Chapter 7: Redox Reactions

Assign oxidation numbers to every atom on both sides of the equation.

Reactants:

• In $$\mathrm{H_2S}$$: H is $$+1$$ and S is $$-2$$.
• In $$\mathrm{Cl_2}$$ (elemental): Cl is $$0$$.

Products:

• In $$\mathrm{HCl}$$: H is $$+1$$ and Cl is $$-1$$.
• In $$\mathrm{S}$$ (elemental): S is $$0$$.

Compare the changes:

Sulphur: $$-2 \rightarrow 0$$. The oxidation number of S increases (loss of electrons), therefore $$\mathrm{H_2S}$$ is oxidised.

Chlorine: $$0 \rightarrow -1$$. The oxidation number of Cl decreases (gain of electrons), therefore $$\mathrm{Cl_2}$$ is reduced.

Assign oxidation numbers using $$\mathrm{O} = -2$$ for every oxide.

Reactants:

• In $$\mathrm{Fe_3O_4}$$: $$3x + 4(-2) = 0$$ gives an average $$\mathrm{Fe} = +8/3$$. (Structurally $$\mathrm{Fe_3O_4 \equiv FeO\cdot Fe_2O_3}$$ — one Fe(II) and two Fe(III), averaging $$+8/3$$.)
• In elemental $$\mathrm{Al}$$: $$\mathrm{Al} = 0$$.

Products:

• In elemental $$\mathrm{Fe}$$: $$\mathrm{Fe} = 0$$.
• In $$\mathrm{Al_2O_3}$$: $$\mathrm{Al} = +3$$, $$\mathrm{O} = -2$$.

Aluminium: $$0 \rightarrow +3$$ — its oxidation number increases, so $$\mathrm{Al}$$ is oxidised.

Iron: $$+8/3 \rightarrow 0$$ — its oxidation number decreases, so $$\mathrm{Fe_3O_4}$$ is reduced.

Oxidation numbers:

• Elemental $$\mathrm{Na}$$: $$0$$.
• Elemental $$\mathrm{H_2}$$: $$0$$.
• In $$\mathrm{NaH}$$, sodium is far more electropositive than hydrogen, so this is a saline hydride: $$\mathrm{Na} = +1$$ and $$\mathrm{H} = -1$$.

Sodium: $$0 \rightarrow +1$$ — increase, hence $$\mathrm{Na}$$ is oxidised.

Hydrogen: $$0 \rightarrow -1$$ — decrease, hence $$\mathrm{H_2}$$ is reduced.

Two elemental gases combine to give a single product, with oxidation numbers changing:

$$\mathrm{N_2}$$ ($$\mathrm{N} = 0$$) $$+\;\mathrm{O_2}$$ ($$\mathrm{O} = 0$$) $$\rightarrow \mathrm{NO}$$ ($$\mathrm{N} = +2$$, $$\mathrm{O} = -2$$).

Since two reactants combine into one product with simultaneous oxidation ($$\mathrm{N}: 0 \rightarrow +2$$) and reduction ($$\mathrm{O}: 0 \rightarrow -2$$), this is a combination redox reaction.

A single compound breaks up into several products — that is the signature of a decomposition reaction. Check the oxidation numbers:

• $$\mathrm{Pb(NO_3)_2}$$: $$\mathrm{Pb} = +2$$, $$\mathrm{N} = +5$$, $$\mathrm{O} = -2$$.
• $$\mathrm{PbO}$$: $$\mathrm{Pb} = +2$$, $$\mathrm{O} = -2$$.
• $$\mathrm{NO_2}$$: $$\mathrm{N} = +4$$, $$\mathrm{O} = -2$$.
• $$\mathrm{O_2}$$: $$\mathrm{O} = 0$$.

Nitrogen: $$+5 \rightarrow +4$$ (reduced).

Oxygen: $$-2 \rightarrow 0$$ (oxidised, only the O that becomes $$\mathrm{O_2}$$).

One reactant decomposes with simultaneous oxidation and reduction. This is a decomposition redox reaction.

Assign oxidation numbers, taking H as $$-1$$ in the saline hydride $$\mathrm{NaH}$$ and $$+1$$ in $$\mathrm{H_2O}$$ and $$\mathrm{NaOH}$$:

• $$\mathrm{NaH}$$: Na $$+1$$, H $$-1$$.
• $$\mathrm{H_2O}$$: H $$+1$$, O $$-2$$.
• $$\mathrm{NaOH}$$: Na $$+1$$, O $$-2$$, H $$+1$$.
• $$\mathrm{H_2}$$: H $$0$$.

The hydridic H ($$-1$$, from $$\mathrm{NaH}$$) is oxidised to $$0$$ (in $$\mathrm{H_2}$$), and the protonic H ($$+1$$, from $$\mathrm{H_2O}$$) is reduced to $$0$$. The two hydrogens in the product $$\mathrm{H_2}$$ come from different starting materials.

An atom (here, hydridic H of $$\mathrm{NaH}$$) displaces another atom (protonic H of $$\mathrm{H_2O}$$) and the displaced atom appears in elemental form $$\mathrm{H_2}$$. This is a displacement (single-displacement) redox reaction.

Look at the oxidation number of nitrogen on both sides:

• $$\mathrm{NO_2}$$: $$\mathrm{N} = +4$$.
• $$\mathrm{NO_2^-}$$: $$x + 2(-2) = -1 \Rightarrow \mathrm{N} = +3$$.
• $$\mathrm{NO_3^-}$$: $$x + 3(-2) = -1 \Rightarrow \mathrm{N} = +5$$.

Of the two $$\mathrm{NO_2}$$ molecules, one is reduced ($$+4 \rightarrow +3$$ in $$\mathrm{NO_2^-}$$) and the other is oxidised ($$+4 \rightarrow +5$$ in $$\mathrm{NO_3^-}$$). The same element (N), starting from a single intermediate oxidation state ($$+4$$), simultaneously increases and decreases its oxidation number — the defining feature of disproportionation.

Take the standard oxidation numbers: $$\mathrm{Na} = +1$$, $$\mathrm{H} = +1$$, $$\mathrm{O} = -2$$. Let oxidation number of P be $$x$$. The compound is neutral, so the sum is zero:

$$(+1) + 2(+1) + x + 4(-2) = 0$$

$$1 + 2 + x - 8 = 0$$

$$x = +5$$

With $$\mathrm{Na} = +1$$, $$\mathrm{H} = +1$$, $$\mathrm{O} = -2$$, let $$\mathrm{S} = x$$.

$$(+1) + (+1) + x + 4(-2) = 0$$

$$2 + x - 8 = 0$$

$$x = +6$$

With $$\mathrm{H} = +1$$, $$\mathrm{O} = -2$$, and the molecule neutral, let $$\mathrm{P} = x$$:

$$4(+1) + 2x + 7(-2) = 0$$

$$4 + 2x - 14 = 0$$

$$2x = 10 \Rightarrow x = +5$$

(Pyrophosphoric acid has both P atoms in the $$+5$$ state.)

