Oxidation and Reduction

Q: What is the difference between oxidation and reduction?

Oxidation is the loss of electrons, seen as a rise in oxidation number; reduction is the gain of electrons, seen as a fall. They always happen together because electrons lost by one species must be taken by another. The species that is reduced is the oxidising agent.

Q: How do you find the oxidation number of an element in a compound?

Fix the atoms whose values are known: F is minus 1, alkali metals plus 1, alkaline earth metals plus 2, H plus 1 and O minus 2 unless an exception applies. Set the sum of all oxidation numbers equal to zero for a molecule or to the charge for an ion, and solve for the unknown.

Q: Why is the oxidation number of oxygen minus one in hydrogen peroxide?

In H-O-O-H each oxygen takes the O-H bond pair from hydrogen, which is less electronegative, but the O-O bond pair is shared equally because both atoms are identical. Each oxygen therefore gains only one electron in the bookkeeping, giving minus 1 instead of the usual minus 2.

Q: What is a disproportionation reaction?

It is a redox reaction in which one element in one oxidation state is oxidised and reduced at the same time. For example, in 2H2O2 giving 2H2O and O2, oxygen at minus 1 goes to minus 2 and to 0. The element must be able to show at least three oxidation states.

Q: Can an oxidation number be a fraction?

Yes, but a fraction is only an average. In Fe3O4 the average is plus 8/3, yet one iron is plus 2 and two are plus 3. In S4O6 2- the average is plus 2.5, while the real values are plus 5, 0, 0 and plus 5. The structure always reveals the real whole numbers.

Q: What does a negative standard electrode potential mean?

A negative standard electrode potential means the couple is a stronger reducing agent than the hydrogen couple, so its reduced form gives up electrons more easily than hydrogen gas. Zinc at minus 0.76 V displaces hydrogen from acids, while copper at plus 0.34 V does not.

Q: Which questions from oxidation and reduction are asked in NEET?

NEET questions stay close to NCERT: finding oxidation numbers, identifying the oxidant and reductant, classifying combination, decomposition, displacement and disproportionation reactions, Stock notation, and comparing oxidising or reducing strength from standard electrode potentials. One or two questions from this chapter appear most years.

Q: How is oxidation number tested in JEE Main and JEE Advanced?

JEE favours structure-based traps: chromium in CrO5 and sulphur in H2SO5 or H2S2O8 are plus 6 because of peroxide links, and average values in S4O6 2- or Fe3O4 differ from the real ones. It also asks which species can disproportionate and uses electrode potentials to predict feasible reactions.

ChemistryRedox ReactionsFor NEET aspirants

Oxidation and reduction always occur together: whenever one species loses electrons (oxidation), another gains them (reduction), which is why the pair is called a redox reaction. This page builds the idea in three steps, from the old oxygen-hydrogen rule to electron transfer and finally oxidation numbers, and then uses oxidation numbers to find the oxidant, the reductant and the type of redox reaction. It ends with electrode potentials, which rank oxidation and reduction strength. NEET and JEE Main ask oxidation-number and oxidant-identification questions every year.

On this page1Three definitions2Electron transfer3Metal activity4Oxidation number rules5Fractional ON6ON and agents7Types of redox8Electrode potentials

Key Formulas - Quick Reference

★ Must learnOxidation = loss of electrons = rise in oxidation number; reduction = gain of electrons = fall in oxidation number.
★ Must learnThe oxidising agent takes electrons and is itself reduced; the reducing agent gives electrons and is itself oxidised.
★ Must learnSum of oxidation numbers $= 0$ in a molecule and $=$ the charge in an ion.
Fixed values: F $- 1$ ; group 1 $+ 1$ ; group 2 $+ 2$ ; Al $+ 3$ ; H $+ 1$ ( $- 1$ in metal hydrides); O $- 2$ ( $- 1$ peroxide, $- \frac{1}{2}$ superoxide, $+ 2$ in $O F_{2}$ ).
Highest oxidation number $=$ group number (groups 1, 2) or group number $- 10$ (groups 13-17).
★ Must learnA fractional oxidation number is an average; the structure gives the real values ( $S_{4} O_{6}^{2 -}$ : $+ 5, 0, 0, + 5$ ).
Disproportionation: one element in one oxidation state goes to a higher and a lower state at the same time.
★ Must learnHigher $E^{\circ}$ = stronger oxidant ( $F_{2}$ , $+ 2.87$ V); lower $E^{\circ}$ = stronger reductant (Li, $- 3.05$ V).
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ ; a positive value means the reaction is feasible.

1. Three Ways to Define Oxidation and Reduction

The word oxidation first meant combining with oxygen. Air is about 20% dioxygen, which is why most elements occur on Earth as oxides. Burning magnesium, sulphur or methane all add oxygen to a substance:

2Mg + O_{2} 2MgO

C H_{4} + 2 O_{2} C O_{2} + 2 H_{2} O

In methane burning, hydrogen is also taken away from carbon. Chemists then noticed that removing hydrogen, or adding another electronegative element such as F, Cl or S, behaves exactly like adding oxygen. The definitions grew step by step.

Classical rule	Oxidation example	Reduction example
Oxygen added / removed	$2Mg$ + $O_{2}$ → $2MgO$	$2HgO$ → $2Hg$ + $O_{2}$ (on heating)
Hydrogen removed / added	$2 H_{2} S$ + $O_{2}$ → $2S$ + $2 H_{2} O$	$C H_{2} = C H_{2}$ + $H_{2}$ → $C H_{3} - C H_{3}$
Electronegative element added / removed	$Mg$ + $C l_{2}$ → $MgC l_{2}$	$2FeC l_{3}$ + $H_{2}$ → $2FeC l_{2}$ + $2HCl$
Electropositive element removed / added	$2 K_{4} [Fe (CN)_{6}]$ + $H_{2} O_{2}$ → $2 K_{3} [Fe (CN)_{6}]$ + $2KOH$	$2HgC l_{2}$ + $SnC l_{2}$ → $H g_{2} C l_{2}$ + $SnC l_{4}$

Look again at the last reduction: while $HgC l_{2}$ is reduced (mercury is added), $SnC l_{2}$ is oxidised to $SnC l_{4}$ (chlorine is added). Every example hides such a partner, so oxidation and reduction were joined into one word: redox.

Figure 1: Oxidation and reduction were redefined twice. The oxidation-number view covers even covalent reactions such as

H_{2} + C l_{2} \to 2HCl

, where no ions form.

