Redox Reactions · Lesson 8 of 8

Chapter Summary and Practice

“Connect the whole chapter by moving from observed changes to electron accounting, balanced equations, and useful applications.”

Learning Objectives

• Translate among classical, electron-transfer, and oxidation-number descriptions of redox. • Recognise reaction patterns while checking the actual oxidation-state changes. • Choose and verify a balancing method for acidic or basic medium. • Use balanced stoichiometry in calculations and interpretation of observations. • Connect oxidising and reducing strength with titrations, cells, and electrode processes. • Explain the limits of oxidation numbers and standard-potential comparisons.

Use one chain of reasoning across the chapter

A redox problem may begin with a burning fuel, a coloured solution, an unfamiliar formula, or an electrode diagram. The same sequence of questions helps in each case. Identify the species before and after the change, decide which atoms gain or lose electron density, account for the electron-equivalent change, and check the complete equation. Then connect that equation to what is observed or measured.

This lesson brings the chapter together through that sequence. The aim is to choose a useful representation and explain the reasoning, rather than attach a reaction name and stop. A strong explanation connects the visible observation, the symbolic equation, and the behaviour of electrons or oxidation states.

Question to ask	Tool from the chapter	What the tool establishes
What chemical change is visible?	Oxygen/hydrogen transfer; colour or deposit observations	A starting description of oxidation and reduction.
Who gives and receives electron equivalents?	Electron transfer and oxidation-number changes	Oxidant, reductant, and paired changes.
How are reactants reorganised?	Combination, decomposition, displacement, disproportionation	The reaction pattern, after confirming redox.
In what proportions do species react?	Oxidation-number or half-reaction balancing	Conservation of atoms, charge, and electron equivalents.
How is the reaction measured or used?	Titration stoichiometry, cells, and potential comparisons	Concentrations, relative tendencies, and energy or product interpretations.

Three descriptions of the same partnership

Classical descriptions are useful when an element such as oxygen or hydrogen clearly changes partners. Electron transfer is most direct for ionic changes. Oxidation numbers extend the accounting to covalent compounds, but they do not turn every shared electron into a freely transferred particle. These descriptions support one another when applied carefully.

Description	Oxidation	Reduction
Classical	Gain of oxygen/electronegative element, or loss of hydrogen/electropositive element.	The reverse kinds of change.
Electron transfer	Loss of electrons.	Gain of electrons.
Oxidation number	Increase in oxidation number.	Decrease in oxidation number.
Electron density	Decrease around the atom or atoms concerned.	Increase around the atom or atoms concerned.

The reducing agent undergoes oxidation; the oxidising agent undergoes reduction. Remembering only “oxidation means loss” is not enough unless you specify that the loss is of electrons. The classical definition can instead involve a gain of oxygen. Clear wording prevents these two descriptions from seeming contradictory.

Example — A reaction with no oxygen

Problem
Analyse $4BC l_{3} + 3LiAl H_{4} \to 2 B_{2} H_{6} + 3LiCl + 3AlC l_{3}$ .

1.Boron is +3 in boron trichloride. In this chapter’s oxidation-number convention for boranes, hydrogen is assigned +1 and boron is -3 in diborane.
2.Hydrogen in lithium aluminium hydride is -1 and becomes +1 in diborane, so it is oxidised. Boron decreases from +3 to -3 and is reduced.
3.Four boron atoms each decrease by 6, giving 24 electron equivalents. Twelve hydrogen atoms each increase by 2, also giving 24.
4.The hydride-containing reagent is the reductant, and boron trichloride is the oxidant. The conclusion comes from state changes rather than oxygen transfer.

Distinguish a formula average from a reaction pathway

The formula sum may give a useful average without identifying each atomic environment. Peroxide bonds and mixed oxidation states are reminders to inspect structure. Even a correctly balanced net equation may hide which reactant supplied a particular product atom. Redox bookkeeping and mechanism are related questions, but they are not the same question.

For photosynthesis, $6C O_{2} + 12 H_{2} O \to C_{6} H_{12} O_{6} + 6 H_{2} O + 6 O_{2}$ keeps the water consumed and formed visible before cancelling. Isotopic labelling can trace oxygen atoms and show that the evolved oxygen comes from water. Similarly, in $O_{3} + H_{2} O_{2} \to H_{2} O + O_{2} + O_{2}$ , writing the two oxygen products separately can help frame the question of atom origins. A labelled-oxygen experiment, not the balanced equation alone, investigates the pathway.

