Redox Reactions · Lesson 7 of 8

Redox in Titrations and Electrochemical Cells

“Connect redox stoichiometry with colour endpoints, electrode reactions, and electrical energy.”

Learning Objectives

• Explain how redox titrations measure an oxidant or reductant. • Distinguish self-indicators, redox indicators, and the iodine–starch endpoint. • Describe redox couples and the operation of a Daniell cell. • Track electrons in wires and ions in the salt bridge, distinguishing electron flow from conventional current. • Use standard reduction potentials to compare oxidants, reductants, and reaction feasibility. • Recognise how active and inert electrodes affect simple electrolysis products.

Electron transfer becomes a measurement

A balanced redox equation tells us how much oxidant reacts with a given amount of reductant. If the end of their reaction can be detected, this ratio lets us determine an unknown concentration. A redox titration uses this idea: one solution of known concentration is added carefully to another until the reacting amount is reached. A colour change often provides the signal.

Definition

Equivalence point

The point at which oxidant and reductant have been mixed in the exact proportions required by the balanced reaction. It is not necessarily an equal number of moles of the two substances.

Definition

Endpoint

The observed signal used to stop a titration, such as a persistent colour change. A suitable signal occurs close to the equivalence point.

The colour does not replace the stoichiometry. It tells us when to use the measured volume in the balanced equation. An endpoint that needs only a very small excess of reagent makes the measured result close to the true equivalence point.

Three ways to see an endpoint

Some oxidants are themselves strongly coloured. Others need a separate substance whose colour changes when it is oxidised or reduced. A third useful method first liberates iodine and then measures that iodine using thiosulphate. Each method detects a chemical change through a different visible signal.

Method	What the student observes	Chemical explanation
Permanganate as self-indicator	A first faint pink colour remains after swirling.	Before the endpoint, added permanganate is consumed by reductant. The first slight excess persists once the reductant is exhausted.
Dichromate with diphenylamine	The indicator develops an intense blue colour near the endpoint.	Dichromate is not used as its own endpoint signal in this procedure. A small oxidant excess changes the redox-sensitive indicator.
Iodine with starch and thiosulphate	Blue iodine–starch colour disappears as iodine is consumed.	Thiosulphate reduces iodine to iodide. Once the iodine is gone, the blue signal is lost.

For acidic permanganate reacting with iron(II), the reduction product is $M n^{2 +}$ . A very small persistent excess of the purple reagent is visible as a faint pink colour, detectable at a permanganate concentration of roughly $1 0^{- 6} mol L^{- 1}$ . Do not look for a strongly purple endpoint: a large colour excess means more reagent has been added than the stoichiometric reaction requires. The same self-indicator principle applies to a suitable oxalate titration.

In an iodine-based determination, copper(II) can oxidise iodide: $2C u^{2 +} + 4 I^{-} \to 2CuI (s) + I_{2}$ . Copper(I) iodide precipitates, helping the reaction proceed. Iodine is held in solution in excess iodide mainly as triiodide, $I_{3}^{-}$ . It can then be titrated with thiosulphate.

Iodine reacts with thiosulphateLaTeX

I_{2} + 2 S_{2} O_{3}^{2 -} \to 2 I^{-} + S_{4} O_{6}^{2 -}

One mole of iodine consumes two moles of thiosulphate. Sulphur is oxidised overall while iodine is reduced.

Starch forms the familiar intense blue colour with iodine in the iodide-containing solution. In a careful titration it is usually added near the endpoint, when the iodine colour has weakened, to make the final disappearance easier to detect. The logical sequence is liberation of iodine, measurement of that iodine, and use of the balanced ratios to infer the original oxidant amount.

For a concentration calculation, let $C$ be the amount concentration in moles per litre, $V$ the solution volume in litres, and $n$ the amount in moles. Their relationship is $n = C V$ . First use this relationship for the measured reagent, then apply the balanced mole ratio, and finally divide the unknown amount by its own sample volume.

Example — A redox titration calculation

Problem
A $20.0 mL$ iron(II) sample requires $15.0 mL$ of $0.0200 mol L^{- 1}$ acidic permanganate. Find the iron(II) concentration.