With $$\mathrm{K} = +1$$, $$\mathrm{O} = -2$$, and the molecule neutral, let $$\mathrm{Mn} = x$$:

$$2(+1) + x + 4(-2) = 0$$

$$2 + x - 8 = 0$$

$$x = +6$$

(Potassium manganate(VI).)

$$\mathrm{CaO_2}$$ is calcium peroxide, not a normal oxide. Ca is a group-2 metal and always has oxidation state $$+2$$. Let $$\mathrm{O} = x$$:

$$(+2) + 2x = 0 \Rightarrow x = -1$$

The two O atoms are joined by an $$\mathrm{O\!-\!O}$$ peroxide bond, and each O carries $$-1$$.

$$\mathrm{NaBH_4}$$ is sodium borohydride — a complex hydride. Sodium is more electropositive than hydrogen, so $$\mathrm{Na} = +1$$. The four hydrogens are attached to boron in the tetrahedral $$\mathrm{[BH_4]^-}$$ anion. Boron (electronegativity $$2.04$$) is less electronegative than hydrogen ($$2.20$$), but for oxidation-state book-keeping in saline/complex hydrides we treat each H as $$-1$$.

Let $$\mathrm{B} = x$$:

$$(+1) + x + 4(-1) = 0$$

$$x = +3$$

$$\mathrm{H_2S_2O_7}$$ is pyrosulphuric (oleum) acid. With $$\mathrm{H} = +1$$, $$\mathrm{O} = -2$$ and neutrality, let $$\mathrm{S} = x$$:

$$2(+1) + 2x + 7(-2) = 0$$

$$2 + 2x - 14 = 0$$

$$2x = 12 \Rightarrow x = +6$$

This is potash alum. The water of crystallisation contributes no charge, so consider only $$\mathrm{KAl(SO_4)_2}$$. With $$\mathrm{K} = +1$$, $$\mathrm{Al} = +3$$, $$\mathrm{O} = -2$$ and the salt neutral, let $$\mathrm{S} = x$$:

$$(+1) + (+3) + 2\,[x + 4(-2)] = 0$$

$$4 + 2x - 16 = 0$$

$$2x = 12 \Rightarrow x = +6$$

(Each S in the sulphate is in the $$+6$$ state.)

With $$\mathrm{K} = +1$$ and the salt neutral, the $$\mathrm{I_3^-}$$ anion must carry $$-1$$. Treating all three I atoms identically:

$$3x = -1 \Rightarrow x = -\tfrac{1}{3}$$

A fractional oxidation state is purely a book-keeping artefact. The real structure of $$\mathrm{I_3^-}$$ is a linear $$\mathrm{I\!-\!I\!-\!I}^-$$ ion that can be thought of as an $$\mathrm{I_2}$$ molecule coordinated to an $$\mathrm{I^-}$$ ion. So two of the iodine atoms are in oxidation state $$0$$ (the $$\mathrm{I_2}$$ part) and the third is in oxidation state $$-1$$ (the $$\mathrm{I^-}$$ part). The average $$0 + 0 + (-1) = -1$$, divided over three atoms, gives $$-\tfrac{1}{3}$$ per atom.

$$\mathrm{H_2S_4O_6}$$ is tetrathionic acid. With $$\mathrm{H} = +1$$ and $$\mathrm{O} = -2$$, let the four S atoms have average oxidation number $$x$$:

$$2(+1) + 4x + 6(-2) = 0$$

$$4x = 10 \Rightarrow x = +\tfrac{5}{2}$$

The fractional value arises because the four S atoms are not equivalent. The structure is

$$\mathrm{HO_3S\!-\!S\!-\!S\!-\!SO_3H}$$

The two terminal sulphur atoms (bonded to three O and one S) carry oxidation state $$+5$$ each; the two central sulphur atoms (in the $$\mathrm{S\!-\!S\!-\!S}$$ chain, bonded only to other S atoms) carry oxidation state $$0$$ each. Average $$= (2\times 5 + 2\times 0)/4 = +\tfrac{5}{2}$$.

With $$\mathrm{O} = -2$$, the average oxidation number of Fe:

$$3x + 4(-2) = 0 \Rightarrow x = +\tfrac{8}{3}$$

This fractional value is again the artefact of averaging non-equivalent atoms. Magnetite is best described as $$\mathrm{FeO\cdot Fe_2O_3}$$ — it contains one $$\mathrm{Fe^{2+}}$$ centre and two $$\mathrm{Fe^{3+}}$$ centres per formula unit. The average $$= (1\times 2 + 2\times 3)/3 = 8/3$$.

For organic compounds, simple algebra on the molecular formula $$\mathrm{C_2H_6O}$$ gives only an average:

$$2x + 6(+1) + (-2) = 0 \Rightarrow x = -2$$ (average per C).

To get the oxidation state of each carbon, use the bonding-based rule: in each C–H bond H is $$+1$$ and C is $$-1$$; in C–O bond O takes the electrons (C is $$+1$$, O is $$-1$$ per bond); in C–C bonds neither atom changes.

Methyl carbon, $$\mathrm{\underline{C}H_3}$$: bonded to 3 H and 1 C. Contribution: $$3(-1) + 0 = -3$$, so this C is at $$\boxed{-3}$$.

Methylene carbon, $$\mathrm{\underline{C}H_2OH}$$: bonded to 2 H, 1 C, 1 O (single bond). Contribution: $$2(-1) + 0 + (+1) = -1$$, so this C is at $$\boxed{-1}$$.

Average $$= \tfrac{1}{2}(-3 + (-1)) = -2$$ ✓.

The molecular formula is $$\mathrm{C_2H_4O_2}$$. Algebra gives an average:

$$2x + 4(+1) + 2(-2) = 0 \Rightarrow x = 0$$ (average per C).

Bonding-based assignment of the two non-equivalent carbons:

Methyl carbon, $$\mathrm{\underline{C}H_3}$$: bonded to 3 H and 1 C. Contribution: $$3(-1) + 0 = -3$$, so this C is at $$\boxed{-3}$$.

Carboxyl carbon, $$\mathrm{\underline{C}OOH}$$: bonded to 1 C, 1 doubly-bonded O ($$=\!\!\mathrm{O}$$), 1 single-bonded O ($$-\!\!\mathrm{OH}$$). Each C–O bond contributes $$+1$$ for C; the C=O double bond contributes $$+2$$. Total: $$0 + 2 + 1 = +3$$, so this C is at $$\boxed{+3}$$.

Average $$= \tfrac{1}{2}(-3 + 3) = 0$$ ✓. (The algebraic average masks the very different bonding environments of the two carbons.)

Assign oxidation numbers:

• $$\mathrm{CuO}$$: Cu $$+2$$, O $$-2$$.
• $$\mathrm{H_2}$$: H $$0$$.
• $$\mathrm{Cu}$$: $$0$$.
• $$\mathrm{H_2O}$$: H $$+1$$, O $$-2$$.

Cu: $$+2 \rightarrow 0$$ (gain of $$2$$e$${}^-$$) — reduction. Hence $$\mathrm{CuO}$$ is the oxidising agent (reduced).

H: $$0 \rightarrow +1$$ (loss of $$1$$e$${}^-$$ per atom; $$2$$e$${}^-$$ per $$\mathrm{H_2}$$) — oxidation. Hence $$\mathrm{H_2}$$ is the reducing agent (oxidised).

Simultaneous oxidation and reduction → redox reaction.

Oxidation numbers:

• $$\mathrm{Fe_2O_3}$$: Fe $$+3$$, O $$-2$$.
• $$\mathrm{CO}$$: C $$+2$$, O $$-2$$.
• $$\mathrm{Fe}$$: $$0$$.
• $$\mathrm{CO_2}$$: C $$+4$$, O $$-2$$.

Fe: $$+3 \rightarrow 0$$ — reduced. $$\mathrm{Fe_2O_3}$$ is the oxidant.

C: $$+2 \rightarrow +4$$ — oxidised. $$\mathrm{CO}$$ is the reductant.