The oxidation state available to a reactant also helps explain its role. A species in an intermediate state may be oxidised or reduced depending on its partner; a species already at its ordinary highest state cannot be oxidised further within that set of states. Availability of states is a first check, not a complete test of feasibility or speed.

Example — Same reducing agent, different products

Problem
Why can thiosulphate give tetrathionate with iodine but sulphate with bromine?

1.With iodine, $2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -} + 2 I^{-}$ . The sulphur oxidation is limited to the tetrathionate product.
2.With bromine in the stated aqueous reaction, $S_{2} O_{3}^{2 -} + 4B r_{2} + 5 H_{2} O \to 2S O_{4}^{2 -} + 8B r^{-} + 10 H^{+}$ .
3.Bromine is a stronger oxidant than iodine in the chapter’s comparison and can oxidise sulphur further to sulphate under these conditions.
4.The different products change the total electron transfer. The two sulphur atoms increase from a total oxidation number of +4 to +12, supplying eight electron equivalents. Four bromine molecules each accept two, so the coefficient 4 is required for sulphate formation.

Reagent amounts can affect how far oxidation proceeds. An excess of reductant may favour a lower final state, whereas sufficient oxidant can produce a higher state when the chemistry permits. For example, limited versus excess chlorine can give iron(II) versus iron(III) chloride, tin(II) versus tin(IV) chloride, and phosphorus(III) versus phosphorus(V) chloride under suitable controlled conditions. These are condition-dependent possibilities, not a universal rule that excess reagent overrides chemical stability.

Experimental observations also help us compare actual reagent systems. The following reactions show several source applications in a form that can be analysed directly. Look for the changing centre, rather than assuming that all atoms in a reacting compound undergo the same change.

System	Balanced change	Changing centres
Silver bromide and hydroquinone	$2AgBr + C_{6} H_{6} O_{2} \to 2Ag + 2HBr + C_{6} H_{4} O_{2}$	Silver +1 to 0; the organic compound loses two hydrogen atoms and is oxidised to the quinone.
Formaldehyde with silver–ammonia reagent	$HCHO + 2 [Ag (N H_{3})_{2}]^{+} + 3O H^{-} \to 2Ag + HCO O^{-} + 4N H_{3} + 2 H_{2} O$	Carbon 0 to +2; silver +1 to 0.
Formaldehyde with alkaline copper(II)	$HCHO + 2C u^{2 +} + 5O H^{-} \to C u_{2} O + HCO O^{-} + 3 H_{2} O$	Carbon 0 to +2; copper +2 to +1.
Hydrazine and hydrogen peroxide	$N_{2} H_{4} + 2 H_{2} O_{2} \to N_{2} + 4 H_{2} O$	Nitrogen -2 to 0; peroxide oxygen -1 to -2.
Lead storage-cell discharge	$Pb + Pb O_{2} + 2 H_{2} S O_{4} \to 2PbS O_{4} + 2 H_{2} O$	Lead metal 0 to +2; lead in dioxide +4 to +2.

Hypophosphorous acid, $H_{3} P O_{2}$ , is oxidised to $H_{3} P O_{4}$ by either silver(I) or the stated copper(II) reagent. Phosphorus changes from +1 to +5, supplying four electron equivalents: enough to form four silver atoms or two copper atoms. In contrast, benzaldehyde, $C_{6} H_{5} CHO$ , gives a reaction with the silver–ammonia reagent but no observed change with the stated alkaline copper(II) reagent. These observations show selective oxidising behaviour under the test conditions; complex formation and reaction kinetics mean we should not replace this evidence with an unqualified free-ion ranking.

The balanced acid equations are $H_{3} P O_{2} + 4AgN O_{3} + 2 H_{2} O \to H_{3} P O_{4} + 4Ag + 4HN O_{3}$ and $H_{3} P O_{2} + 2CuS O_{4} + 2 H_{2} O \to H_{3} P O_{4} + 2Cu + 2 H_{2} S O_{4}$ . The sulphuric-acid coefficient is 2 because two sulphate groups enter. For benzaldehyde, the positive silver test can be written $C_{6} H_{5} CHO + 2 [Ag (N H_{3})_{2}]^{+} + 3O H^{-} \to C_{6} H_{5} CO O^{-} + 2Ag + 4N H_{3} + 2 H_{2} O$ .