1.First balance the reacting ions: $Mn O_{4}^{-} + 8 H^{+} + 5F e^{2 +} \to M n^{2 +} + 5F e^{3 +} + 4 H_{2} O$ . One permanganate ion consumes five iron(II) ions.
2.Convert the titrant volume to litres: $15.0 mL = 0.0150 L$ . Its amount is $n = C V = 0.0200 \times 0.0150 = 3.00 \times 1 0^{- 4} mol$ .
3.The iron(II) amount is $5 (3.00 \times 1 0^{- 4}) = 1.50 \times 1 0^{- 3} mol$ .
4.Divide by sample volume: $C = (1.50 \times 1 0^{- 3}) /0.0200 = 0.0750 mol L^{- 1}$ . The five-to-one ratio, not a one-to-one assumption, controls the result.

Separate the two halves to obtain electrical energy

Zinc placed directly in copper sulphate transfers electrons to copper ions at the reacting surfaces, and heat is released. If we arrange the same partners in separate compartments and provide an external wire, the electrons can travel through the wire instead. The reaction can then supply electrical energy to an external circuit.

Definition

Redox couple

The oxidised and reduced forms of a substance participating in a half reaction, conventionally written with the oxidised form first, such as zinc(II)/zinc or copper(II)/copper.

The couples $Z n^{2 +} /Zn$ and $C u^{2 +} /Cu$ each involve a metal and its ions at an electrode–solution interface. A Daniell cell places a zinc electrode in zinc sulphate and a copper electrode in copper sulphate. The electrodes are connected by a wire, and the solutions by a salt bridge. A complete circuit needs both paths.

A couple can also be written with a vertical line, such as $Z n^{2 +} ∣Zn$ , to mark a phase boundary. The oxidised form is named before the reduced form in the couple. This notation identifies the two forms; whether oxidation or reduction actually occurs depends on how the half reaction is used in the complete cell.

Daniell cell: two paths complete one circuit— Electrons travel from zinc to copper through the wire. Ions move through the salt bridge to prevent charge build-up in the solutions.

At the zinc electrode, oxidation releases electrons and adds positive zinc ions to solution. At the copper electrode, reduction consumes electrons and removes positive copper ions from solution. Without compensating ion movement, charge would build up and stop the sustained current. The salt bridge allows ions to migrate while limiting bulk mixing of the solutions.

Anode reactionLaTeX

Zn (s) \to Z n^{2 +} (a q) + 2 e^{-}

Oxidation occurs at the anode. In this galvanic cell, the zinc anode is negative.

Cathode reactionLaTeX

C u^{2 +} (a q) + 2 e^{-} \to Cu (s)

Reduction occurs at the cathode. In this galvanic cell, the copper cathode is positive.

A salt bridge may contain an inert electrolyte held in a gel, for example potassium chloride or ammonium nitrate when compatible with the solutions. Anions compensate the accumulating positive charge in the zinc compartment; cations compensate the removal of copper ions in the copper compartment. Electrons travel through the external metal circuit, not through the salt bridge. Conventional current in the wire is opposite to electron flow.

With the circuit open, the cell does not sustain a current or the overall transfer reaction. Closing the circuit permits the paired electrode processes and ion migration; an ammeter in the external wire can detect the resulting current. Oxidation at an electrode is called an anode process and reduction a cathode process; these definitions remain useful beyond this particular cell.

Electrode potentials compare redox tendencies

An electrode has a potential associated with its redox couple. Only a potential difference is directly measured, so electrode potentials are quoted relative to a reference. By convention, the standard hydrogen electrode has a standard potential of zero volts. A table of standard reduction potentials then allows different couples to be compared on one scale.

For the standard comparison used here, temperature is $298 K$ and dissolved species are represented at standard activity, often approximated by $1 mol L^{- 1}$ for an introductory treatment. The thermodynamic gas reference pressure is $1 bar$ ; introductory descriptions sometimes use one atmosphere as an approximation. The key requirement is to use the stated reference conditions rather than treat a standard potential as the potential at arbitrary concentrations and pressures.

Definition

Standard reduction potential

The electrode potential for a reduction half reaction under standard conditions, relative to the standard hydrogen electrode. It is written as $E^{\circ}$ and measured in volts.