Both oxidation and reduction occur, so it is a redox reaction (this is the basic process in iron extraction in a blast furnace).

Track the oxidation numbers of every element. Take $$\mathrm{Cl} = -1$$, $$\mathrm{Li} = +1$$. In $$\mathrm{LiAlH_4}$$, H is hydridic ($$-1$$) so Al $$= +3$$.

The interesting element is hydrogen in $$\mathrm{B_2H_6}$$. Boron (Pauling EN $$2.04$$) is less electronegative than hydrogen ($$2.20$$), so in $$\mathrm{B_2H_6}$$ we treat H as $$+1$$, which gives

$$2\,\mathrm{B} + 6(+1) = 0 \Rightarrow \mathrm{B} = -3$$

(In the alternative convention where H in $$\mathrm{B_2H_6}$$ is taken as $$-1$$, B is $$+3$$. NCERT adopts the first convention so that the hydrogen change becomes visible — see below.)

Summary of oxidation states:

Boron is reduced ($$+3 \rightarrow -3$$) and hydrogen is oxidised ($$-1 \rightarrow +1$$). Hence the reaction is redox.

$$\mathrm{BCl_3}$$ is the oxidising agent (gains electrons through B) and $$\mathrm{LiAlH_4}$$ is the reducing agent (supplies hydridic H).

Oxidation numbers:

• $$\mathrm{K}$$ (elemental): $$0$$.
• $$\mathrm{F_2}$$ (elemental): $$0$$.
• $$\mathrm{K^+F^-}$$: K $$+1$$, F $$-1$$.

K: $$0 \rightarrow +1$$ (loss of one electron per atom) — oxidation. $$\mathrm{K}$$ is the reducing agent.

F: $$0 \rightarrow -1$$ (gain of one electron per atom) — reduction. $$\mathrm{F_2}$$ is the oxidising agent.

Both oxidation and reduction occur — redox.

Oxidation numbers:

• $$\mathrm{NH_3}$$: N $$-3$$, H $$+1$$.
• $$\mathrm{O_2}$$: O $$0$$.
• $$\mathrm{NO}$$: N $$+2$$, O $$-2$$.
• $$\mathrm{H_2O}$$: H $$+1$$, O $$-2$$.

N: $$-3 \rightarrow +2$$ — oxidation (loss of $$5$$e$${}^-$$ per N). $$\mathrm{NH_3}$$ is the reducing agent.

O: $$0 \rightarrow -2$$ — reduction (gain of $$2$$e$${}^-$$ per O). $$\mathrm{O_2}$$ is the oxidising agent.

H stays at $$+1$$. Redox confirmed. (This is the first step of Ostwald's process for nitric acid manufacture.)

Mercury(II) means $$\mathrm{Hg} = +2$$. Chloride is $$\mathrm{Cl^-}$$. To balance one $$\mathrm{Hg^{2+}}$$ we need two $$\mathrm{Cl^-}$$: $$\mathrm{HgCl_2}$$.

Nickel(II) is $$\mathrm{Ni^{2+}}$$; sulphate is $$\mathrm{SO_4^{2-}}$$. The charges $$+2$$ and $$-2$$ already balance, so one of each gives a neutral compound: $$\mathrm{NiSO_4}$$.

Tin(IV) is $$\mathrm{Sn^{4+}}$$; oxide is $$\mathrm{O^{2-}}$$. To balance $$\mathrm{Sn^{4+}}$$ we need two $$\mathrm{O^{2-}}$$: $$\mathrm{SnO_2}$$.

Thallium(I) is $$\mathrm{Tl^+}$$; sulphate is $$\mathrm{SO_4^{2-}}$$. Two $$\mathrm{Tl^+}$$ balance one $$\mathrm{SO_4^{2-}}$$: $$\mathrm{Tl_2SO_4}$$.

Iron(III) is $$\mathrm{Fe^{3+}}$$; sulphate is $$\mathrm{SO_4^{2-}}$$. Cross-multiplying charges: two $$\mathrm{Fe^{3+}}$$ ($$+6$$) balance three $$\mathrm{SO_4^{2-}}$$ ($$-6$$): $$\mathrm{Fe_2(SO_4)_3}$$.

Chromium(III) is $$\mathrm{Cr^{3+}}$$; oxide is $$\mathrm{O^{2-}}$$. Cross-multiplying: two $$\mathrm{Cr^{3+}}$$ ($$+6$$) balance three $$\mathrm{O^{2-}}$$ ($$-6$$): $$\mathrm{Cr_2O_3}$$.

Two issues decide the choice of medium for oxidising toluene with $$\mathrm{KMnO_4}$$.

(i) Miscibility. Toluene is a non-polar organic liquid and is immiscible with water. Aqueous (acidic or alkaline) $$\mathrm{KMnO_4}$$ and toluene exist as two separate layers, so reaction is slow and inefficient. Adding an alcohol (e.g. ethanol) gives a homogeneous solution that dissolves both the $$\mathrm{KMnO_4}$$ (slightly) and the toluene, allowing efficient contact.

(ii) Strength control. Aqueous acidic $$\mathrm{KMnO_4}$$ ($$\mathrm{Mn(VII) \rightarrow Mn(II)}$$, $$5$$e$${}^-$$ change) is a powerful oxidant; aqueous alkaline $$\mathrm{KMnO_4}$$ ($$\mathrm{Mn(VII) \rightarrow Mn(IV)}$$, $$3$$e$${}^-$$ change) is milder but the reaction is heterogeneous. Alcoholic $$\mathrm{KMnO_4}$$ provides moderate, controlled oxidation that stops cleanly at the carboxylic acid stage (–CH$$_3$$ $$\rightarrow$$ –COOH) without burning the ring.

Balanced equation. Toluene is oxidised to potassium benzoate; permanganate is reduced to manganese dioxide:

$$\mathrm{C_6H_5CH_3 + 2\,KMnO_4 \rightarrow C_6H_5COOK + 2\,MnO_2 + KOH + H_2O}$$

Acidifying with dilute mineral acid liberates benzoic acid:

$$\mathrm{C_6H_5COOK + HCl \rightarrow C_6H_5COOH + KCl}$$

Check (Mn): $$\mathrm{Mn} : +7 \rightarrow +4$$ (gain $$3$$e$${}^-$$ × $$2 = 6$$e$${}^-$$). Methyl C: $$-3 \rightarrow +3$$ in $$\mathrm{COO^-}$$ (loss of $$6$$e$${}^-$$). Electrons balance.

Concentrated $$\mathrm{H_2SO_4}$$ is both a strong acid and a moderately strong oxidising agent.

Chloride: $$\mathrm{Cl^-}$$ is a weak reducing agent — the standard reduction potential $$E^\circ(\mathrm{Cl_2/Cl^-}) = +1.36\,\mathrm{V}$$ is high enough that conc. $$\mathrm{H_2SO_4}$$ cannot oxidise $$\mathrm{Cl^-}$$ to $$\mathrm{Cl_2}$$. Only the acid–base reaction takes place, producing colourless, pungent $$\mathrm{HCl}$$ gas:

$$\mathrm{NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl\!\uparrow}$$

Bromide: $$\mathrm{Br^-}$$ is a much stronger reducing agent ($$E^\circ(\mathrm{Br_2/Br^-}) = +1.09\,\mathrm{V}$$, lower than $$\mathrm{Cl_2/Cl^-}$$). Conc. $$\mathrm{H_2SO_4}$$ first liberates $$\mathrm{HBr}$$, but the $$\mathrm{HBr}$$ produced is then oxidised by the same conc. $$\mathrm{H_2SO_4}$$ to give red-brown $$\mathrm{Br_2}$$ vapours:

$$\mathrm{2\,NaBr + 2\,H_2SO_4 \rightarrow Na_2SO_4 + 2\,H_2O + SO_2 + Br_2\!\uparrow}$$

Equivalently: $$\mathrm{2\,HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2\,H_2O}$$.