Solubility and contact can matter as well as oxidising strength. Toluene mixes poorly with water, so a suitable solvent arrangement brings it into contact with permanganate; the choice is not proof that aqueous permanganate lacks oxidising power. A balanced net equation for oxidation to benzoate with manganese dioxide formation is $C_{6} H_{5} C H_{3} + 2Mn O_{4}^{-} \to C_{6} H_{5} CO O^{-} + 2Mn O_{2} + O H^{-} + H_{2} O$ . Acidifying benzoate produces benzoic acid in a separate acid–base step.

Concentrated sulphuric acid with a chloride salt can release hydrogen chloride by an acid–base process without changing oxidation states. Bromide is more readily oxidised: $2HBr + H_{2} S O_{4} \to B r_{2} + S O_{2} + 2 H_{2} O$ . Here bromine rises from -1 to 0 and sulphur falls from +6 to +4. The red-brown bromine observation therefore reflects a redox change in addition to acid behaviour.

A balanced equation becomes a quantitative tool

Both balancing methods enforce the same conservation laws. Oxidation-number balancing compares total rises and falls, while half-reaction balancing makes electrons explicit. After balancing, the coefficients become mole ratios. They can be used in a titration or in a limiting-reactant calculation, but not as direct ratios of masses.

Example — From electron balance to maximum product

Problem
Ammonia is oxidised to nitric oxide and steam. What maximum mass of nitric oxide can form from $10.00 g$ ammonia and $20.00 g$ oxygen? Use molar masses 17.0, 32.0, and 30.0 grams per mole for ammonia, oxygen, and nitric oxide.

1.Balance the reaction: $4N H_{3} + 5 O_{2} \to 4NO + 6 H_{2} O$ . Nitrogen rises from -3 to +2, and oxygen falls from 0 to -2.
2.Ammonia amount is $10.00/17.0 = 0.5882 mol$ . Oxygen amount is $20.00/32.0 = 0.6250 mol$ .
3.The available reaction extents are $0.5882/4 = 0.1471$ and $0.6250/5 = 0.1250$ . Oxygen gives the smaller value and is limiting.
4.The nitric oxide amount is $4 (0.1250) = 0.5000 mol$ , corresponding to $0.5000 \times 30.0 = 15.0 g$ .
5.The result assumes complete conversion by the stated reaction and no side losses. Comparing 10 g and 20 g directly would not identify the limiting reactant.

Example — One state splits into two products

Problem
Balance the disproportionation of manganese(III) into manganese(II) and manganese dioxide in water.

1.Manganese(III) becomes +2 by accepting one electron, or +4 by losing one electron. The two manganese products therefore occur in a one-to-one ratio.
2.The reduction half is $M n^{3 +} + e^{-} \to M n^{2 +}$ . The oxidation half is $M n^{3 +} + 2 H_{2} O \to Mn O_{2} + 4 H^{+} + e^{-}$ .
3.Adding gives $2M n^{3 +} + 2 H_{2} O \to M n^{2 +} + Mn O_{2} + 4 H^{+}$ .
4.Check total charge: +6 on the left, +2 + 4 = +6 on the right. This combines pattern recognition, half reactions, and medium balancing.

Connect reagent strength with observations and cells

Zinc deposits copper from its ions because zinc can supply the electrons required for copper reduction under the stated conditions. Separating those two changes gives the Daniell cell. The same electron-equivalent ratios govern chemical analysis: a titration converts the endpoint volume into an amount using a balanced equation. These are different experimental arrangements of the same conservation principles.

Standard reduction potentials compare specified couples, concentrations, and reference conditions. They identify favourable standard tendencies when one half reaction is reversed and paired with another. They do not directly state the reaction rate, and they cannot ignore precipitation, complex formation, solvent, or electrode conditions. Selective observations with silver–ammonia reagent and alkaline copper(II) reagent must therefore be interpreted for those particular systems.

Example — Compare two electron acceptors

Problem
Under standard comparison conditions, can iron(III) oxidise copper metal? Can it oxidise silver metal?

1.Use $E^{\circ} (F e^{3 +} /F e^{2 +}) = 0.77 V$ , $E^{\circ} (C u^{2 +} /Cu) = 0.34 V$ , and $E^{\circ} (A g^{+} /Ag) = 0.80 V$ .
2.For copper oxidation paired with iron(III) reduction, the difference is $0.77 - 0.34 = + 0.43 V$ . The standard-direction tendency favours $2F e^{3 +} + Cu \to 2F e^{2 +} + C u^{2 +}$ .
3.For silver oxidation, the difference is $0.77 - 0.80 = - 0.03 V$ . The proposed standard direction is not favoured.
4.This agrees with the need to identify the actual oxidised and reduced forms. Electrode potentials are not multiplied by balancing coefficients.