The table writes every half reaction in the reduction direction. A more positive value means a greater tendency for the oxidised form on the left to accept electrons under the stated conditions. The reduced form on the right is then a weaker electron donor. Conversely, a very negative reduction potential identifies a reduced form that is a strong reducing agent compared with hydrogen.

Reduction half reaction at 298 K	Standard potential / V
$F_{2} (g) + 2 e^{-} \to 2 F^{-}$	+2.87
$C o^{3 +} + e^{-} \to C o^{2 +}$	+1.81
$H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O$	+1.78
$Mn O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$	+1.51
$A u^{3 +} + 3 e^{-} \to Au (s)$	+1.40
$C l_{2} (g) + 2 e^{-} \to 2 C l^{-}$	+1.36
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$	+1.33
$O_{2} (g) + 4 H^{+} + 4 e^{-} \to 2 H_{2} O$	+1.23
$Mn O_{2} (s) + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O$	+1.23
$B r_{2} + 2 e^{-} \to 2 B r^{-}$	+1.09
$N O_{3}^{-} + 4 H^{+} + 3 e^{-} \to NO (g) + 2 H_{2} O$	+0.97
$2 H g^{2 +} + 2 e^{-} \to H g_{2}^{2 +}$	+0.92
$A g^{+} + e^{-} \to Ag (s)$	+0.80
$F e^{3 +} + e^{-} \to F e^{2 +}$	+0.77
$O_{2} (g) + 2 H^{+} + 2 e^{-} \to H_{2} O_{2}$	+0.68
$I_{2} (s) + 2 e^{-} \to 2 I^{-}$	+0.54
$C u^{+} + e^{-} \to Cu (s)$	+0.52
$C u^{2 +} + 2 e^{-} \to Cu (s)$	+0.34
$AgCl (s) + e^{-} \to Ag (s) + C l^{-}$	+0.22
$AgBr (s) + e^{-} \to Ag (s) + B r^{-}$	+0.10
$2 H^{+} + 2 e^{-} \to H_{2} (g)$	0.00
$P b^{2 +} + 2 e^{-} \to Pb (s)$	-0.13
$S n^{2 +} + 2 e^{-} \to Sn (s)$	-0.14
$N i^{2 +} + 2 e^{-} \to Ni (s)$	-0.25
$F e^{2 +} + 2 e^{-} \to Fe (s)$	-0.44
$C r^{3 +} + 3 e^{-} \to Cr (s)$	-0.74
$Z n^{2 +} + 2 e^{-} \to Zn (s)$	-0.76
$2 H_{2} O + 2 e^{-} \to H_{2} (g) + 2 O H^{-}$	-0.83
$A l^{3 +} + 3 e^{-} \to Al (s)$	-1.66
$M g^{2 +} + 2 e^{-} \to Mg (s)$	-2.36
$N a^{+} + e^{-} \to Na (s)$	-2.71
$C a^{2 +} + 2 e^{-} \to Ca (s)$	-2.87
$K^{+} + e^{-} \to K (s)$	-2.93
$L i^{+} + e^{-} \to Li (s)$	-3.05

Read the left-hand species as possible oxidants, becoming stronger toward more positive reduction potentials. Read the right-hand species as possible reductants, becoming stronger toward more negative reduction potentials. Be precise about the species: a positive copper(II)/copper value refers to the couple, not a claim that copper metal is a strong oxidising agent.

Example — Predicting a reaction from the table

Problem
Can iron(III) oxidise iodide under standard comparison conditions?

1.The reduction $F e^{3 +} + e^{-} \to F e^{2 +}$ has $E^{\circ} = + 0.77 V$ . The iodine/iodide reduction has $E^{\circ} = + 0.54 V$ .
2.Iron(III) is the stronger electron acceptor in this pair, so it can accept electrons while iodide undergoes the reverse of the iodine reduction half reaction.
3.The balanced change is $2F e^{3 +} + 2 I^{-} \to 2F e^{2 +} + I_{2}$ . The difference of reduction potentials is $0.77 - 0.54 = + 0.23 V$ for this direction.
4.This establishes a favourable standard tendency, not a promise about the speed of every mixture. Multiplying a half reaction to balance electrons does not multiply its electrode potential.