So the difference comes from the relative reducing strengths $$\mathrm{I^- > Br^- > Cl^- > F^-}$$ — conc. $$\mathrm{H_2SO_4}$$ is strong enough to oxidise $$\mathrm{Br^-}$$ (and $$\mathrm{I^-}$$) but not $$\mathrm{Cl^-}$$.

$$\mathrm{C_6H_6O_2}$$ is hydroquinone (1,4-dihydroxybenzene) and $$\mathrm{C_6H_4O_2}$$ is benzoquinone (1,4-cyclohexadienedione). This is the classical photographic developer reaction.

Oxidation numbers of interest:

• Ag: $$+1$$ in $$\mathrm{AgBr}$$ $$\rightarrow$$ $$0$$ in metallic $$\mathrm{Ag}$$. Reduction.
• Hydroquinone loses two H atoms (–OH $$\rightarrow$$ =O) — the two phenolic O atoms go from $$-2$$ (with H) to $$-2$$ (as C=O), but the two carbons that carry those O atoms move from $$0$$ to $$+1$$ each, releasing $$2$$e$${}^-$$ overall. So hydroquinone is oxidised.

Therefore:

• Oxidised: $$\mathrm{C_6H_6O_2}$$ (hydroquinone) — it is the reducing agent.
• Reduced: $$\mathrm{AgBr}$$ (Ag goes $$+1 \rightarrow 0$$) — it is the oxidising agent.

This is Tollens' test on formaldehyde.

Oxidation numbers of interest:

• Ag in the diamminesilver complex $$\mathrm{[Ag(NH_3)_2]^+}$$: $$+1$$. Free metallic $$\mathrm{Ag}$$: $$0$$. So Ag is reduced ($$+1 \rightarrow 0$$).
• C in $$\mathrm{HCHO}$$: $$0$$ (formula $$\mathrm{CH_2O}$$: $$x + 2(+1) + (-2) = 0$$). C in $$\mathrm{HCOO^-}$$ (formate): $$+1 + x + 2(-2) = -1 \Rightarrow x = +2$$. So C is oxidised ($$0 \rightarrow +2$$).

Therefore:

• Oxidised: $$\mathrm{HCHO}$$ — reducing agent.
• Reduced: $$\mathrm{[Ag(NH_3)_2]^+}$$ — oxidising agent.

This is Fehling's test on formaldehyde.

Oxidation numbers of interest:

• Cu: $$+2$$ in $$\mathrm{Cu^{2+}}$$ $$\rightarrow +1$$ in $$\mathrm{Cu_2O}$$ (red precipitate). Reduction.
• C in $$\mathrm{HCHO}$$: $$0 \rightarrow +2$$ in $$\mathrm{HCOO^-}$$. Oxidation.

Therefore:

• Oxidised: $$\mathrm{HCHO}$$ — reducing agent.
• Reduced: $$\mathrm{Cu^{2+}}$$ — oxidising agent.

Oxidation numbers of interest:

• N in $$\mathrm{N_2H_4}$$ (hydrazine): $$2x + 4(+1) = 0 \Rightarrow x = -2$$. N in $$\mathrm{N_2}$$: $$0$$. So N: $$-2 \rightarrow 0$$ — oxidation.
• O in $$\mathrm{H_2O_2}$$ (peroxide): $$-1$$. O in $$\mathrm{H_2O}$$: $$-2$$. So O: $$-1 \rightarrow -2$$ — reduction.

Therefore:

• Oxidised: $$\mathrm{N_2H_4}$$ — reducing agent.
• Reduced: $$\mathrm{H_2O_2}$$ — oxidising agent.

This is the discharging reaction of a lead-acid battery. Oxidation numbers of interest:

• Pb (metal): $$0 \rightarrow +2$$ in $$\mathrm{PbSO_4}$$ — oxidation. So elemental $$\mathrm{Pb}$$ is oxidised.
• Pb in $$\mathrm{PbO_2}$$: $$+4 \rightarrow +2$$ in $$\mathrm{PbSO_4}$$ — reduction. So $$\mathrm{PbO_2}$$ is reduced.
• H, S, O in $$\mathrm{H_2SO_4}/\mathrm{H_2O}$$: unchanged.

Interesting case: lead acts as both the oxidised and the reduced species, but in different starting forms ($$\mathrm{Pb^0}$$ vs $$\mathrm{Pb^{IV}}$$) — this is a comproportionation.

• Oxidised: $$\mathrm{Pb}$$ — reducing agent.
• Reduced: $$\mathrm{PbO_2}$$ — oxidising agent.

Oxidation numbers:

• P in $$\mathrm{H_3PO_2}$$ (hypophosphorous acid): $$3(+1) + x + 2(-2) = 0 \Rightarrow x = +1$$.
• P in $$\mathrm{H_3PO_4}$$ (phosphoric acid): $$3 + x - 8 = 0 \Rightarrow x = +5$$.
• Ag in $$\mathrm{AgNO_3}$$: $$+1$$. Ag (metal): $$0$$.
• N in $$\mathrm{NO_3^-}$$ (both reactant and product): $$+5$$ — unchanged.

P: $$+1 \rightarrow +5$$ (loss of $$4$$e$${}^-$$). Ag: $$+1 \rightarrow 0$$ (gain of $$1$$e$${}^-$$ × 4 atoms = $$4$$e$${}^-$$). Electrons balance.

Observation: $$\mathrm{H_3PO_2}$$ reduces $$\mathrm{Ag^+}$$ to metallic Ag. So $$\mathrm{Ag^+}$$ acts as an oxidising agent in the presence of the strong reducing agent $$\mathrm{H_3PO_2}$$.

Oxidation numbers:

• P: $$+1$$ (in $$\mathrm{H_3PO_2}$$) $$\rightarrow +5$$ (in $$\mathrm{H_3PO_4}$$). Oxidation.
• Cu: $$+2$$ (in $$\mathrm{CuSO_4}$$) $$\rightarrow 0$$ (metallic Cu). Reduction.
• S, O, H stay put in spectator $$\mathrm{SO_4^{2-}}$$.

(Note: the printed equation balances atoms only if the right side reads $$2\,\mathrm{H_2SO_4}$$ rather than $$\mathrm{H_2SO_4}$$ — that does not affect which species is oxidised vs reduced.)

Observation: $$\mathrm{H_3PO_2}$$ reduces $$\mathrm{Cu^{2+}}$$ to metallic Cu — so $$\mathrm{Cu^{2+}}$$ can act as an oxidising agent in the presence of $$\mathrm{H_3PO_2}$$.

This is Tollens' test on benzaldehyde. Track the carbon of the –CHO group and the Ag of the complex:

• C of $$\mathrm{C_6H_5CHO}$$ (–CHO): 1 H ($$-1$$), 1 C of ring ($$0$$), 1 $$=\!\!\mathrm{O}$$ ($$+2$$) gives C $$= +1$$. In $$\mathrm{C_6H_5COO^-}$$, the carboxyl C: 1 ring C ($$0$$), 1 $$=\!\!\mathrm{O}$$ ($$+2$$), 1 $$\mathrm{O^-}$$ ($$+1$$) gives C $$= +3$$.
• Ag: $$+1$$ (in $$\mathrm{[Ag(NH_3)_2]^+}$$) $$\rightarrow 0$$ (in $$\mathrm{Ag}$$).