2. Redox as Electron Transfer

Sodium chloride, sodium oxide and sodium sulphide are ionic: $N a^{+} C l^{-}$ , $(N a^{+})_{2} O^{2 -}$ , $(N a^{+})_{2} S^{2 -}$ . So when sodium reacts, it actually hands electrons to the non-metal. Each reaction splits into two half reactions:

2Na oxidation 2N a^{+} + 2 e^{-}

C l_{2} + 2 e^{-} reduction 2C l^{-}

Adding the halves cancels the electrons and gives $2Na + C l_{2} \to 2NaCl$ . The same view explains $2Na + H_{2} \to 2NaH$ , which the classical rules cannot: NaH is $N a^{+} H^{-}$ , so hydrogen gains an electron and is reduced.

Oxidation: loss of electron(s) by any species. Reduction: gain of electron(s). Oxidising agent (oxidant): electron acceptor. Reducing agent (reductant): electron donor.

Figure 2: Every redox reaction is two half reactions. The species that loses electrons (Na) is oxidised and acts as the reductant; the one that gains them (

C l_{2}

) is reduced and acts as the oxidant.

Exam Trick OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). Then flip it for the agents: an agent does to others what is done in reverse to itself. The oxidising agent oxidises its partner and is reduced.

Common oxidising agents

$O_{2}$ , $O_{3}$ , $F_{2}$ , $C l_{2}$ , $KMn O_{4}$ , $K_{2} C r_{2} O_{7}$ , $HN O_{3}$ , hot conc. $H_{2} S O_{4}$ , $H_{2} O_{2}$ , $F e^{3 +}$ . All are high up in oxidation number or very electronegative.

Common reducing agents

$H_{2}$ , C, CO, active metals (Na, Mg, Al, Zn), $H_{2} S$ , $S O_{2}$ , $SnC l_{2}$ , HI, $F e^{2 +}$ , $LiAl H_{4}$ , $NaB H_{4}$ . All hold electrons loosely or sit in a low oxidation number.

2.1 Competitive electron transfer

Put a zinc strip in copper nitrate solution. In about an hour the blue colour fades and the zinc gets a red-brown coat of copper. The solution now holds $Z n^{2 +}$ (it gives white ZnS with $H_{2} S$ once made alkaline with ammonia).

Zn + C u^{2 +} Z n^{2 +} + Cu

Reverse the test with a copper strip in zinc sulphate: nothing happens, and even passing $H_{2} S$ (which would give black CuS from the tiniest trace of $C u^{2 +}$ ) finds no copper ions. The equilibrium lies almost fully on the product side. Copper in silver nitrate turns the solution blue as silver deposits, while cobalt in nickel sulphate reaches an equilibrium with both ions present in moderate amounts:

Cu + 2A g^{+} C u^{2 +} + 2Ag

Co + N i^{2 +} ⇌ C o^{2 +} + Ni

Figure 3: A metal gives electrons only to the ion of a metal below it in reducing power, so Zn reduces

C u^{2 +}

, Cu reduces

A g^{+}

, but Cu cannot reduce

Z n^{2 +}

Ranking metals by how readily they release electrons gives the metal activity series (electrochemical series), just as acids are ranked by how readily they release protons. The same competition drives galvanic cells (section 6).

Key idea

Redox = electron transfer. A metal reduces the ions of any metal below it in the activity series: Zn > Cu > Ag.

3. Oxidation Number

In $2 H_{2} + O_{2} \to 2 H_{2} O$ no ions form, yet hydrogen clearly moves from a neutral state towards a positive one and oxygen towards a negative one, because the O-H bond electrons shift towards oxygen. To track such partial shifts, chemists use a bookkeeping number.

Oxidation number (ON) is the charge an atom would carry if every bond pair were given to the more electronegative atom. A pair shared by two identical atoms is split equally. Oxidation number and oxidation state mean the same thing.

Figure 4: Oxidation number is the charge an atom would carry if every bond pair went to the more electronegative partner. A bond between identical atoms adds nothing, which is why peroxide oxygen is

- 1

, not

- 2

2 H 0_{2} + O 0_{2} ⟶ 2 H + 1_{2} O - 2

H 0_{2} + Cl 0_{2} ⟶ 2 H + 1 Cl - 1

C - 4 H + 1_{4} + 4 Cl 0_{2} ⟶ C + 4 Cl - 1_{4} + 4 H + 1 Cl - 1

The full transfer is imagined only for bookkeeping, but it makes every redox reaction, ionic or covalent, look the same.

3.1 Rules for oxidation numbers

Free elements are 0 in any form: $H_{2}$ , $O_{2}$ , $O_{3}$ , $P_{4}$ , $S_{8}$ , Na, Mg, Al.
Monatomic ions carry their charge: $N a^{+}$ $+ 1$ , $M g^{2 +}$ $+ 2$ , $F e^{3 +}$ $+ 3$ , $C l^{-}$ $- 1$ , $O^{2 -}$ $- 2$ . In compounds, alkali metals are always $+ 1$ , alkaline earth metals $+ 2$ , aluminium $+ 3$ .
Fluorine is always $- 1$ . Cl, Br, I are $- 1$ as halides but positive when bonded to O or to a lighter halogen ( $HCl O_{4}$ : Cl $+ 7$ ; ICl: I $+ 1$ ).
Hydrogen is $+ 1$ , except $- 1$ in binary metal hydrides (LiH, NaH, $Ca H_{2}$ , $LiAl H_{4}$ ).
Oxygen is $- 2$ , except $- 1$ in peroxides ( $H_{2} O_{2}$ , $N a_{2} O_{2}$ , $Ba O_{2}$ ), $- \frac{1}{2}$ in superoxides ( $K O_{2}$ , $Rb O_{2}$ ), $+ 2$ in $O F_{2}$ and $+ 1$ in $O_{2} F_{2}$ .
Sum rule: the oxidation numbers add up to 0 in a neutral species and to the charge in an ion. In $C O_{3}^{2 -}$ : $x + 3 (- 2) = - 2$ , so C $= + 4$ .

Exam Trick When rules clash, apply them in this priority order: F, then group 1/2/Al, then H, then O, then everything else.

O F_{2}

: F wins, so O

= + 2

. NaH: Na wins, so H

= - 1

K O_{2}

: K wins, so O

= - \frac{1}{2}

Carbon in organic compounds, the fast way: count each bond from the carbon. A bond to H gives

- 1

, a bond to O, N or a halogen gives

+ 1

, a C-C bond gives 0. In

C H_{3} C H_{2} OH

the

C H_{3}

carbon is

- 3

and the

C H_{2} OH

carbon is

- 3 + 2 = - 1

(average

- 2

). In

C H_{3} COOH

the COOH carbon is

+ 3

(average 0).

3.2 Highest oxidation number and the periodic table

Metals show only positive oxidation numbers; non-metals show both signs; transition metals show several positive states. For main-group elements the maximum equals the number of valence electrons, so it rises across a period.

