S is the reducing agent.

Board level - balancing permanganate oxidising bromide in basic medium

Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation (extending NCERT Problem 7.9, using the oxidation number method).

Step 1. Write the skeletal equation and assign oxidation numbers: $\text{MnO}_4^-(aq) + \text{Br}^-(aq) \rightarrow \text{MnO}_2(s) + \text{BrO}_3^-(aq)$. Manganese falls from $+7$ to $+4$ (a fall of $3$); bromine rises from $-1$ to $+5$ (a rise of $6$).

Step 2. Equalise rise and fall. One bromine rising by $6$ matches two manganese atoms each falling by $3$, so $\text{MnO}_4^-$ and $\text{MnO}_2$ take coefficient two: $2\text{MnO}_4^-(aq) + \text{Br}^-(aq) \rightarrow 2\text{MnO}_2(s) + \text{BrO}_3^-(aq)$.

Step 3. Balance ionic charge in basic medium with $\text{OH}^-$. Left charge is $2(-1)+(-1)=-3$; right charge is $-1$. Adding $2\text{OH}^-$ to the right matches both sides: $2\text{MnO}_4^-(aq) + \text{Br}^-(aq) \rightarrow 2\text{MnO}_2(s) + \text{BrO}_3^-(aq) + 2\text{OH}^-(aq)$.

Step 4. Balance hydrogen with water. The right side has $2$ hydrogens and the left has none, so one $\text{H}_2\text{O}$ goes on the left: $2\text{MnO}_4^-(aq) + \text{Br}^-(aq) + \text{H}_2\text{O}(l) \rightarrow 2\text{MnO}_2(s) + \text{BrO}_3^-(aq) + 2\text{OH}^-(aq)$. Oxygen checks out at $9$ on each side. Answer: 2MnO4−(aq) + Br−(aq) + H2O(l) to 2MnO2(s) + BrO3−(aq) + 2OH−(aq), fully balanced for atoms and charge.

Competitive level (JEE / NEET) - limiting-reagent stoichiometry in the Ostwald process

In the Ostwald process for manufacturing nitric acid, the first step oxidises ammonia gas with oxygen gas to give nitric oxide gas and steam, $4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)$. Starting with $10.00\ \text{g}$ of ammonia and $20.00\ \text{g}$ of oxygen, find the maximum mass of nitric oxide obtainable (extending NCERT Exercise 7.25; molar masses $\text{NH}_3 = 17.0\ \text{g/mol}$, $\text{O}_2 = 32.0\ \text{g/mol}$, $\text{NO} = 30.0\ \text{g/mol}$).

Step 1. Convert both masses to moles. Moles of $\text{NH}_3 = 10.00 / 17.0 = 0.588\ \text{mol}$. Moles of $\text{O}_2 = 20.00 / 32.0 = 0.625\ \text{mol}$.

Step 2. Identify the limiting reagent using the stoichiometric ratio $4\text{NH}_3 : 5\text{O}_2$. The oxygen required to consume all the ammonia is $0.588 \times (5/4) = 0.735\ \text{mol}$, but only $0.625\ \text{mol}$ is available, so oxygen is insufficient and limiting (confirmed since the ammonia needed for $0.625\ \text{mol}$ of $\text{O}_2$ is only $0.500\ \text{mol}$, less than the $0.588\ \text{mol}$ present).

Step 3. Compute nitric oxide from the limiting reagent, oxygen, using the ratio $5\text{O}_2 : 4\text{NO}$. Moles of $\text{NO} = 0.625 \times (4/5) = 0.500\ \text{mol}$.

Step 4. Convert to mass: mass of $\text{NO} = 0.500 \times 30.0 = 15.0\ \text{g}$. Answer: oxygen is the limiting reagent, and the maximum mass of nitric oxide obtainable is 15.0 g. Timing note: about ninety seconds. The fast route is to convert both masses to moles first, then compare the mole ratio actually available (0.588 : 0.625, roughly 0.94 : 1) against the required stoichiometric ratio (4 : 5, exactly 0.80 : 1); since the available ammonia-to-oxygen ratio is larger than required, oxygen runs out first. Do not skip this comparison step and assume the reagent with the smaller mass or smaller mole count is automatically limiting, mole count alone is a trap here, since ammonia has fewer grams but more moles relative to what stoichiometry demands.

Mixing up the oxidising agent with the species that is oxidised. The oxidising agent is the reagent that itself gets reduced while it oxidises something else; students commonly write "the oxidising agent is oxidised," which is backwards. Ask: does this species's own oxidation number go up or down? Down means oxidant (reduced); up means reductant (oxidised).

Forgetting the two exceptions for oxygen's oxidation number. Oxygen is not always $-2$; in peroxides such as $\text{H}_2\text{O}_2$ it is $-1$, in superoxides such as $\text{KO}_2$ it is $-\tfrac{1}{2}$, and in $\text{OF}_2$ or $\text{O}_2\text{F}_2$ it turns positive. Applying $-2$ blindly to $\text{H}_2\text{O}_2$ gives an impossible oxidation number for hydrogen. Similarly, a fraction like $+\tfrac{4}{3}$ for carbon in $\text{C}_3\text{O}_2$ is only an average of whole-number states borne by chemically distinct atoms, never a physically real fractional charge.

Assuming every decomposition or combination reaction is redox. $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ is decomposition but not redox, since no oxidation number changes; only decompositions and combinations involving at least one elemental species are guaranteed redox.

Forgetting charge balance, and balancing in the wrong medium. A common error in half-reaction balancing is adding water to fix oxygen and hydrogen while leaving ionic charges on the two sides unequal; charge must be balanced with electrons, and the net charge must match on both sides of the final equation. A related slip is using $\text{H}^+$ to balance an equation explicitly stated to occur in basic solution (or vice versa); always convert with the extra $\text{OH}^-$ (or $\text{H}^+$) step shown in the derivation above, since $\text{H}^+$ and $\text{OH}^-$ cannot coexist in appreciable amounts.

Reading the electrochemical series upside down. A more negative $E^{\circ}$ means a stronger reducing agent (the metal itself is easily oxidised), not a stronger oxidising agent; students frequently reverse this and predict the wrong direction for a displacement reaction.

• State the classical, electron-transfer and oxidation-number definitions of oxidation and reduction, and translate freely between them for any given reaction.
• Apply all six rules to assign the oxidation number of any atom in a neutral compound or a polyatomic ion, including the peroxide, superoxide and oxygen-fluoride exceptions.
• Explain why a fractional oxidation number is only an average, and identify the whole-number states hidden inside species like C3O2 or S4O6 two minus.
• Classify any given redox reaction as combination, decomposition, displacement or disproportionation, and correctly identify when a combination or decomposition is not redox at all.
• Balance a redox equation in acidic or basic medium using either the oxidation number method or the half reaction (ion-electron) method, checking both atom balance and charge balance at the end.
• Identify the oxidising agent and reducing agent in any redox equation without confusing an agent's name with its own fate.
• Read a table of standard electrode potentials to rank oxidising and reducing strength, and predict whether a given displacement or redox reaction is spontaneous.
• Describe how the Daniell cell physically separates oxidation and reduction, and define a redox couple and a standard electrode potential precisely.
• Explain how self-indicating, externally-indicated and iodometric redox titrations each locate their end point.

Ask the tutor: if you would like to work through more balancing practice, such as chlorine reacting with hot alkali or the disproportionation of Mn3+, ask the tutor to walk you through it step by step using the same oxidation number and half reaction methods from this chapter.