Part 1 · Chapter 8

Redox Reactions

The give and take of electrons — oxidation numbers, oxidising and reducing agents, balancing in acid and base, and the equivalents behind every titration

Fundamentals of Chemistry Prof. Mithun Mondal Reading time ≈ 44 min
i What you'll learn
  • Oxidation and reduction by the oxygen/hydrogen and the modern electron-transfer pictures.
  • The rules for assigning an oxidation number and reading change from it.
  • How to spot the oxidising agent and reducing agent in any reaction.
  • The five types of redox change, and disproportionation vs comproportionation.
  • Balancing redox equations by the oxidation-number and ion-electron methods in acid and base.
  • The n-factor, equivalent weight, and the law of equivalence behind redox titrations.
Section 8-1

Two Ways to See Redox

The word redox is a contraction of reduction and oxidation — two halves of a single event that always occur together. The classical picture talks of oxygen and hydrogen: oxidation is gain of oxygen or loss of hydrogen; reduction is loss of oxygen or gain of hydrogen. The modern and far more powerful picture is purely about electrons.

Oxidation is loss of electrons; reduction is gain of electrons. The two are inseparable — electrons lost by one species are gained by another, so a redox reaction is simply an electron transfer. The mnemonics OIL RIG (Oxidation Is Loss, Reduction Is Gain) and LEO the lion says GER (Lose Electrons = Oxidation, Gain Electrons = Reduction) capture it.

reducing agent gets oxidised oxidising agent gets reduced e⁻ electrons flow; both halves happen at once
Redox is one electron transfer seen from two sides
A redox reaction split into half-reactions
\[ \underset{\text{oxidation}}{\ce{Zn -> Zn^2+ + 2e^-}}\qquad\quad \underset{\text{reduction}}{\ce{Cu^2+ + 2e^- -> Cu}} \]
Adding the halves gives \(\ce{Zn + Cu^2+ -> Zn^2+ + Cu}\): zinc loses electrons, copper gains them.
Section 8-2

Oxidation Number

To track electrons without drawing every bond, chemists use the oxidation number (or oxidation state) — the charge an atom would carry if every bond were treated as fully ionic. A rise in oxidation number is oxidation; a fall is reduction. The rules below are applied in order of priority.

RuleOxidation number
Free element (\(\ce{O2},\ \ce{Na},\ \ce{P4}\))\(0\)
Monatomic ionequal to its charge
Group 1 / Group 2 metals\(+1\ /\ +2\)
Fluorine in compounds\(-1\) always
Hydrogen\(+1\) (but \(-1\) in metal hydrides, e.g. \(\ce{NaH}\))
Oxygen\(-2\) (but \(-1\) in peroxides, \(+2\) in \(\ce{OF2}\))
Sum over a speciesequals its overall charge
Apply rules top-down. When rules conflict, the higher one wins. In \(\ce{H2O2}\), oxygen is \(-1\) (peroxide), not \(-2\); in \(\ce{NaH}\), hydrogen is \(-1\) because the Group-1 rule fixes \(\ce{Na}=+1\) first.
Reading an oxidation state
\[ \ce{KMnO4}:\ (+1)+x+4(-2)=0\ \Rightarrow\ x_{\ce{Mn}}=+7 \]
The same balance gives \(\ce{Cr}=+6\) in \(\ce{K2Cr2O7}\) and \(\ce{S}=+6\) in \(\ce{H2SO4}\).
Section 8-3

Oxidising & Reducing Agents

The species that causes oxidation is the oxidising agent — and to do so it must accept electrons, so it is itself reduced. The species that causes reduction is the reducing agent, which donates electrons and is itself oxidised. The roles always cross.

SpeciesDoes what?Happens to it
Oxidising agentremoves electrons from othersis reduced (gains e⁻)
Reducing agentsupplies electrons to othersis oxidised (loses e⁻)
Where to look. Elements in their highest oxidation state (\(\ce{KMnO4},\ \ce{K2Cr2O7},\ \ce{HNO3}\)) tend to be strong oxidisers; those in their lowest state (\(\ce{H2S},\ \ce{Na},\ \ce{C2O4^2-}\)) tend to be reducers. An element in an intermediate state can act either way.
Section 8-4

Types of Redox Reactions

Most familiar reactions sort into a handful of redox families. Recognising the family often tells you what to expect of the products.