Figure 5: The maximum oxidation number of a main-group element equals its number of valence electrons, so it climbs by one across the period: Na

+ 1

to Cl

+ 7

3.3 Stock notation

German chemist Alfred Stock wrote the metal's oxidation number as a Roman numeral in brackets. It replaces the old -ous (lower state) and -ic (higher state) names and shows at once which form is reduced: $H g_{2} (I) C l_{2}$ is the reduced form of $Hg (II) C l_{2}$ .

Old name	Formula	Stock name
aurous chloride	AuCl	gold(I) chloride, Au(I)Cl
auric chloride	$AuC l_{3}$	gold(III) chloride, $Au (III) C l_{3}$
stannous chloride	$SnC l_{2}$	tin(II) chloride
stannic chloride	$SnC l_{4}$	tin(IV) chloride
ferrous / ferric chloride	$FeC l_{2}$ / $FeC l_{3}$	iron(II) / iron(III) chloride
cuprous / cupric oxide	$C u_{2} O$ / CuO	copper(I) / copper(II) oxide
mercurous / mercuric chloride	$H g_{2} C l_{2}$ / $HgC l_{2}$	mercury(I) / mercury(II) chloride

3.4 Fractional oxidation numbers

Electrons are never transferred in fractions, so a fractional oxidation number is always an average over atoms of the same element sitting in different environments. The structure shows the real values.

Figure 6: A fractional oxidation number is only an average over atoms in different environments:

C_{3} O_{2}

averages

+ \frac{4}{3}

S_{4} O_{6}^{2 -}

+ 2.5

B r_{3} O_{8}

+ \frac{16}{3}

F e_{3} O_{4}

+ \frac{8}{3}

The mixed oxides $F e_{3} O_{4}$ (FeO· $F e_{2} O_{3}$ ), $M n_{3} O_{4}$ and $P b_{3} O_{4}$ (2PbO· $Pb O_{2}$ ) behave the same way. A true fraction exists only when all atoms are equivalent, as in $O_{2}^{+}$ ( $+ \frac{1}{2}$ each) and $O_{2}^{-}$ ( $- \frac{1}{2}$ each).

Figure 7: Whenever the formula method gives an oxidation number above the group maximum, look for O-O (peroxide) links. With two peroxide O in

H_{2} S O_{5}

and four in

Cr O_{5}

, both central atoms are

+ 6

JEE Advanced When the formula method fails, draw the structure. An answer above the group maximum signals a peroxide link:

Cr O_{5}

+ 6

(two O-O groups),

H_{2} S O_{5}

and

H_{2} S_{2} O_{8}

+ 6

(one O-O group each). Two different environments of one element:

N H_{4} N O_{3}

has N at

- 3

and

+ 5

;

CaOC l_{2}

(Ca(OCl)Cl) has Cl at

+ 1

and

- 1

;

K I_{3}

(

K^{+} I_{3}^{-}

I_{2}

I^{-}

) has two I at 0 and one at

- 1

. Mixed oxides react as their parts:

P b_{3} O_{4} + 4HN O_{3} \to 2Pb (N O_{3})_{2} + Pb O_{2} + 2 H_{2} O

, because only the PbO part is basic.

Key idea

Oxidation number is bookkeeping: bond pairs to the more electronegative atom, sum equals charge, and fractions are averages.

Quick Recall: tap to check

Oxidation number of O in

O F_{2}

H_{2} O_{2}

and

K O_{2}

+ 2

- 1

and

- \frac{1}{2}

Why is H

- 1

Ca H_{2}

Calcium is less electronegative than hydrogen, so the bond pair is counted with H (binary metal hydride).

Average and real ON of S in

N a_{2} S_{4} O_{6}

Average

+ 2.5

; real

+ 5, 0, 0, + 5

(two S-S bonded S at 0).

4. Oxidation, Reduction and Agents by Oxidation Number

Oxidation: increase in the oxidation number of an element. Reduction: decrease. Oxidant: a reagent that raises another element's oxidation number (its own falls). Reductant: a reagent that lowers another element's oxidation number (its own rises). Redox reaction: any reaction in which oxidation numbers change.

2 Cu + 1_{2} O - 2 + Cu + 1_{2} S - 2 ⟶ 6 Cu 0 + S + 4 O - 2_{2}

Here copper falls from $+ 1$ to 0 (reduced) in both reactants, and sulphur of $C u_{2} S$ rises from $- 2$ to $+ 4$ (oxidised). So Cu(I) is the oxidant, and the sulphur of $C u_{2} S$ is the reductant.

Figure 8: Moving up a ladder is oxidation, moving down is reduction. A species at the top (

HN O_{3}

C O_{2}

) can only be reduced, one at the bottom (

N H_{3}

C H_{4}

) can only be oxidised, and middle states can go either way.

The ladders explain a classic question. A species with its element at the highest state can only be reduced, so it acts only as an oxidant ( $HN O_{3}$ , $KMn O_{4}$ , $O_{3}$ ). One at the lowest state acts only as a reductant ( $N H_{3}$ , $H_{2} S$ , $I^{-}$ ). Species in middle states, such as $S O_{2}$ (S $+ 4$ ), $H_{2} O_{2}$ (O $- 1$ ) and $HN O_{2}$ (N $+ 3$ ), can act either way.

Figure 9: The four questions that settle every 'is it redox, and who is the oxidant?' problem. The agent is the whole species, not the atom: in

C u_{2} O + C u_{2} S

C u_{2} S

is the reductant.

Limits of the idea. Oxidation number is a convention, and the idea of redox is still evolving. Today oxidation is often described as a decrease in electron density around an atom and reduction as an increase, which also covers bonds with only partial charge shift.

5. Types of Redox Reactions

Figure 10: The four classes of redox reactions. Decomposition is redox only if an element forms:

KCl O_{3}

qualifies,

CaC O_{3}

does not.

5.1 Combination

$A + B \to C$ , where A or B (or both) is an element. Every combustion in dioxygen is a combination redox reaction.

C 0 + O 0_{2} Δ C + 4 O - 2_{2}

3 Mg 0 + N 0_{2} Δ Mg + 2_{3} N - 3_{2}

5.2 Decomposition

The reverse: a compound breaks into parts, at least one of them an element. Not every element must change: potassium stays $+ 1$ below.

2 H + 1_{2} O - 2 Δ 2 H 0_{2} + O 0_{2}

2 Na + 1 H - 1 Δ 2 Na 0 + H 0_{2}

2 K + 1 Cl + 5 O - 2_{3} Δ 2 K + 1 Cl - 1 + 3 O 0_{2}

Not all decompositions are redox. In

CaC O_{3} Δ CaO + C O_{2}

every element keeps its oxidation number (Ca

+ 2

, C

+ 4

, O

- 2

5.3 Displacement

$X + YZ \to XZ + Y$ : an atom or ion in a compound is replaced by another element. The displacing element must be the stronger reducing (or oxidising) agent.