Common mistake

Keep the scope of every rule clear. Average oxidation numbers are not necessarily individual atomic states. Disproportionation needs one initial state, not simply two products. A balanced equation does not reveal a complete mechanism. A favourable standard potential does not guarantee a rapid reaction or fix all electrolysis products.

Quick check

In $CuO + H_{2} \to Cu + H_{2} O$ , which statement connects the classical and electronic descriptions?

Quick check

Which observation most directly supports competition for electrons?

Quick check

Which statement about the two balancing methods is correct?

Quick check

What does a disappearing blue starch signal indicate in a thiosulphate titration?

Quick check

Which pair correctly describes the Daniell cell charge paths?

Quick check

A reaction begins with oxygen at -1 and forms oxygen at both -2 and 0. What pattern is present?

Quick check

For $4N H_{3} + 5 O_{2} \to 4NO + 6 H_{2} O$ , what comparison identifies the limiting reactant?

Practice Problems

Assign oxidation numbers to P in $Na H_{2} P O_{4}, H_{4} P_{2} O_{7}$ , S in $NaHS O_{4}, H_{2} S_{2} O_{7}, KAl (S O_{4})_{2} \cdot 12 H_{2} O$ , Mn in $K_{2} Mn O_{4}$ , O in $Ca O_{2}$ , and B in $NaB H_{4}$ . State the exception used in each relevant case.
Explain the averages for iodine in $K I_{3}$ , sulphur in $H_{2} S_{4} O_{6}$ , and iron in $F e_{3} O_{4}$ . Then use bonding to calculate the individual carbon states in $C H_{3} C H_{2} OH$ and $C H_{3} COOH$ .
Justify the redox character of $CuO + H_{2} \to Cu + H_{2} O$ , $F e_{2} O_{3} + 3CO \to 2Fe + 3C O_{2}$ , $4BC l_{3} + 3LiAl H_{4} \to 2 B_{2} H_{6} + 3LiCl + 3AlC l_{3}$ , $2K + F_{2} \to 2KF$ , and $4N H_{3} + 5 O_{2} \to 4NO + 6 H_{2} O$ .
For $H_{2} O + F_{2} \to HF + HOF$ , assign oxygen and fluorine states, identify the redox partners, and distinguish the result from fluorine disproportionation.
Compare the oxidation-number calculations for S in $H_{2} S O_{5}$ , Cr in $C r_{2} O_{7}^{2 -}$ , and N in $N O_{3}^{-}$ . Explain why structural information about a peroxide bond changes the first calculation.
Write formulas for mercury(II) chloride, nickel(II) sulphate, tin(IV) oxide, thallium(I) sulphate, iron(III) sulphate, and chromium(III) oxide.
Give suitable compounds illustrating carbon oxidation states from -4 to +4 and nitrogen states from -3 to +5. Use structures where an average would be ambiguous.
Explain why sulphur dioxide and hydrogen peroxide can each serve as an oxidant or reductant. Contrast the usual roles of ozone and nitric acid in the reactions considered here.
Explain why retaining water on both sides of the photosynthesis equation helps formulate an atom-tracing question. Describe how oxygen-isotope labelling can distinguish atom sources in photosynthesis and the ozone–peroxide reaction.
Explain why an unstable silver(II) compound such as $Ag F_{2}$ can be a strong oxidant when conversion to the more stable silver(I) state is available.
Use suitable equations for iron, tin, and phosphorus with chlorine to explain how limited versus excess oxidant can affect the final oxidation state. State why conditions and product stability still matter.
Explain why a solvent that permits contact with toluene is useful in its permanganate oxidation to benzoate or benzoic acid, even though aqueous acidic and alkaline permanganate are both oxidants. Balance a suitable net redox equation. Separately, explain why concentrated sulphuric acid gives hydrogen chloride with a chloride salt but can oxidise bromide to bromine.
Identify the oxidised and reduced substances and both agents in silver bromide reduction by hydroquinone, formaldehyde oxidation by silver–ammonia reagent, formaldehyde oxidation by alkaline copper(II), hydrazine reacting with peroxide, and $Pb + Pb O_{2} + 2 H_{2} S O_{4} \to 2PbS O_{4} + 2 H_{2} O$ . In each case state the changing atom and its initial and final states.
Balance thiosulphate oxidation by iodine to tetrathionate and by bromine to sulphate. Explain the different products using oxidising strength and verify the bromine coefficient.
Use halogen displacement equations to compare fluorine, chlorine, bromine, and iodine as oxidants. Explain why iodide is more readily oxidised than chloride, relating this to the reducing behaviour of hydroiodic acid.
