Ch 1 · Redox processes: electron transfer and changes in oxidation number · AS & A Level Chemistry 9701
Topic 1 of 1 in Electrochemistry — AS & A Level Chemistry 9701 — 58 lessons.
Introduction
Redox chemistry underlies electrochemical cells, extraction of metals and much of inorganic and organic chemistry. Cambridge 9701 expects two equivalent descriptions of the same process: electron transfer, and change in oxidation number. Fluency in assigning oxidation numbers lets you identify what is oxidised and what is reduced in any equation, name compounds unambiguously with Roman numerals, and balance redox equations systematically.
Key ideas — Rules for assigning oxidation numbers
The oxidation number of an atom is the charge it would carry if the compound were fully ionic. Standard rules, applied in order of priority: the oxidation number of an uncombined element is 0; the oxidation number of a simple ion equals its charge; the sum of oxidation numbers in a neutral compound is 0, and in an ion equals the ion's overall charge; Group 1 metals are always +1 and Group 2 metals always +2 in compounds; hydrogen is +1 except in metal hydrides, where it is −1; oxygen is −2 except in peroxides (−1) and in OF_2 (+2); fluorine is always −1. Any other element is found by making the oxidation numbers in the species sum to the correct total. For example, in mathrmCr_2O_7^2-, oxygen contributes 7 × (-2) = -14, so the two chromium atoms must total -2 - (-14) = +12, giving chromium an oxidation number of +6.
🧑🏫 Teacher's explanation
Think of oxidation numbers as a 'bookkeeping' system for electrons. In a real chemical bond, electrons might be shared or held tightly, but we pretend they are fully transferred to one side or the other to make the math easy. It is like tracking a debt: if you owe someone \$10 but you have \$15, your net balance is +\$5. Oxidation number is that net balance.
📖 Definitions
🔍 Worked example — Easy
Calculate the oxidation number of Sulfur (S) in H_2SO_4.
🔍 Worked example — Standard
Determine the oxidation number of Manganese (Mn) in the ion MnO_4^-.
🔍 Worked example — Exam-style
In the reaction 3Cl_2 + 6OH^- arrow 5Cl^- + ClO_3^- + 3H_2O, calculate the oxidation number of Chlorine (Cl) in ClO_3^-.
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🎯 Exam question
Key ideas — Oxidation, reduction and redox
In terms of electron transfer, oxidation is the loss of electrons and reduction is the gain of electrons; a mnemonic is OIL RIG (Oxidation Is Loss, Reduction Is Gain). In terms of oxidation number, oxidation is an increase in oxidation number and reduction is a decrease in oxidation number. A redox reaction is one in which both oxidation and reduction occur simultaneously, since electrons lost by one species must be gained by another; the two half-reactions can be written separately, each showing electrons explicitly, for example:
🧑🏫 Teacher's explanation
Think of redox reactions like a financial transaction involving electrons. If you give away money, your balance decreases (oxidation); if you receive money, your balance increases (reduction). In chemistry, electrons are the 'currency'. When an atom loses an electron, it becomes more positive because it has fewer negative charges to balance its protons. When it gains an electron, it becomes more negative.
📖 Definitions
🔍 Worked example — Easy
Determine the oxidation number of Sulfur in H_2SO_4.
🔍 Worked example — Standard
In the reaction CuO(s) + H_2(g) ightarrow Cu(s) + H_2O(g), identify the oxidising agent and the species being reduced.
🔍 Worked example — Exam-style
Balance the following redox half-equation in acidic conditions: MnO_4^-(aq) ightarrow Mn^2+(aq).
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🎯 Exam question
Key ideas — Oxidising agents and reducing agents
An oxidising agent is a species that oxidises another species by accepting electrons from it; in doing so, the oxidising agent itself is reduced, and its own oxidation number decreases. A reducing agent is a species that reduces another species by donating electrons to it; the reducing agent itself is oxidised, and its own oxidation number increases. In the equations above, MnO_4^- is the oxidising agent (manganese falls from +7 to +2) and mathrmFe^2+ is the reducing agent (iron rises from +2 to +3).
🧑🏫 Teacher's explanation
Think of electrons like money in a transaction. In a redox reaction, electrons are being 'spent' or 'transferred' between particles. If a particle is 'greedy' and takes electrons from others, it is an oxidising agent. Because it gained negative charge, its oxidation number drops. If a particle is 'generous' and gives electrons away, it is a reducing agent. Because it lost negative charge, its oxidation number rises.