Type	Reaction	Point to remember
Metal displacement	$CuS O_{4}$ + $Zn$ → $Cu$ + $ZnS O_{4}$	Zn is the better reductant
Metal displacement (metallurgy)	$V_{2} O_{5}$ + $5Ca$ → $2V$ + $5CaO$ ; $TiC l_{4}$ + $2Mg$ → $Ti$ + $2MgC l_{2}$	pure metals from ores
Aluminothermy	$C r_{2} O_{3}$ + $2Al$ → $A l_{2} O_{3}$ + $2Cr$	Al is a strong reductant
H from cold water	$2Na$ + $2 H_{2} O$ → $2NaOH$ + $H_{2}$ ; $Ca$ + $2 H_{2} O$ → $Ca (OH)_{2}$ + $H_{2}$	alkali metals, Ca, Sr, Ba
H from hot water / steam	$Mg$ + $2 H_{2} O$ → $Mg (OH)_{2}$ + $H_{2}$ ; $3Fe$ + $4 H_{2} O$ → $F e_{3} O_{4}$ + $4 H_{2}$	Mg, Fe (less active)
H from acids	$Zn$ + $2HCl$ → $ZnC l_{2}$ + $H_{2}$	rate Mg > Zn > Fe; Cd, Sn also react; Ag, Au do not
O from water	$2 H_{2} O$ + $2 F_{2}$ → $4HF$ + $O_{2}$	$F_{2}$ is so reactive it attacks water
Halogen displacement	$C l_{2}$ + $2KBr$ → $2KCl$ + $B r_{2}$ ; $C l_{2}$ + $2KI$ → $2KCl$ + $I_{2}$ ; $B r_{2}$ + $2 I^{-}$ → $2B r^{-}$ + $I_{2}$	oxidising power $F_{2}$ > $C l_{2}$ > $B r_{2}$ > $I_{2}$

Chlorine displacing bromide and iodide is the basis of the layer test: the freed $B r_{2}$ colours a $CC l_{4}$ layer orange-brown and $I_{2}$ colours it violet. Recovering halogens from halides is an oxidation, $2 X^{-} \to X_{2} + 2 e^{-}$ . Chemical oxidants can do this for $C l^{-}$ , $B r^{-}$ and $I^{-}$ , but nothing is a stronger oxidant than fluorine, so $F^{-}$ can be turned into $F_{2}$ only by electrolysis.

5.4 Disproportionation

In disproportionation one element in one oxidation state is oxidised and reduced at the same time. The element must be able to exist in at least three oxidation states, and the reactant must hold it in a middle state.

2 H + 1_{2} O - 1_{2} ⟶ 2 H + 1_{2} O - 2 + O 0_{2}

P 0_{4} + 3 O H^{-} + 3 H_{2} O ⟶ P - 3 H_{3} + 3 H_{2} P + 1 O_{2}^{-}

S 0_{8} + 12 O H^{-} ⟶ 4 S^{2 -} + 2 S + 2_{2} O_{3}^{2 -} + 6 H_{2} O

Cl 0_{2} + 2 O H^{-} ⟶ Cl + 1 O^{-} + Cl^{-} + H_{2} O

The last reaction makes household bleach: $Cl O^{-}$ oxidises coloured stains to colourless products. Bromine and iodine behave like chlorine, but fluorine does not. With alkali it gives $2 F_{2} + 2O H^{-} \to 2 F^{-} + O F_{2} + H_{2} O$ , where F is only reduced (to $- 1$ ) and O is oxidised (to $+ 2$ ). Being the most electronegative element, fluorine has no positive oxidation state to go up to.

Figure 11: In disproportionation one element in one state moves both ways at once; the reactant must sit in an intermediate state. Comproportionation is the reverse: +5 and

- 1

iodine meet at 0 as

I_{2}

Disproportionation

One element, one state, goes up and down. Example: $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ (O: $- 1 \to - 2$ and 0).

Comproportionation

Two states of one element meet at a middle state. Example: $2 H_{2} S + S O_{2} \to 3S + 2 H_{2} O$ (S: $- 2$ and $+ 4 \to 0$ ).

Top and bottom cannot split. A species can disproportionate only if its element can go both up and down from where it is.

Cl O_{4}^{-}

(Cl

+ 7

, top) and

F_{2}

(F cannot be positive) never disproportionate;

Cl O^{-}

Cl O_{2}^{-}

Cl O_{3}^{-}

H_{2} O_{2}

N O_{2}

and

C u^{+}

can.

Key idea

Four types: combination, decomposition, displacement, disproportionation. Each is redox only if some oxidation number changes.

Quick Recall: tap to check

CaC O_{3} \to CaO + C O_{2}

a redox reaction?

No. Ca, C and O keep

+ 2

+ 4

and

- 2

Why can

Cl O_{4}^{-}

not disproportionate?

Cl is already at its highest state,

+ 7

, so it can only be reduced.

Classify

NaH + H_{2} O \to NaOH + H_{2}

Displacement: hydride (

- 1

) displaces H from water (

+ 1

); both end at 0 in

H_{2}

6. Redox Couples and Electrode Potentials

When a zinc rod sits in copper sulphate solution, electrons pass directly from Zn to $C u^{2 +}$ and the energy appears as heat. If zinc and copper ions are kept in separate beakers, the same electrons can be made to travel through a wire.

A redox couple is the oxidised and reduced form of the same species taking part in a half reaction, written oxidised form / reduced form:

Z n^{2 +} /Zn

C u^{2 +} /Cu

Dip a Zn rod in $ZnS O_{4}$ solution and a Cu rod in $CuS O_{4}$ solution. Join the rods by a wire with a voltmeter and switch, and join the solutions by a salt bridge (a U-tube of KCl or $N H_{4} N O_{3}$ set in agar jelly), which lets ions move without the solutions mixing. This is the Daniell cell.

Figure 12: In a Daniell cell the same Zn/Cu redox reaction runs indirectly: electrons travel through the wire from anode (Zn,

-

) to cathode (Cu,

+

), ions carry charge through the salt bridge, and

E_{cell}^{\circ} = 1.10

With the switch off, nothing happens in either beaker.
With the switch on, electrons leave Zn (oxidation at the anode) and travel through the wire to Cu, where $C u^{2 +}$ is reduced (at the cathode). Conventional current flows the opposite way.
Inside the cell, ions moving through the salt bridge complete the circuit.
Current flows only because there is a potential difference between the two electrodes.

The potential of each electrode is its electrode potential. With every species at unit concentration (gases at 1 atm) and 298 K, it is the standard electrode potential, $E^{\circ}$ . By convention $E^{\circ}$ of the hydrogen electrode, $H^{+} / H_{2}$ , is 0.00 V. A negative $E^{\circ}$ means the couple is a stronger reducing agent than $H^{+} / H_{2}$ ; a positive $E^{\circ}$ means a weaker one.