So the aldehydic C is oxidised ($$+1 \rightarrow +3$$), and Ag is reduced ($$+1 \rightarrow 0$$).

Observation: $$\mathrm{[Ag(NH_3)_2]^+}$$ (Tollens' reagent) oxidises an aromatic aldehyde — i.e. $$\mathrm{Ag^+}$$ acts as an oxidising agent even toward $$\mathrm{C_6H_5CHO}$$ (silver-mirror formed).

No reaction takes place. $$\mathrm{Cu^{2+}}$$ (Fehling's reagent) is unable to oxidise an aromatic aldehyde such as benzaldehyde — there is no red $$\mathrm{Cu_2O}$$ precipitate.

The reason is that the aldehydic C in $$\mathrm{C_6H_5CHO}$$ is stabilised by conjugation with the benzene ring (the lone-pair / π electrons of the ring delocalise into the C=O). $$\mathrm{Cu^{2+}}$$ is a weaker oxidant than $$\mathrm{Ag^+}$$ (cf. $$E^\circ(\mathrm{Cu^{2+}/Cu}) = +0.34\,\mathrm{V}$$ vs $$E^\circ(\mathrm{Ag^+/Ag}) = +0.80\,\mathrm{V}$$) and cannot overcome that stabilisation.

Tabulate the four observations:

$$\mathrm{Ag^+}$$ is reduced by both $$\mathrm{H_3PO_2}$$ (a strong reductant) and $$\mathrm{C_6H_5CHO}$$ (a weaker, aromatic aldehyde). $$\mathrm{Cu^{2+}}$$ is reduced only by the strong reductant $$\mathrm{H_3PO_2}$$ but not by the weaker $$\mathrm{C_6H_5CHO}$$.

This is consistent with the standard reduction potentials:

$$E^\circ(\mathrm{Ag^+/Ag}) = +0.80\,\mathrm{V}$$  >  $$E^\circ(\mathrm{Cu^{2+}/Cu}) = +0.34\,\mathrm{V}$$.

Inference: $$\mathrm{Ag^+}$$ is a stronger oxidising agent than $$\mathrm{Cu^{2+}}$$. It can oxidise both aliphatic-type reductants (like $$\mathrm{H_3PO_2}$$, $$\mathrm{HCHO}$$) and aromatic ones ($$\mathrm{C_6H_5CHO}$$). $$\mathrm{Cu^{2+}}$$, being weaker, only oxidises easier substrates.

Oxidation-number changes: Mn $$+7 \rightarrow +4$$ (gain of $$3$$e$${}^-$$); I $$-1 \rightarrow 0$$ (loss of $$1$$e$${}^-$$ per atom).

Step 1: write the half-reactions and balance them in acidic form first.

Reduction: $$\mathrm{MnO_4^- + 4\,H^+ + 3\,e^- \rightarrow MnO_2 + 2\,H_2O}$$

Oxidation: $$\mathrm{2\,I^- \rightarrow I_2 + 2\,e^-}$$

Step 2: balance electrons. LCM of 3 and 2 is 6. Multiply reduction by $$2$$ and oxidation by $$3$$:

$$\mathrm{2\,MnO_4^- + 8\,H^+ + 6\,e^- \rightarrow 2\,MnO_2 + 4\,H_2O}$$

$$\mathrm{6\,I^- \rightarrow 3\,I_2 + 6\,e^-}$$

Step 3: add and cancel.

$$\mathrm{2\,MnO_4^- + 6\,I^- + 8\,H^+ \rightarrow 2\,MnO_2 + 3\,I_2 + 4\,H_2O}$$

Step 4: convert to basic medium. Add $$8\,\mathrm{OH^-}$$ to both sides (to neutralise the $$8\,\mathrm{H^+}$$):

$$\mathrm{2\,MnO_4^- + 6\,I^- + 8\,H_2O \rightarrow 2\,MnO_2 + 3\,I_2 + 4\,H_2O + 8\,OH^-}$$

Cancel $$4\,\mathrm{H_2O}$$ from both sides:

$$\mathrm{2\,MnO_4^- + 6\,I^- + 4\,H_2O \rightarrow 2\,MnO_2 + 3\,I_2 + 8\,OH^-}$$

Check: Mn 2 = 2; I 6 = 6; O $$8+4=12 = 4+8$$; H $$8=8$$; charge $$-2-6 = -8 = -8$$. Balanced.

Oxidation-number changes: Mn $$+7 \rightarrow +2$$ (gain of $$5$$e$${}^-$$); S $$+4 \rightarrow +6$$ (loss of $$2$$e$${}^-$$).

Reduction half (acidic):

$$\mathrm{MnO_4^- + 8\,H^+ + 5\,e^- \rightarrow Mn^{2+} + 4\,H_2O}$$

Oxidation half (acidic): $$\mathrm{SO_2 \rightarrow HSO_4^-}$$. Balance O with $$\mathrm{H_2O}$$ and H with $$\mathrm{H^+}$$:

$$\mathrm{SO_2 + 2\,H_2O \rightarrow HSO_4^- + 3\,H^+ + 2\,e^-}$$

Check: S 1=1; O $$2+2 = 4 = 4$$; H $$4 = 1+3$$; charge $$0 \rightarrow -1+3-2 = 0$$.

Balance electrons: LCM(5,2) = 10. Multiply reduction by $$2$$ and oxidation by $$5$$:

$$\mathrm{2\,MnO_4^- + 16\,H^+ + 10\,e^- \rightarrow 2\,Mn^{2+} + 8\,H_2O}$$

$$\mathrm{5\,SO_2 + 10\,H_2O \rightarrow 5\,HSO_4^- + 15\,H^+ + 10\,e^-}$$

Add and cancel:

$$\mathrm{2\,MnO_4^- + 5\,SO_2 + 16\,H^+ + 10\,H_2O \rightarrow 2\,Mn^{2+} + 8\,H_2O + 5\,HSO_4^- + 15\,H^+}$$

Net $$\mathrm{H^+}$$: $$16 - 15 = 1$$ remains on left. Net $$\mathrm{H_2O}$$: $$10 - 8 = 2$$ remains on left.

$$\mathrm{2\,MnO_4^- + 5\,SO_2 + 2\,H_2O + H^+ \rightarrow 2\,Mn^{2+} + 5\,HSO_4^-}$$

Check: Mn 2=2; S 5=5; O $$8+10+2 = 20 = 20$$; H $$1+4=5 = 5$$; charge $$-2+1 = -1$$ vs $$+4-5 = -1$$. Balanced.

Oxidation-number changes: O in $$\mathrm{H_2O_2}$$ goes $$-1 \rightarrow -2$$ (in $$\mathrm{H_2O}$$ — gain of $$1$$e$${}^-$$ per O, $$2$$e$${}^-$$ per $$\mathrm{H_2O_2}$$); Fe goes $$+2 \rightarrow +3$$ (loss of $$1$$e$${}^-$$).