TypeExample
Combination\(\ce{C + O2 -> CO2}\)
Decomposition\(\ce{2KClO3 -> 2KCl + 3O2}\)
Displacement (metal)\(\ce{Zn + CuSO4 -> ZnSO4 + Cu}\)
Displacement (non-metal)\(\ce{Cl2 + 2KBr -> 2KCl + Br2}\)
Disproportionation\(\ce{2H2O2 -> 2H2O + O2}\)
Section 8-5

Disproportionation & Comproportionation

In a disproportionation, a single element in one intermediate oxidation state is simultaneously oxidised and reduced — part of it goes up, part goes down. The element must have an accessible higher and lower state for this to occur.

Cl₂ (0) Cl⁻ (−1) ↓ reduced ClO⁻ (+1) ↑ oxidised
Disproportionation: one state splits into a higher and a lower one
Chlorine in cold dilute alkali
\[ \ce{Cl2 + 2OH^- -> Cl^- + ClO^- + H2O} \]
Chlorine \((0)\) becomes \(\ce{Cl-}\,(-1)\) and \(\ce{ClO-}\,(+1)\) at once.

The reverse process, comproportionation, takes two different oxidation states of the same element and merges them into one intermediate state — for example \(\ce{2H2S + SO2 -> 3S + 2H2O}\), where sulphur at \(-2\) and \(+4\) both meet at \(0\).

Section 8-6

Balancing: Oxidation-Number Method

The first balancing technique bookkeeps the change in oxidation number. The total increase (oxidation) must equal the total decrease (reduction), because the electrons lost equal the electrons gained.

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Oxidation-number method — the steps
Assign states → find ↑ and ↓ per atom → equalise total change → balance the rest

After matching the electron change with suitable coefficients, balance remaining atoms by inspection, then balance \(\ce{O}\) with \(\ce{H2O}\) and \(\ce{H}\) with \(\ce{H+}\) (acidic) or \(\ce{OH-}\) (basic).

Section 8-7

Balancing: Ion-Electron (Half-Reaction) Method

The ion-electron method splits the reaction into an oxidation half and a reduction half, balances each completely, then recombines them. It is the cleaner route for ionic equations and the one that mirrors what happens in a cell.

StepAcidic mediumBasic medium
1. SplitWrite the oxidation and reduction half-reactions
2. Balance atoms (not O, H)Balance the main element first
3. Balance Oadd \(\ce{H2O}\)add \(\ce{H2O}\)
4. Balance Hadd \(\ce{H+}\)add \(\ce{H+}\), then neutralise with \(\ce{OH-}\)
5. Balance chargeadd electrons to the deficient side
6. Combinescale so electrons cancel, then add
Permanganate oxidising oxalate (acidic)
\[ \ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O}\qquad \ce{C2O4^2- -> 2CO2 + 2e^-} \]
\[ \ce{2MnO4^- + 16H+ + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2} \]
Cross-multiply to make 10 electrons on each side, then add — charge and atoms now balance.
Section 8-8

Equivalents & the n-Factor

Redox stoichiometry is often handled with equivalents rather than moles, because in a titration the equivalents of oxidant and reductant are always equal. The bridge is the n-factor — the number of electrons exchanged per formula unit.

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Equivalent weight and the law of equivalence
\(\text{Eq. wt}=\dfrac{\text{molar mass}}{n\text{-factor}},\qquad N_1V_1=N_2V_2\)

Here the n-factor is the electrons gained or lost per formula unit. Equivalents of oxidising agent equal equivalents of reducing agent at the end point.