Figure 13: The higher the

E^{\circ}

, the stronger the oxidised form as an oxidant; the lower the

E^{\circ}

, the stronger the reduced form as a reductant. An oxidant (left) reacts with any reductant (right) lying below it:

C u^{2 +}

is reduced by Zn, not by Ag.

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

6.1 Using $E^{\circ}$ to predict a reaction

Pick the couple that must be reduced (the oxidant's couple) and the couple that must be oxidised. The reaction is feasible when

E_{cell}^{\circ} = E_{oxidant couple}^{\circ} - E_{reductant couple}^{\circ} > 0

$F e^{3 +}$ with $I^{-}$ : $0.77 - 0.54 = + 0.23$ V, so $F e^{3 +}$ oxidises iodide to iodine. Ag with $F e^{3 +}$ : $0.77 - 0.80 = - 0.03$ V, so no reaction. For the Daniell cell, $0.34 - (- 0.76) = 1.10$ V.

Exam Trick Upper-left eats lower-right. In a table of reduction potentials written with the oxidised form on the left, a species on the left reacts with any species on the right that lies below it.

C l_{2}

(1.36) oxidises

B r^{-}

(1.09) and

I^{-}

(0.54), but

I_{2}

cannot oxidise

B r^{-}

Key idea

High

E^{\circ}

= strong oxidant (top left); low

E^{\circ}

= strong reductant (bottom right).

E_{cell}^{\circ} > 0

means feasible.

Quick Recall: tap to check

Which electrode is negative in a Daniell cell?

The zinc anode, where electrons are released by oxidation.

Can Zn reduce

M g^{2 +}

No:

E_{cell}^{\circ} = - 2.36 - (- 0.76) = - 1.60

V, negative.

Strongest oxidant and strongest reductant in the series?

F_{2}

(

+ 2.87

V) and Li (

- 3.05

V).

6.2 The whole concept at a glance

Figure 14: The whole concept on one page. Revise from the centre outwards.

7. Solved Examples

Solved Example 1

Identify the species oxidised and reduced: (i)

H_{2} S

C l_{2}

→

2HCl

S

(ii)

3F e_{3} O_{4}

8Al

→

9Fe

4A l_{2} O_{3}

(iii)

2Na

H_{2}

→

2NaH

Solution:

(i) $H_{2} S$ is oxidised: hydrogen (electropositive) is removed from S, and S goes from $- 2$ to 0. $C l_{2}$ is reduced: hydrogen is added to it (Cl goes from 0 to $- 1$ ).

(ii) Al is oxidised (it gains oxygen, 0 to $+ 3$ ). $F e_{3} O_{4}$ is reduced (it loses oxygen; iron goes from $+ \frac{8}{3}$ to 0).

(iii) Classical rules are ambiguous here. Electronegativity settles it: H (2.1) is more electronegative than Na (0.9), so NaH is $N a^{+} H^{-}$ . Sodium is oxidised and hydrogen is reduced.

Solved Example 2

Show that

2Na

H_{2}

→

2NaH

is a redox change.

Solution:

NaH is ionic, $N a^{+} H^{-}$ , so the reaction splits into two half reactions:

2Na 2N a^{+} + 2 e^{-}

H_{2} + 2 e^{-} 2 H^{-}

Sodium loses electrons (oxidised) and hydrogen gains them (reduced), so it is a redox reaction.

Solved Example 3

Write in Stock notation:

HAuC l_{4}

T l_{2} O

, FeO,

F e_{2} O_{3}

, CuI, CuO, MnO,

Mn O_{2}

Solution:

Metal oxidation numbers: Au $+ 3$ , Tl $+ 1$ , Fe $+ 2$ , Fe $+ 3$ , Cu $+ 1$ , Cu $+ 2$ , Mn $+ 2$ , Mn $+ 4$ . Stock names: $HAu (III) C l_{4}$ , $T l_{2} (I) O$ , Fe(II)O, $F e_{2} (III) O_{3}$ , Cu(I)I, Cu(II)O, Mn(II)O, $Mn (IV) O_{2}$ .

Solved Example 4

Show that

2C u_{2} O

C u_{2} S

→

6Cu

S O_{2}

is a redox reaction. Name the species oxidised and reduced, the oxidant and the reductant.

Solution:

2 Cu + 1_{2} O - 2 + Cu + 1_{2} S - 2 ⟶ 6 Cu 0 + S + 4 O - 2_{2}

Copper is reduced ( $+ 1 \to 0$ ) and sulphur is oxidised ( $- 2 \to + 4$ ), so the reaction is redox. Cu(I), in both $C u_{2} O$ and $C u_{2} S$ , is the oxidant; the sulphur of $C u_{2} S$ is the reductant, since it lowers the oxidation number of copper in both compounds. This self-reduction is how copper is extracted.

Solved Example 5

Which of

Cl O^{-}

Cl O_{2}^{-}

Cl O_{3}^{-}

and

Cl O_{4}^{-}

cannot disproportionate, and why? Write the disproportionation of the others.

Solution:

$Cl O_{4}^{-}$ cannot: Cl is at its highest state, $+ 7$ . The others:

3 Cl + 1 O^{-} ⟶ 2 Cl^{-} + Cl + 5 O_{3}^{-}

6 Cl + 3 O_{2}^{-} light 4 Cl + 5 O_{3}^{-} + 2 Cl^{-}

4 Cl + 5 O_{3}^{-} ⟶ Cl^{-} + 3 Cl + 7 O_{4}^{-}

Solved Example 6

Classify: (a)

N_{2}

O_{2}

→

2NO

(b)

2Pb (N O_{3})_{2}

→

2PbO

4N O_{2}

O_{2}

(c)

NaH

H_{2} O

→

NaOH

H_{2}

(d)

2N O_{2}

2O H^{-}

→

N O_{2}^{-}

N O_{3}^{-}

H_{2} O

Solution:

(a) Combination: two elements form nitric oxide.

(b) Decomposition: lead nitrate breaks into three products, one of them the element $O_{2}$ .

(d) Disproportionation: N in $N O_{2}$ ( $+ 4$ ) goes to $+ 3$ in $N O_{2}^{-}$ and $+ 5$ in $N O_{3}^{-}$ .

Solved Example 7

Why do these reactions go differently?

P b_{3} O_{4}

8HCl

→

3PbC l_{2}

C l_{2}

4 H_{2} O

and

P b_{3} O_{4}

4HN O_{3}

→

2Pb (N O_{3})_{2}

Pb O_{2}

2 H_{2} O

Solution:

$P b_{3} O_{4}$ is a fixed mixture, 2PbO + $Pb O_{2}$ . PbO (Pb $+ 2$ ) is a basic oxide; $Pb O_{2}$ (Pb $+ 4$ ) is an oxidant, since $+ 2$ is the stable state of lead.