Reduction half (acidic). $$\mathrm{H_2O_2}$$ has 2 O atoms; balance O with $$\mathrm{H_2O}$$ and H with $$\mathrm{H^+}$$:

$$\mathrm{H_2O_2 + 2\,H^+ + 2\,e^- \rightarrow 2\,H_2O}$$

Oxidation half.

$$\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$$

Balance electrons. Multiply the oxidation half by $$2$$:

$$\mathrm{2\,Fe^{2+} \rightarrow 2\,Fe^{3+} + 2\,e^-}$$

Add the two halves:

$$\mathrm{H_2O_2 + 2\,Fe^{2+} + 2\,H^+ \rightarrow 2\,Fe^{3+} + 2\,H_2O}$$

Check: H $$2+2 = 4 = 4$$; O $$2 = 2$$; Fe $$2 = 2$$; charge — left $$0 + 4 + 2 = +6$$, right $$+6 + 0 = +6$$. Balanced. ($$\mathrm{H_2O_2}$$ is the oxidising agent, $$\mathrm{Fe^{2+}}$$ the reducing agent.)

Oxidation-number changes: Cr goes $$+6 \rightarrow +3$$ (gain of $$3$$e$${}^-$$ per Cr, i.e. $$6$$e$${}^-$$ per $$\mathrm{Cr_2O_7^{2-}}$$); S goes $$+4 \rightarrow +6$$ (loss of $$2$$e$${}^-$$).

Reduction half (acidic). Balance O with $$\mathrm{H_2O}$$, H with $$\mathrm{H^+}$$:

$$\mathrm{Cr_2O_7^{2-} + 14\,H^+ + 6\,e^- \rightarrow 2\,Cr^{3+} + 7\,H_2O}$$

Oxidation half (acidic). $$\mathrm{SO_2 \rightarrow SO_4^{2-}}$$:

$$\mathrm{SO_2 + 2\,H_2O \rightarrow SO_4^{2-} + 4\,H^+ + 2\,e^-}$$

Balance electrons. Multiply the oxidation half by $$3$$ so both halves involve $$6$$e$${}^-$$:

$$\mathrm{3\,SO_2 + 6\,H_2O \rightarrow 3\,SO_4^{2-} + 12\,H^+ + 6\,e^-}$$

Add and cancel.

$$\mathrm{Cr_2O_7^{2-} + 3\,SO_2 + 14\,H^+ + 6\,H_2O \rightarrow 2\,Cr^{3+} + 7\,H_2O + 3\,SO_4^{2-} + 12\,H^+}$$

Cancel $$6\,\mathrm{H_2O}$$ and $$12\,\mathrm{H^+}$$ from both sides:

$$\mathrm{Cr_2O_7^{2-} + 3\,SO_2 + 2\,H^+ \rightarrow 2\,Cr^{3+} + 3\,SO_4^{2-} + H_2O}$$

Check: Cr $$2 = 2$$; S $$3 = 3$$; O $$7 + 6 = 13 = 12 + 1$$; H $$2 = 2$$; charge — left $$-2 + 0 + 2 = 0$$, right $$+6 - 6 = 0$$. Balanced.

This is a disproportionation. Oxidation states of P: $$0$$ in $$\mathrm{P_4}$$; $$-3$$ in $$\mathrm{PH_3}$$ ($$x + 3(+1) = 0$$); $$+2$$ in $$\mathrm{HPO_2^-}$$ ($$(+1) + x + 2(-2) = -1$$). So P is simultaneously reduced ($$0 \rightarrow -3$$) and oxidised ($$0 \rightarrow +2$$).

Ion–electron method.

Reduction half (balance in acidic form first): $$\mathrm{P_4 \rightarrow 4\,PH_3}$$. Add $$12\,\mathrm{H^+}$$ for the 12 H atoms, then $$12$$e$${}^-$$:

$$\mathrm{P_4 + 12\,H^+ + 12\,e^- \rightarrow 4\,PH_3}$$

Convert to basic — add $$12\,\mathrm{OH^-}$$ to both sides ($$\mathrm{H^+ + OH^- \rightarrow H_2O}$$):

$$\mathrm{P_4 + 12\,H_2O + 12\,e^- \rightarrow 4\,PH_3 + 12\,OH^-}$$

Oxidation half (acidic form): $$\mathrm{P_4 \rightarrow 4\,HPO_2^-}$$. Add $$8\,\mathrm{H_2O}$$ for the 8 O atoms, then $$12\,\mathrm{H^+}$$ to balance H, then $$8$$e$${}^-$$:

$$\mathrm{P_4 + 8\,H_2O \rightarrow 4\,HPO_2^- + 12\,H^+ + 8\,e^-}$$

Convert to basic — add $$12\,\mathrm{OH^-}$$ to both sides and cancel $$8\,\mathrm{H_2O}$$:

$$\mathrm{P_4 + 12\,OH^- \rightarrow 4\,HPO_2^- + 4\,H_2O + 8\,e^-}$$

Balance electrons: $$\mathrm{LCM}(12, 8) = 24$$. Multiply the reduction half by $$2$$ and the oxidation half by $$3$$:

$$\mathrm{2\,P_4 + 24\,H_2O + 24\,e^- \rightarrow 8\,PH_3 + 24\,OH^-}$$

$$\mathrm{3\,P_4 + 36\,OH^- \rightarrow 12\,HPO_2^- + 12\,H_2O + 24\,e^-}$$

Add and cancel $$24\,\mathrm{OH^-}$$ and $$12\,\mathrm{H_2O}$$:

$$\mathrm{5\,P_4 + 12\,OH^- + 12\,H_2O \rightarrow 8\,PH_3 + 12\,HPO_2^-}$$

Oxidation-number method (cross-check). Each reduced P gains $$3$$e$${}^-$$ ($$0 \rightarrow -3$$); each oxidised P loses $$2$$e$${}^-$$ ($$0 \rightarrow +2$$). For the electrons to balance, (P reduced) $$\times 3 = $$ (P oxidised) $$\times 2$$, i.e. reduced : oxidised $$= 2 : 3$$. So out of every $$5$$ P atoms, $$2$$ form $$\mathrm{PH_3}$$ and $$3$$ form $$\mathrm{HPO_2^-}$$. Taking $$20$$ P atoms ($$= 5\,\mathrm{P_4}$$): $$8 \rightarrow \mathrm{PH_3}$$ and $$12 \rightarrow \mathrm{HPO_2^-}$$ — the same equation.

Atom check: P $$20 = 8 + 12$$; O $$12 + 12 = 24 = 12 \times 2$$; H $$12 + 24 = 36 = 24 + 12$$; charge $$-12 = -12$$. Balanced.

$$\mathrm{P_4}$$ acts as both the oxidising agent and the reducing agent (disproportionation).

Oxidation states: N in $$\mathrm{N_2H_4}$$ is $$-2$$ ($$2x + 4(+1) = 0$$); N in $$\mathrm{NO}$$ is $$+2$$. Cl in $$\mathrm{ClO_3^-}$$ is $$+5$$ ($$x + 3(-2) = -1$$); Cl in $$\mathrm{Cl^-}$$ is $$-1$$. So N is oxidised ($$-2 \rightarrow +2$$, loss of $$4$$e$${}^-$$ per N) and Cl is reduced ($$+5 \rightarrow -1$$, gain of $$6$$e$${}^-$$).

Ion–electron method.