Oxidant / reductantChangen-factorEq. wt
\(\ce{KMnO4}\) (acidic)\(+7 \to +2\)5\(158/5=31.6\)
\(\ce{KMnO4}\) (neutral)\(+7 \to +4\)3\(158/3=52.7\)
\(\ce{KMnO4}\) (strongly basic)\(+7 \to +6\)1\(158\)
\(\ce{K2Cr2O7}\) (acidic)\(+6 \to +3\) (×2 Cr)6\(294/6=49\)
\(\ce{H2C2O4}\) (reductant)\(+3 \to +4\) (×2 C)2\(90/2=45\)
Section 8-9

Redox Titrations

A redox titration uses an oxidant–reductant pair to find an unknown concentration. Permanganometry uses \(\ce{KMnO4}\), which is self-indicating — the first faint pink persists once the reductant is used up. Iodometry/iodimetry uses the \(\ce{I2}/\ce{I-}\) couple with starch as indicator. Dichromatometry uses \(\ce{K2Cr2O7}\) with a redox indicator such as diphenylamine.

One line does the arithmetic. At the end point, \((NV)_{\text{oxidant}}=(NV)_{\text{reductant}}\), or equivalently \(\dfrac{n_1}{x_1}=\dfrac{n_2}{x_2}\) in moles weighted by n-factors. Decide each n-factor from the medium before you compute.
Section 8-10

A Glimpse of Electrode Potential

Why does zinc reduce \(\ce{Cu^2+}\) but not the reverse? Each half-reaction has a tendency to gain electrons measured by its standard electrode potential \(E^\circ\), referenced to the standard hydrogen electrode at \(0\ \text{V}\). Arranged by \(E^\circ\), these form the electrochemical series.

🔋
Which way will it go?
A species with higher \(E^\circ\) is the better oxidising agent and is reduced; \(E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}\)

A positive \(E^\circ_{\text{cell}}\) means the reaction is spontaneous as written. The full machinery — Nernst equation, cells, and the link \(\Delta G^\circ=-nFE^\circ\) — is developed in Chapter 10, Electrochemistry.

Worked Examples

Putting It to Work

1 Oxidation number

Problem. Find the oxidation number of \(\ce{Cr}\) in \(\ce{K2Cr2O7}\) and of \(\ce{Mn}\) in \(\ce{MnO4^-}\).

Solution. Set the sum equal to the charge:

Working
\[ 2(+1)+2x+7(-2)=0\ \Rightarrow\ x_{\ce{Cr}}=+6;\qquad y+4(-2)=-1\ \Rightarrow\ y_{\ce{Mn}}=+7 \]
2 Identify the agents

Problem. In \(\ce{Zn + 2HCl -> ZnCl2 + H2}\), name the species oxidised and reduced and the two agents.

Solution. Track oxidation numbers: \(\ce{Zn}\) goes \(0 \to +2\) (oxidised), \(\ce{H}\) goes \(+1 \to 0\) (reduced).

Working
\[ \ce{Zn}\ \text{= reducing agent},\qquad \ce{HCl}\,(\ce{H+})\ \text{= oxidising agent} \]
3 Classify the reaction

Problem. Classify \(\ce{2H2O2 -> 2H2O + O2}\) and state the oxidation-state changes.

Solution. Oxygen in \(\ce{H2O2}\) is \(-1\); part goes to \(-2\) (in \(\ce{H2O}\)) and part to \(0\) (in \(\ce{O2}\)).

Working
\[ -1 \to -2\ (\text{reduced}),\quad -1 \to 0\ (\text{oxidised})\ \Rightarrow\ \textbf{disproportionation} \]
4 Balance in acidic medium

Problem. Balance \(\ce{MnO4^- + Fe^2+ -> Mn^2+ + Fe^3+}\) in acid by the ion-electron method.

Solution. Build each half, then equalise electrons (5 from \(\ce{MnO4-}\), 1 from each \(\ce{Fe^2+}\)):

Working
\[ \ce{MnO4^- + 8H+ + 5Fe^2+ -> Mn^2+ + 5Fe^3+ + 4H2O} \]
5 Balance in basic medium

Problem. Balance \(\ce{MnO4^- + I^- -> MnO2 + I2}\) in basic medium.