📖 Definitions
🔍 Worked example — Easy
In the reaction 2Fe^2+ + MnO_4^- + 8H^+ arrow 2Fe^3+ + Mn^2+ + 4H_2O, identify the oxidising agent and the reducing agent.
🔍 Worked example — Standard
Calculate the oxidation number of Sulfur in the sulfate ion, SO_4^2-.
🔍 Worked example — Exam-style
In the reaction Cl_2 + 2OH^- arrow Cl^- + ClO^- + H_2O, identify the type of redox reaction and the oxidation state change for Chlorine.
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🎯 Exam question
Key ideas — Disproportionation
Disproportionation is a redox reaction in which atoms of the same element, in the same species, are simultaneously oxidised and reduced, so that one portion of the element ends up with a higher oxidation number and another portion with a lower oxidation number. For example, when chlorine reacts with cold, dilute sodium hydroxide:
🧑🏫 Teacher's explanation
Imagine you are at a party where a group of friends decides to split up. Instead of everyone staying the same, half the group decides to become 'super-energetic' (losing energy/electrons) and the other half becomes 'super-relaxed' (gaining energy/electrons). Even though they all started as the same type of person, they have split into two different states.
📖 Definitions
🔍 Worked example — Easy
In the reaction ceCl2(aq) + 2NaOH(aq) arrow NaCl(aq) + NaOCl(aq) + H2O(l), determine the change in oxidation number for Chlorine.
🔍 Worked example — Standard
A species X undergoes disproportionation to form X^2+ and X^4+. If the initial oxidation state of X was 0, calculate the total change in oxidation numbers for the products.
🔍 Worked example — Exam-style
Consider the reaction: ceP_4(s) + 3NaOH(aq) + 3H_2O(l) arrow PH_3(g) + NaH_2PO_2(aq). Determine the oxidation states of Phosphorus in the reactants and products.
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Key ideas — Roman numeral (stock) notation
A Roman numeral in brackets after an element's name specifies its oxidation number in that compound, which is essential where an element commonly shows more than one oxidation state. For example, iron(II) chloride, FeCl_2, contains mathrmFe^2+, while iron(III) chloride, FeCl_3, contains mathrmFe^3+; sodium chlorate(I), NaOCl, contains chlorine at +1, while sodium chlorate(V), NaClO_3, contains chlorine at +5. Roman numerals are not written for elements that show only one common oxidation state, such as sodium or oxygen.
🧑🏫 Teacher's explanation
Think of oxidation numbers as a 'bookkeeping' system for electrons. In a molecule, atoms aren't always neutral; they 'share' electrons unevenly. The oxidation number is a way of tracking where those electrons seem to be 'hanging out' at any given moment. It is like tracking the debt or credit of an atom in a business transaction.
📖 Definitions
🔍 Worked example — Easy
Determine the oxidation number of sulfur in H_2SO_4.
🔍 Worked example — Standard
Write the correct name for Fe_2O_3 using Roman numeral notation.
🔍 Worked example — Exam-style
Calculate the oxidation number of Manganese in KMnO_4.
⚠️ Common errors
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🎯 Exam question
Worked example
Question. Manganate(VII) ions oxidise iron(II) ions in acidic solution. Use changes in oxidation number to deduce the balanced ionic equation for the reaction between MnO_4^- and mathrmFe^2+ in the presence of H^+, given that manganese is reduced from +7 to +2 and the products include mathrmMn^2+ and H_2O. [4]
🔍 Worked example — Easy
Determine the oxidation number of Sulfur in the sulfate ion, SO_4^2-.
🔍 Worked example — Standard
In the reaction a extMnO_4^- + b extFe^2+ + c extH^+ ightarrow d extMn^2+ + e extFe^3+ + f extH_2 extO, find the stoichiometric coefficients a, b, c, d, e, f using the electron transfer method.
🔍 Worked example — Exam-style
A reaction involves the disproportionation of Chlorine gas in hot, concentrated alkali: $ ext{Cl}_2 + 2 ext{OH}^- ightarrow ext{Cl}^- + ext{ClO}_3^- + ext{H}_2 ext{O}$. Calculate the change in oxidation state for Chlorine in both products.
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Practice
Answers