Oxidation half (acidic form): $$\mathrm{N_2H_4 \rightarrow 2\,NO}$$ — each N loses $$4$$e$${}^-$$, so $$8$$e$${}^-$$ in all. Balance O with $$\mathrm{H_2O}$$, H with $$\mathrm{H^+}$$:

$$\mathrm{N_2H_4 + 2\,H_2O \rightarrow 2\,NO + 8\,H^+ + 8\,e^-}$$

Reduction half (acidic form): $$\mathrm{ClO_3^- \rightarrow Cl^-}$$:

$$\mathrm{ClO_3^- + 6\,H^+ + 6\,e^- \rightarrow Cl^- + 3\,H_2O}$$

Balance electrons: $$\mathrm{LCM}(8, 6) = 24$$. Multiply the oxidation half by $$3$$ and the reduction half by $$4$$:

$$\mathrm{3\,N_2H_4 + 6\,H_2O \rightarrow 6\,NO + 24\,H^+ + 24\,e^-}$$

$$\mathrm{4\,ClO_3^- + 24\,H^+ + 24\,e^- \rightarrow 4\,Cl^- + 12\,H_2O}$$

Add and cancel $$24\,\mathrm{H^+}$$ and $$6\,\mathrm{H_2O}$$:

$$\mathrm{3\,N_2H_4 + 4\,ClO_3^- \rightarrow 6\,NO + 4\,Cl^- + 6\,H_2O}$$

Every $$\mathrm{H^+}$$ cancelled, so the equation is identical in acidic and basic medium — no $$\mathrm{OH^-}$$ needs to be added.

Oxidation-number method (cross-check). Electrons lost $$=$$ electrons gained: $$3\,\mathrm{N_2H_4}$$ supply $$6\,\mathrm{N} \times 4 = 24$$e$${}^-$$; $$4\,\mathrm{ClO_3^-}$$ accept $$4 \times 6 = 24$$e$${}^-$$. Balanced.

Atom check: N $$6 = 6$$; H $$12 = 12$$; Cl $$4 = 4$$; O $$12 = 6 + 6$$; charge $$-4 = -4$$.

Oxidising agent: $$\mathrm{ClO_3^-}$$ (Cl is reduced). Reducing agent: $$\mathrm{N_2H_4}$$ (N is oxidised).

Oxidation states: Cl in $$\mathrm{Cl_2O_7}$$ is $$+7$$ ($$2x + 7(-2) = 0$$); Cl in $$\mathrm{ClO_2^-}$$ is $$+3$$ ($$x + 2(-2) = -1$$). O in $$\mathrm{H_2O_2}$$ is $$-1$$ (peroxide); O in $$\mathrm{O_2}$$ is $$0$$. So Cl is reduced ($$+7 \rightarrow +3$$, gain of $$4$$e$${}^-$$ per Cl, i.e. $$8$$e$${}^-$$ per $$\mathrm{Cl_2O_7}$$) and the peroxide O is oxidised ($$-1 \rightarrow 0$$, loss of $$1$$e$${}^-$$ per O, $$2$$e$${}^-$$ per $$\mathrm{H_2O_2}$$).

Ion–electron method.

Reduction half (acidic form): $$\mathrm{Cl_2O_7 \rightarrow 2\,ClO_2^-}$$. Balance O with $$3\,\mathrm{H_2O}$$ on the right, then H with $$6\,\mathrm{H^+}$$ on the left, then $$8$$e$${}^-$$:

$$\mathrm{Cl_2O_7 + 6\,H^+ + 8\,e^- \rightarrow 2\,ClO_2^- + 3\,H_2O}$$

Oxidation half:

$$\mathrm{H_2O_2 \rightarrow O_2 + 2\,H^+ + 2\,e^-}$$

Balance electrons: multiply the oxidation half by $$4$$:

$$\mathrm{4\,H_2O_2 \rightarrow 4\,O_2 + 8\,H^+ + 8\,e^-}$$

Add and cancel $$6\,\mathrm{H^+}$$:

$$\mathrm{Cl_2O_7 + 4\,H_2O_2 \rightarrow 2\,ClO_2^- + 4\,O_2 + 3\,H_2O + 2\,H^+}$$

This is the form matching the given skeleton. To express it in basic medium, add $$2\,\mathrm{OH^-}$$ to both sides and combine $$\mathrm{H^+ + OH^- \rightarrow H_2O}$$:

$$\mathrm{Cl_2O_7 + 4\,H_2O_2 + 2\,OH^- \rightarrow 2\,ClO_2^- + 4\,O_2 + 5\,H_2O}$$

Check (basic form): Cl $$2 = 2$$; O $$7 + 8 + 2 = 17 = 4 + 8 + 5$$; H $$8 + 2 = 10 = 10$$; charge $$-2 = -2$$. Balanced.

Oxidation-number method (cross-check). $$\mathrm{Cl_2O_7}$$ gains $$8$$e$${}^-$$; $$4\,\mathrm{H_2O_2}$$ lose $$4 \times 2 = 8$$e$${}^-$$. Balanced.

Oxidising agent: $$\mathrm{Cl_2O_7}$$ (Cl is reduced). Reducing agent: $$\mathrm{H_2O_2}$$ (peroxide O is oxidised).

An element shows only a negative oxidation state if it is so electronegative that no other element can pull electrons away from it. Fluorine is the most electronegative element of all — nothing can oxidise it. Hence in every compound F is $$-1$$ (and $$0$$ as the free element $$\mathrm{F_2}$$); it never shows a positive oxidation state.

Caesium is an alkali metal with configuration $$[\mathrm{Xe}]6s^1$$. It has a very low ionisation enthalpy and readily loses its single valence electron to give $$\mathrm{Cs^+}$$ ($$+1$$). It has essentially no tendency to gain electrons, so it never shows a negative oxidation state — only $$+1$$ (and $$0$$ as the free metal).

Iodine is a halogen ($$[\mathrm{Kr}]4d^{10}5s^25p^5$$). With more electropositive elements it gains an electron and shows the negative state $$-1$$ (e.g. in $$\mathrm{KI}$$ and $$\mathrm{HI}$$). With the more electronegative oxygen and fluorine it shows positive states $$+1, +3, +5, +7$$ (e.g. in $$\mathrm{ICl}$$, $$\mathrm{HIO_3}$$, $$\mathrm{HIO_4}$$, $$\mathrm{IF_7}$$). Hence iodine exhibits both positive and negative oxidation states.

Neon is a noble gas with a completely filled, very stable octet ($$[\mathrm{He}]2s^22p^6$$). It has neither a tendency to lose electrons (very high ionisation enthalpy) nor to gain them (positive electron-gain enthalpy). It forms no normal compounds, so it shows only the oxidation state $$0$$ — neither positive nor negative.

A non-metal can disproportionate only if it can exist in an intermediate oxidation state — one from which it can move both up and down. Among the p-block non-metals:

• Phosphorus — e.g. $$\mathrm{P_4}$$ (state $$0$$, intermediate between $$-3$$ and $$+5$$): $$\mathrm{P_4 + 3\,OH^- + 3\,H_2O \rightarrow PH_3 + 3\,H_2PO_2^-}$$.
• Sulphur — intermediate states between $$-2$$ and $$+6$$ are available.
• Chlorine, bromine and iodine — they have intermediate states ($$0, +1, +3, +5$$) between $$-1$$ and $$+7$$, e.g. $$\mathrm{Cl_2 + 2\,OH^- \rightarrow Cl^- + ClO^- + H_2O}$$.

Fluorine cannot disproportionate — being the most electronegative element it shows only $$-1$$ and $$0$$, with no higher (positive) state to move to.