Analyse $Xe O_{6}^{4 -} + 2 F^{-} + 6 H^{+} \to Xe O_{3} + F_{2} + 3 H_{2} O$ . Identify the xenon states and redox partners. What does this exceptional stated reaction show about the limits of an unqualified claim that fluoride can never be oxidised chemically?
Given that hypophosphorous acid reduces both silver(I) and copper(II), while benzaldehyde gives a reaction with silver–ammonia reagent but no observed change with the stated alkaline copper(II) reagent, infer the selective oxidising behaviour of those test systems. Why is this not simply a comparison of two free-ion standard potentials?
Use half reactions to balance permanganate with iodide in base; permanganate with sulphur dioxide to hydrogen sulphate in acid; peroxide with iron(II) in acid; and dichromate with sulphur dioxide in acid. Identify the agents.
Use both balancing methods for phosphorus forming phosphine and hypophosphite in base, $P_{4} + O H^{-} \to P H_{3} + H_{2} P O_{2}^{-}$ ; hydrazine with chlorate forming nitric oxide and chloride in base; and dichlorine heptoxide with peroxide forming chlorite and oxygen in base. Check every atom and charge.
For $(CN)_{2} + 2O H^{-} \to C N^{-} + CN O^{-} + H_{2} O$ , calculate carbon oxidation states, classify the reaction, and explain the information available from its coefficients.
Derive the manganese(III) disproportionation equation without looking at the worked example, then show separate electron and charge checks.
Compare Cs, Ne, I, and F: identify which normally exhibits positive states only in compounds, negative states only in compounds, both signs in compounds, or neither in the ordinary chemistry considered here. Explain the role of the uncombined state.
Balance the aqueous removal of excess chlorine by sulphur dioxide, and connect the equation to the roles of oxidant and reductant.
Using oxidation-state ranges and suitable examples, select non-metals and three metals that can show disproportionation. Explain why an intermediate state is necessary but not sufficient to establish favourable reaction under all conditions.
Recalculate the maximum nitric oxide from 10.00 g ammonia and 20.00 g oxygen. Show the mole-ratio comparison that identifies the limiting reactant and calculate unused ammonia.
Use the standard-potential table to test the proposed reactions between iron(III) and iodide, silver(I) and copper, iron(III) and copper, silver metal and iron(III), and bromine and iron(II). Write the favourable balanced changes.
Predict the usual products for aqueous silver nitrate with silver electrodes, silver nitrate with platinum electrodes, dilute sulphuric acid with platinum electrodes, and copper chloride with platinum electrodes. State the conditions needed for an unambiguous chloride prediction.
Arrange Al, Cu, Fe, Mg, and Zn from strongest to weakest electron donor. Explain how this predicts displacement from salt solutions.
Given reduction potentials in volts of potassium -2.93, silver +0.80, mercury(II)/mercury +0.79, magnesium -2.37, and chromium(III)/chromium -0.74, arrange the metals in increasing reducing power. Explain why the order reverses the order of reduction potentials.
Depict the zinc–silver galvanic cell for $Zn + 2A g^{+} \to Z n^{2 +} + 2Ag$ . Identify the negative electrode, both half reactions, the external electron direction, and the charge carriers inside the cell.

Key Takeaways

Redox is a coupled redistribution of electrons or electron density, described by classical changes, half reactions, or oxidation numbers. The reductant is oxidised and the oxidant reduced. Structure explains exceptions and distinguishes formal averages from individual atomic environments. Reaction patterns become reliable only when checked against state changes. Balanced equations conserve atoms and charge and match all electron equivalents. Their coefficients connect redox chemistry with titrations, limiting-reactant calculations, and electrode processes. Potential comparisons, experimental observations, and knowledge of conditions together explain what a redox system can do.

Previous · Lesson 7

Redox in Titrations and Electrochemical Cells

End of chapter

Use one chain of reasoning across the chapter

Three descriptions of the same partnership

Distinguish a formula average from a reaction pathway

A balanced equation becomes a quantitative tool

Connect reagent strength with observations and cells

Quiz

Practice Problems

Key Takeaways