Electrolysis reverses the energy arrangement

A galvanic cell supplies electrical energy from a favourable redox reaction. Electrolysis uses an external supply to drive electrode reactions. Reduction still occurs at the cathode and oxidation at the anode, but in an electrolytic cell the supply makes the cathode negative and anode positive. In water, dissolved ions and water may compete, and an electrode itself may react.

Aqueous system	Cathode product	Anode product or change
Silver nitrate; silver electrodes	Silver deposits from silver ions.	Silver electrode dissolves to silver ions.
Silver nitrate; platinum electrodes	Silver deposits.	Water is oxidised to oxygen; nitrate is usually a spectator.
Dilute sulphuric acid; platinum electrodes	Hydrogen gas forms.	Oxygen gas forms.
Copper chloride; platinum electrodes	Copper normally deposits.	Chlorine is the usual stated product for sufficiently concentrated chloride; oxygen competition depends on conditions.

Platinum is treated as an inert electrode in these introductory examples. Silver is active: it can supply silver ions by oxidation. Product predictions therefore require the solution, electrode material, concentration, and operating conditions. A table of standard potentials helps organise the possibilities, but does not by itself include every kinetic effect in electrolysis.

Example — A zinc–silver galvanic cell

Problem
Describe a cell for $Zn + 2A g^{+} \to Z n^{2 +} + 2Ag$ .

1.At the zinc anode, $Zn \to Z n^{2 +} + 2 e^{-}$ . At the silver cathode, $2A g^{+} + 2 e^{-} \to 2Ag$ .
2.The zinc electrode is negative in this galvanic cell. Electrons move through the wire from zinc to silver, while ions in the bridge maintain charge balance between solutions.
3.Using the table, the potential difference is $0.80 - (- 0.76) = 1.56 V$ under the stated standard comparison. The silver potential is not doubled when the half reaction is multiplied by 2.

Common mistake

Electrons do not pass through the salt bridge. Do not confuse anode/cathode definitions with a fixed sign in every kind of cell: oxidation is always at the anode, while the electrode signs differ between galvanic and electrolytic operation.

Quick check

Why does a faint pink colour persist at the endpoint of the permanganate titration?

Quick check

What is the iodine:thiosulphate mole ratio?

Quick check

Which carries charge through a salt bridge?

Quick check

Which statement describes the Daniell cell?

Quick check

What does a more positive standard reduction potential normally indicate?

Quick check

Why do silver electrodes change the electrolysis of silver nitrate compared with platinum?

Practice Problems

A $25.0 mL$ iron(II) sample consumes $12.0 mL$ of $0.0200 mol L^{- 1}$ permanganate in acid. Calculate its concentration.
An iodine solution consumes $10.0 mL$ of $0.100 mol L^{- 1}$ thiosulphate. Find the iodine amount and the amount of copper(II) that could have liberated it by the reaction given above.
Draw a Daniell cell and label the electrode signs, electron direction, current direction, half reactions, and directions of compensating bridge-ion movement.
Use the reduction-potential table to predict the standard-direction tendency for silver ions with copper, iron(III) with copper, silver metal with iron(III), and bromine with iron(II).
Arrange aluminium, copper, iron, magnesium, and zinc from greatest to least reducing power using the table.
Compare electrolysis of silver nitrate using silver and platinum anodes, explaining why the products differ.
Explain why copper chloride electrolysis cannot be predicted fully without considering chloride concentration and electrode conditions.

Key Takeaways

A titration endpoint signals that measured volumes can be interpreted using the balanced mole ratio. Permanganate can act as its own indicator; other methods use a redox indicator or the iodine–starch colour. A galvanic cell separates oxidation and reduction so electrons travel through an external circuit. Ions in the salt bridge complete the internal charge path. More positive reduction potentials identify stronger oxidising forms under the stated conditions; more negative values identify stronger reducing forms on the other side of the couple. Electrolysis uses an external supply, and electrode materials and solution conditions affect the products.

Previous · Lesson 6

Balancing Redox Equations by Half Reactions

Next · Lesson 8

Chapter Summary and Practice

Electron transfer becomes a measurement

Three ways to see an endpoint

Separate the two halves to obtain electrical energy

Electrode potentials compare redox tendencies

Electrolysis reverses the energy arrangement

Quiz

Practice Problems

Key Takeaways