With HCl both parts react: an acid-base reaction and a redox reaction in which $Pb O_{2}$ oxidises $C l^{-}$ to $C l_{2}$ .

2PbO + 4HCl 2PbC l_{2} + 2 H_{2} O

Pb + 4 O_{2} + 4 H Cl - 1 ⟶ Pb + 2 Cl_{2} + Cl 0_{2} + 2 H_{2} O

$HN O_{3}$ is itself an oxidant, so $Pb O_{2}$ cannot oxidise it. Only the acid-base part occurs, $2PbO + 4HN O_{3} \to 2Pb (N O_{3})_{2} + 2 H_{2} O$ , and $Pb O_{2}$ is left unchanged.

Solved Example 8

The oxidation number of sulphur in peroxodisulphuric acid,

H_{2} S_{2} O_{8}

, is
(A)

+ 7

(B)

+ 6

(C)

+ 8

(D)

+ 5

Solution:

Answer: (B). The formula alone gives $2 (+ 1) + 2 x + 8 (- 2) = 0$ , so $x = + 7$ , above the group-16 maximum of $+ 6$ . The structure HO3S-O-O-SO3H has one peroxide link: two O at $- 1$ and six at $- 2$ . Then $2 + 2 x - 2 - 12 = 0$ gives $x = + 6$ .

Solved Example 9

In which reaction does

H_{2} O_{2}

act as a reducing agent?
(A)

2FeC l_{2}

2HCl

H_{2} O_{2}

→

2FeC l_{3}

2 H_{2} O

(B)

C l_{2}

H_{2} O_{2}

→

2HCl

O_{2}

(C)

2HI

H_{2} O_{2}

→

2 H_{2} O

I_{2}

(D)

H_{2} S O_{3}

H_{2} O_{2}

→

H_{2} S O_{4}

H_{2} O

Solution:

Answer: (B). A reductant is oxidised. Only in (B) does peroxide oxygen rise, from $- 1$ to 0 in $O_{2}$ , while Cl falls from 0 to $- 1$ . In (A), (C) and (D) oxygen falls to $- 2$ in water, so $H_{2} O_{2}$ is the oxidant.

Solved Example 10

Find the oxidation numbers: (a) Fe in

K_{4} [Fe (CN)_{6}]

(b) N in

N H_{4} N O_{3}

C_{6} H_{12} O_{6}

(d) Cl in bleaching powder,

CaOC l_{2}

Solution:

(a) $K_{4} [Fe (CN)_{6}]$ : $4 (+ 1) + x + 6 (- 1) = 0$ (cyanide is $- 1$ ), so Fe $= + 2$ .

(b) $N H_{4}^{+}$ : $x + 4 = + 1$ , N $= - 3$ ; $N O_{3}^{-}$ : $x - 6 = - 1$ , N $= + 5$ . The average $+ 1$ is not the real value of either N.

(d) Ca(OCl)Cl: Cl in $OC l^{-}$ is $+ 1$ and Cl in $C l^{-}$ is $- 1$ .

Solved Example 11

Using

E^{\circ}

values, show why acidified

KMn O_{4}

should not be used with hydrochloric acid.

Solution:

$Mn O_{4}^{-} /M n^{2 +}$ is $+ 1.51$ V and $C l_{2} /C l^{-}$ is $+ 1.36$ V. For permanganate oxidising chloride, $E_{cell}^{\circ} = 1.51 - 1.36 = + 0.15$ V, which is positive, so $Mn O_{4}^{-}$ oxidises $C l^{-}$ to $C l_{2}$ :

2Mn O_{4}^{-} + 16 H^{+} + 10C l^{-} 2M n^{2 +} + 5C l_{2} + 8 H_{2} O

Part of the permanganate is wasted on the acid, so dilute $H_{2} S O_{4}$ is used instead.

Solved Example 12

Name the type of reaction and the oxidant and reductant:

3C l_{2}

6NaOH

→

5NaCl

NaCl O_{3}

3 H_{2} O

(hot, concentrated alkali).

Solution:

3 Cl 0_{2} + 6 Na OH hot, conc. 5 Na Cl - 1 + Na Cl + 5 O_{3} + 3 H_{2} O

Chlorine goes from 0 to $- 1$ (five atoms) and to $+ 5$ (one atom), so it is disproportionation and $C l_{2}$ is both the oxidant and the reductant. Electrons balance: $5 \times 1 = 1 \times 5$ .