Solution. Reduction: \(\ce{MnO4- + 2H2O + 3e^- -> MnO2 + 4OH-}\); oxidation: \(\ce{2I^- -> I2 + 2e^-}\). Scale ×2 and ×3 for 6 electrons:

Working
\[ \ce{2MnO4^- + 4H2O + 6I^- -> 2MnO2 + 8OH^- + 3I2} \]
6 Equivalents in a titration

Problem. What volume of \(0.02\ \text{M}\ \ce{KMnO4}\) (acidic) is needed to oxidise \(20\ \text{mL}\) of \(0.10\ \text{M}\) oxalic acid?

Solution. n-factors: \(\ce{KMnO4}=5\), \(\ce{H2C2O4}=2\). Equate equivalents \((N=M\times n\text{-factor})\):

Working
\[ (0.02\times5)\,V = (0.10\times2)\times20 \ \Rightarrow\ 0.1\,V = 4 \ \Rightarrow\ V = 40\ \text{mL} \]
Review

Chapter Summary

Definition

Oxidation = loss of electrons (state ↑); reduction = gain (state ↓); they always pair up.

Oxidation number

Assign by priority rules; F \(=-1\), O \(=-2\) (except peroxides), H \(=+1\) (except hydrides).

Agents

Oxidising agent is reduced; reducing agent is oxidised — the roles always cross.

Disproportionation

One intermediate state goes both up and down; comproportionation is the reverse.

Balancing

Ion-electron method: balance atoms, O with \(\ce{H2O}\), H with \(\ce{H+}\)/\(\ce{OH-}\), charge with e⁻.

Equivalents

\(\text{Eq. wt}=\text{M}/n\); \(N_1V_1=N_2V_2\) at the end point of a redox titration.

Practice

Problems

Assign oxidation numbers first, then decide which species moves up and which moves down. For titration questions, fix every n-factor from the medium before calculating. Difficulty rises down the list.

  1. Find the oxidation number of \(\ce{S}\) in \(\ce{H2SO4},\ \ce{Na2S2O3}\) and \(\ce{SO4^2-}\).
  2. In \(\ce{Cl2 + 2KI -> 2KCl + I2}\), name the species oxidised, reduced, and the two agents.
  3. Classify each: \(\ce{C + O2 -> CO2}\); \(\ce{2KClO3 -> 2KCl + 3O2}\); \(\ce{Cl2 + 2NaOH -> NaCl + NaOCl + H2O}\).
  4. Give the n-factor and equivalent weight of \(\ce{K2Cr2O7}\) acting as an oxidant in acid \((M=294)\).
  5. Balance in acidic medium: \(\ce{Cr2O7^2- + Fe^2+ -> Cr^3+ + Fe^3+}\).
  6. Balance in basic medium: \(\ce{Cl2 -> Cl^- + ClO3^-}\) (a disproportionation).
  7. Show that \(\ce{2H2S + SO2 -> 3S + 2H2O}\) is a comproportionation, naming the states involved.
  8. Balance in acidic medium: \(\ce{MnO4^- + C2O4^2- -> Mn^2+ + CO2}\).
  9. \(25\ \text{mL}\) of \(\ce{FeSO4}\) needs \(20\ \text{mL}\) of \(0.02\ \text{M}\ \ce{KMnO4}\) (acidic) to reach the end point. Find the molarity of \(\ce{FeSO4}\).
  10. Explain why \(\ce{KMnO4}\) has different equivalent weights in acidic, neutral and strongly basic media.
  11. Using \(E^\circ_{\ce{Cu^2+/Cu}}=+0.34\ \text{V}\) and \(E^\circ_{\ce{Zn^2+/Zn}}=-0.76\ \text{V}\), find \(E^\circ_{\text{cell}}\) for the Daniell cell and state whether it is spontaneous.
  12. A sample of \(\ce{H2O2}\) is labelled "10 volume." Outline how a permanganate titration in acid would let you verify its strength, naming the n-factor of \(\ce{H2O2}\) as a reductant.
Tip: every redox problem is electron bookkeeping. Assign oxidation numbers, separate the rise from the fall, make electrons lost equal electrons gained, and only then balance \(\ce{O}\) and \(\ce{H}\). For titrations, the n-factor — set by the medium — is the single number that turns moles into equivalents.