Practice Questions

Assign the oxidation number of the named element: (a) P in $Na H_{2} P O_{4}$ (b) S in $NaHS O_{4}$ (c) P in $H_{4} P_{2} O_{7}$ (d) Mn in $K_{2} Mn O_{4}$ (e) O in $Ca O_{2}$ (f) B in $NaB H_{4}$ (g) S in $H_{2} S_{2} O_{7}$ (h) S in $KAl (S O_{4})_{2} \cdot 12 H_{2} O$ .Answer: (a) $+ 5$ (b) $+ 6$ (c) $+ 5$ (d) $+ 6$ (e) $- 1$ (f) $+ 3$ (g) $+ 6$ (h) $+ 6$ .
Find and rationalise the oxidation numbers: (a) I in $K I_{3}$ (b) S in $H_{2} S_{4} O_{6}$ (c) Fe in $F e_{3} O_{4}$ (d) C in $C H_{3} C H_{2} OH$ (e) C in $C H_{3} COOH$ .Answer: (a) average $- \frac{1}{3}$ ; real: $I_{3}^{-}$ = $I_{2}$ (0, 0) + $I^{-}$ ( $- 1$ ). (b) average $+ 2.5$ ; real $+ 5, 0, 0, + 5$ . (c) average $+ \frac{8}{3}$ ; real one $+ 2$ , two $+ 3$ . (d) average $- 2$ ; real $- 3$ ( $C H_{3}$ ) and $- 1$ ( $C H_{2} OH$ ). (e) average 0; real $- 3$ and $+ 3$ .
Show that these are redox: (a) $CuO$ + $H_{2}$ → $Cu$ + $H_{2} O$ (b) $F e_{2} O_{3}$ + $3CO$ → $2Fe$ + $3C O_{2}$ (c) $4BC l_{3}$ + $3LiAl H_{4}$ → $2 B_{2} H_{6}$ + $3LiCl$ + $3AlC l_{3}$ (d) $2K$ + $F_{2}$ → $2KF$ (e) $4N H_{3}$ + $5 O_{2}$ → $4NO$ + $6 H_{2} O$ Answer: (a) Cu $+ 2 \to 0$ , H $0 \to + 1$ . (b) Fe $+ 3 \to 0$ , C $+ 2 \to + 4$ . (c) By the textbook convention (H $+ 1$ in $B_{2} H_{6}$ ): B $+ 3 \to - 3$ and hydride H $- 1 \to + 1$ . (d) K $0 \to + 1$ , F $0 \to - 1$ . (e) N $- 3 \to + 2$ , O $0 \to - 2$ .
Fluorine reacts with ice: $H_{2} O$ + $F_{2}$ → $HF$ + $HOF$ . Show it is redox.Answer: F falls from 0 to $- 1$ (in HF and in HOF); O rises from $- 2$ in $H_{2} O$ to 0 in HOF (H $+ 1$ , F $- 1$ ).
Find the oxidation number of S, Cr and N in $H_{2} S O_{5}$ , $C r_{2} O_{7}^{2 -}$ and $N O_{3}^{-}$ . Suggest structures and explain the fallacy.Answer: Formula method: S $+ 8$ , Cr $+ 6$ , N $+ 5$ . S cannot exceed $+ 6$ : $H_{2} S O_{5}$ has one O-O link, so S is $+ 6$ . $C r_{2} O_{7}^{2 -}$ (Cr-O-Cr bridge, no peroxide) and $N O_{3}^{-}$ have no fallacy.
Write formulas: (a) mercury(II) chloride (b) nickel(II) sulphate (c) tin(IV) oxide (d) thallium(I) sulphate (e) iron(III) sulphate (f) chromium(III) oxide.Answer: (a) $HgC l_{2}$ (b) $NiS O_{4}$ (c) $Sn O_{2}$ (d) $T l_{2} S O_{4}$ (e) $F e_{2} (S O_{4})_{3}$ (f) $C r_{2} O_{3}$ .
List substances in which carbon shows $- 4$ to $+ 4$ and nitrogen $- 3$ to $+ 5$ .Answer: C: $C H_{4}$ $- 4$ , $C_{2} H_{6}$ $- 3$ , $C H_{3} OH$ $- 2$ , $C_{2} H_{2}$ $- 1$ , HCHO 0, CO $+ 2$ , $H_{2} C_{2} O_{4}$ $+ 3$ , $C O_{2}$ $+ 4$ . N: $N H_{3}$ $- 3$ , $N_{2} H_{4}$ $- 2$ , $N H_{2} OH$ $- 1$ , $N_{2}$ 0, $N_{2} O$ $+ 1$ , NO $+ 2$ , $HN O_{2}$ $+ 3$ , $N O_{2}$ $+ 4$ , $HN O_{3}$ $+ 5$ (see Figure 8).
$S O_{2}$ and $H_{2} O_{2}$ act as both oxidants and reductants, but $O_{3}$ and $HN O_{3}$ act only as oxidants. Why?Answer: S ( $+ 4$ , range $- 2$ to $+ 6$ ) and O in $H_{2} O_{2}$ ( $- 1$ , range $- 2$ to 0) sit in middle states. N in $HN O_{3}$ is at its maximum $+ 5$ ; ozone can only gain electrons (O is 0 and goes to $- 2$ ).
$Ag F_{2}$ is unstable, but if formed it is a very strong oxidant. Why?Answer: Silver's stable state is $+ 1$ . Ag(II) grabs an electron at once to return to Ag(I), so it oxidises almost anything.
Identify the substance oxidised, reduced, the oxidant and the reductant: (a) $2AgBr$ + $C_{6} H_{6} O_{2}$ → $2Ag$ + $2HBr$ + $C_{6} H_{4} O_{2}$ (b) $HCHO$ + $2 [Ag (N H_{3})_{2}]^{+}$ + $3O H^{-}$ → $2Ag$ + $HCO O^{-}$ + $4N H_{3}$ + $2 H_{2} O$ (c) $HCHO$ + $2C u^{2 +}$ + $5O H^{-}$ → $C u_{2} O$ + $HCO O^{-}$ + $3 H_{2} O$ (d) $N_{2} H_{4}$ + $2 H_{2} O_{2}$ → $N_{2}$ + $4 H_{2} O$ (e) $Pb$ + $Pb O_{2}$ + $2 H_{2} S O_{4}$ → $2PbS O_{4}$ + $2 H_{2} O$ Answer: Reductant (oxidised) / oxidant (reduced): (a) hydroquinone / AgBr (b) HCHO / $[Ag (N H_{3})_{2}]^{+}$ (c) HCHO / $C u^{2 +}$ (d) $N_{2} H_{4}$ / $H_{2} O_{2}$ (e) Pb / $Pb O_{2}$ .
Show with reactions that fluorine is the best oxidant among halogens and HI the best reductant among hydrohalic acids.Answer: $F_{2}$ oxidises $C l^{-}$ , $B r^{-}$ , $I^{-}$ (and water): $F_{2}$ + $2C l^{-}$ → $2 F^{-}$ + $C l_{2}$ . HI reduces $H_{2} S O_{4}$ to $S O_{2}$ ( $2HI$ + $H_{2} S O_{4}$ → $I_{2}$ + $S O_{2}$ + $2 H_{2} O$ ); HCl cannot. $E^{\circ}$ : $F_{2}$ highest, $I_{2}$ lowest.
Why does $Xe O_{6}^{4 -}$ + $2 F^{-}$ + $6 H^{+}$ → $Xe O_{3}$ + $F_{2}$ + $3 H_{2} O$ occur? What does it say about $N a_{4} Xe O_{6}$ ?Answer: Perxenate (Xe $+ 8$ ) oxidises $F^{-}$ to $F_{2}$ , so $N a_{4} Xe O_{6}$ is an even stronger oxidant than fluorine.
Consider: (a) $H_{3} P O_{2}$ + $4AgN O_{3}$ + $2 H_{2} O$ → $H_{3} P O_{4}$ + $4Ag$ + $4HN O_{3}$ (b) $H_{3} P O_{2}$ + $2CuS O_{4}$ + $2 H_{2} O$ → $H_{3} P O_{4}$ + $2Cu$ + $2 H_{2} S O_{4}$ (c) benzaldehyde reduces $[Ag (N H_{3})_{2}]^{+}$ to Ag (d) benzaldehyde does not reduce $C u^{2 +}$ in alkali. What do you infer about $A g^{+}$ and $C u^{2 +}$ ?Answer: $H_{3} P O_{2}$ is a strong reductant and reduces both. Benzaldehyde, a weaker reductant, reduces only $A g^{+}$ , so $A g^{+}$ is the stronger oxidant ( $E^{\circ}$ 0.80 V vs 0.34 V).
From Cs, Ne, I and F, pick the element that shows (a) only negative (b) only positive (c) both (d) neither oxidation states.Answer: (a) F (b) Cs (c) I (d) Ne.
Using the table of standard electrode potentials, predict feasibility: (a) $F e^{3 +}$ and $I^{-}$ (b) $A g^{+}$ and Cu (c) $F e^{3 +}$ and Cu (d) Ag and $F e^{3 +}$ (e) $B r_{2}$ and $F e^{2 +}$ .Answer: (a) $+ 0.23$ V, yes (b) $+ 0.46$ V, yes (c) $+ 0.43$ V, yes (d) $- 0.03$ V, no (e) $+ 0.32$ V, yes.
Predict the electrolysis products: (i) aqueous $AgN O_{3}$ , silver electrodes (ii) aqueous $AgN O_{3}$ , platinum electrodes (iii) dilute $H_{2} S O_{4}$ , platinum electrodes (iv) aqueous $CuC l_{2}$ , platinum electrodes.Answer: (i) Ag deposits at the cathode, the Ag anode dissolves. (ii) Ag at the cathode, $O_{2}$ at the anode. (iii) $H_{2}$ at the cathode, $O_{2}$ at the anode. (iv) Cu at the cathode, $C l_{2}$ at the anode.
Arrange Al, Cu, Fe, Mg, Zn in the order in which they displace each other from their salt solutions.Answer: Mg > Al > Zn > Fe > Cu (each displaces those after it).
Given $K^{+} /K$ $- 2.93$ , $A g^{+} /Ag$ $0.80$ , $H g^{2 +} /Hg$ $0.79$ , $M g^{2 +} /Mg$ $- 2.37$ , $C r^{3 +} /Cr$ $- 0.74$ V, arrange the metals by increasing reducing power.Answer: Ag < Hg < Cr < Mg < K.
Depict the galvanic cell for $Zn$ + $2A g^{+}$ → $Z n^{2 +}$ + $2Ag$ . Show the negative electrode, the current carriers and each electrode reaction.Answer: Zn | $Z n^{2 +}$ ‖ $A g^{+}$ | Ag. Zn (anode) is negative. Electrons carry current in the wire, ions in the solutions and salt bridge. Anode: $Zn$ → $Z n^{2 +}$ + $2 e^{-}$ ; cathode: $A g^{+}$ + $e^{-}$ → $Ag$ .

Common Mistakes to Avoid

Watch out

Calling the oxidising agent "the species that is oxidised". The oxidant is reduced; the reductant is oxidised.
Naming an atom as the agent. Say " $KMn O_{4}$ is the oxidant", not "Mn" or "+7".
Taking oxygen as $- 2$ in $H_{2} O_{2}$ , $N a_{2} O_{2}$ , $K O_{2}$ or $O F_{2}$ (correct: $- 1$ , $- 1$ , $- \frac{1}{2}$ , $+ 2$ ).
Taking hydrogen as $+ 1$ in NaH, $Ca H_{2}$ or $LiAl H_{4}$ , where it is $- 1$ .
Reporting a fractional oxidation number as the value of every atom: in $S_{4} O_{6}^{2 -}$ no sulphur is actually $+ 2.5$ .
Accepting an oxidation number above the group maximum ( $Cr O_{5}$ $+ 10$ , $H_{2} S O_{5}$ $+ 8$ ) instead of looking for a peroxide link.
Treating every decomposition as redox. $CaC O_{3} \to CaO + C O_{2}$ is not.
Reading $E^{\circ}$ backwards: a large positive $E^{\circ}$ marks a strong oxidant, and electrons flow from anode to cathode in the wire while current flows the other way.

Frequently Asked Questions

What is the difference between oxidation and reduction?

Oxidation is the loss of electrons, seen as a rise in oxidation number; reduction is the gain of electrons, seen as a fall. They always happen together because electrons lost by one species must be taken by another. The species that is reduced is the oxidising agent.

How do you find the oxidation number of an element in a compound?

Fix the atoms whose values are known: F is minus 1, alkali metals plus 1, alkaline earth metals plus 2, H plus 1 and O minus 2 unless an exception applies. Set the sum of all oxidation numbers equal to zero for a molecule or to the charge for an ion, and solve for the unknown.

Why is the oxidation number of oxygen minus one in hydrogen peroxide?

In H-O-O-H each oxygen takes the O-H bond pair from hydrogen, which is less electronegative, but the O-O bond pair is shared equally because both atoms are identical. Each oxygen therefore gains only one electron in the bookkeeping, giving minus 1 instead of the usual minus 2.

What is a disproportionation reaction?

It is a redox reaction in which one element in one oxidation state is oxidised and reduced at the same time. For example, in 2H2O2 giving 2H2O and O2, oxygen at minus 1 goes to minus 2 and to 0. The element must be able to show at least three oxidation states.

Can an oxidation number be a fraction?

Yes, but a fraction is only an average. In Fe3O4 the average is plus 8/3, yet one iron is plus 2 and two are plus 3. In S4O6 2- the average is plus 2.5, while the real values are plus 5, 0, 0 and plus 5. The structure always reveals the real whole numbers.

What does a negative standard electrode potential mean?

A negative standard electrode potential means the couple is a stronger reducing agent than the hydrogen couple, so its reduced form gives up electrons more easily than hydrogen gas. Zinc at minus 0.76 V displaces hydrogen from acids, while copper at plus 0.34 V does not.

Which questions from oxidation and reduction are asked in NEET?

NEET questions stay close to NCERT: finding oxidation numbers, identifying the oxidant and reductant, classifying combination, decomposition, displacement and disproportionation reactions, Stock notation, and comparing oxidising or reducing strength from standard electrode potentials. One or two questions from this chapter appear most years.

How is oxidation number tested in JEE Main and JEE Advanced?

JEE favours structure-based traps: chromium in CrO5 and sulphur in H2SO5 or H2S2O8 are plus 6 because of peroxide links, and average values in S4O6 2- or Fe3O4 differ from the real ones. It also asks which species can disproportionate and uses electrode potentials to predict feasible reactions.

Previous year questions on Oxidation and Reduction

17 questions from past papers, each with a step-by-step solution.

Show all 17 questions

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Practice questions →Take a mock test

1. Three Ways to Define Oxidation and Reduction

2. Redox as Electron Transfer

2.1 Competitive electron transfer

3. Oxidation Number

3.1 Rules for oxidation numbers

3.2 Highest oxidation number and the periodic table

3.3 Stock notation

3.4 Fractional oxidation numbers

4. Oxidation, Reduction and Agents by Oxidation Number

5. Types of Redox Reactions

5.1 Combination

5.2 Decomposition

5.3 Displacement

5.4 Disproportionation

6. Redox Couples and Electrode Potentials

6.1 Using E∘ to predict a reaction

6.2 The whole concept at a glance

7. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is the difference between oxidation and reduction?

How do you find the oxidation number of an element in a compound?

Why is the oxidation number of oxygen minus one in hydrogen peroxide?

What is a disproportionation reaction?

Can an oxidation number be a fraction?

What does a negative standard electrode potential mean?

Which questions from oxidation and reduction are asked in NEET?

How is oxidation number tested in JEE Main and JEE Advanced?

Previous year questions on Oxidation and Reduction

Ready to master Redox Reactions?

6.1 Using $E^{\circ}$ to predict a reaction