A metal disproportionates when it can exist in an intermediate (usually low) oxidation state that is unstable with respect to both a higher and a lower state. Three such metals:

• Copper: $$\mathrm{Cu^+}$$ ($$+1$$) is intermediate between $$\mathrm{Cu}$$ ($$0$$) and $$\mathrm{Cu^{2+}}$$ ($$+2$$): $$\mathrm{2\,Cu^+ \rightarrow Cu + Cu^{2+}}$$.
• Gallium: $$\mathrm{Ga^+}$$ ($$+1$$) is intermediate between $$\mathrm{Ga}$$ ($$0$$) and $$\mathrm{Ga^{3+}}$$ ($$+3$$): $$\mathrm{3\,Ga^+ \rightarrow 2\,Ga + Ga^{3+}}$$.
• Indium: $$\mathrm{In^+}$$ ($$+1$$) is intermediate between $$\mathrm{In}$$ ($$0$$) and $$\mathrm{In^{3+}}$$ ($$+3$$): $$\mathrm{3\,In^+ \rightarrow 2\,In + In^{3+}}$$.

(Manganese is a further example — $$\mathrm{Mn^{3+}}$$ disproportionates to $$\mathrm{Mn^{2+}}$$ and $$\mathrm{MnO_2}$$.)

Relevant standard reduction potentials: $$E^\circ(\mathrm{Fe^{3+}/Fe^{2+}}) = +0.77\ \mathrm{V}$$ and $$E^\circ(\mathrm{I_2/I^-}) = +0.54\ \mathrm{V}$$.

For the proposed reaction, $$\mathrm{Fe^{3+}}$$ is reduced (it is the oxidant) and $$\mathrm{I^-}$$ is oxidised. The cell potential is

$$E^\circ_{\text{cell}} = E^\circ(\text{oxidant couple}) - E^\circ(\text{couple of species oxidised})$$

$$E^\circ_{\text{cell}} = 0.77 - 0.54 = +0.23\ \mathrm{V}$$

Since $$E^\circ_{\text{cell}} > 0$$, the reaction is feasible:

$$\mathrm{2\,Fe^{3+} + 2\,I^- \rightarrow 2\,Fe^{2+} + I_2}$$

Relevant standard reduction potentials: $$E^\circ(\mathrm{Ag^+/Ag}) = +0.80\ \mathrm{V}$$ and $$E^\circ(\mathrm{Cu^{2+}/Cu}) = +0.34\ \mathrm{V}$$.

$$\mathrm{Ag^+}$$ would be reduced (oxidant) and metallic $$\mathrm{Cu}$$ oxidised to $$\mathrm{Cu^{2+}}$$:

$$E^\circ_{\text{cell}} = 0.80 - 0.34 = +0.46\ \mathrm{V}$$

Since $$E^\circ_{\text{cell}} > 0$$, the reaction is feasible:

$$\mathrm{2\,Ag^+ + Cu \rightarrow 2\,Ag + Cu^{2+}}$$

Relevant standard reduction potentials: $$E^\circ(\mathrm{Fe^{3+}/Fe^{2+}}) = +0.77\ \mathrm{V}$$ and $$E^\circ(\mathrm{Cu^{2+}/Cu}) = +0.34\ \mathrm{V}$$.

$$\mathrm{Fe^{3+}}$$ would be reduced to $$\mathrm{Fe^{2+}}$$ (oxidant) and metallic $$\mathrm{Cu}$$ oxidised to $$\mathrm{Cu^{2+}}$$:

$$E^\circ_{\text{cell}} = 0.77 - 0.34 = +0.43\ \mathrm{V}$$

Since $$E^\circ_{\text{cell}} > 0$$, the reaction is feasible:

$$\mathrm{2\,Fe^{3+} + Cu \rightarrow 2\,Fe^{2+} + Cu^{2+}}$$

Here metallic $$\mathrm{Ag}$$ would have to be oxidised to $$\mathrm{Ag^+}$$ while $$\mathrm{Fe^{3+}}$$ is reduced to $$\mathrm{Fe^{2+}}$$. Relevant standard reduction potentials: $$E^\circ(\mathrm{Fe^{3+}/Fe^{2+}}) = +0.77\ \mathrm{V}$$ and $$E^\circ(\mathrm{Ag^+/Ag}) = +0.80\ \mathrm{V}$$.

$$E^\circ_{\text{cell}} = E^\circ(\mathrm{Fe^{3+}/Fe^{2+}}) - E^\circ(\mathrm{Ag^+/Ag}) = 0.77 - 0.80 = -0.03\ \mathrm{V}$$

Since $$E^\circ_{\text{cell}} < 0$$, the reaction is not feasible — metallic silver cannot reduce $$\mathrm{Fe^{3+}}$$.

Relevant standard reduction potentials: $$E^\circ(\mathrm{Br_2/Br^-}) = +1.09\ \mathrm{V}$$ and $$E^\circ(\mathrm{Fe^{3+}/Fe^{2+}}) = +0.77\ \mathrm{V}$$.

$$\mathrm{Br_2}$$ would be reduced to $$\mathrm{Br^-}$$ (oxidant) and $$\mathrm{Fe^{2+}}$$ oxidised to $$\mathrm{Fe^{3+}}$$:

$$E^\circ_{\text{cell}} = 1.09 - 0.77 = +0.32\ \mathrm{V}$$

Since $$E^\circ_{\text{cell}} > 0$$, the reaction is feasible:

$$\mathrm{Br_2 + 2\,Fe^{2+} \rightarrow 2\,Br^- + 2\,Fe^{3+}}$$

The solution contains $$\mathrm{Ag^+}$$, $$\mathrm{NO_3^-}$$ and $$\mathrm{H_2O}$$.

At the cathode (reduction): the contest is between $$\mathrm{Ag^+ + e^- \rightarrow Ag}$$ ($$E^\circ = +0.80\ \mathrm{V}$$) and the reduction of water. $$\mathrm{Ag^+}$$ is much more easily reduced, so silver metal is deposited:

$$\mathrm{Ag^+\,(aq) + e^- \rightarrow Ag\,(s)}$$

At the anode (oxidation): the silver electrode is an active (attackable) electrode. Oxidising the silver of the anode is easier than oxidising water (to $$\mathrm{O_2}$$) or $$\mathrm{NO_3^-}$$, so the silver anode itself dissolves:

$$\mathrm{Ag\,(s) \rightarrow Ag^+\,(aq) + e^-}$$

Net effect: silver simply transfers from the anode to the cathode and the concentration of $$\mathrm{AgNO_3}$$ in solution stays unchanged. (This is the principle of electro-refining/electroplating of silver.)

At the cathode (reduction): $$\mathrm{Ag^+}$$ ($$E^\circ = +0.80\ \mathrm{V}$$) is reduced in preference to water, so silver is deposited:

$$\mathrm{Ag^+\,(aq) + e^- \rightarrow Ag\,(s)}$$

At the anode (oxidation): platinum is an inert electrode and does not dissolve. $$\mathrm{NO_3^-}$$ cannot be oxidised (N is already at its maximum, $$+5$$). Therefore water is oxidised and $$\mathrm{O_2}$$ gas is liberated:

$$\mathrm{2\,H_2O\,(l) \rightarrow O_2\,(g) + 4\,H^+\,(aq) + 4\,e^-}$$

So silver is deposited at the cathode, oxygen gas is evolved at the anode, and the solution gradually becomes acidic (effectively $$\mathrm{HNO_3}$$ is formed).

The solution contains $$\mathrm{H^+}$$, $$\mathrm{SO_4^{2-}}$$ and $$\mathrm{H_2O}$$; the electrodes are inert.

At the cathode (reduction): hydrogen ions are reduced to hydrogen gas:

$$\mathrm{2\,H^+\,(aq) + 2\,e^- \rightarrow H_2\,(g)}$$

At the anode (oxidation): $$\mathrm{SO_4^{2-}}$$ cannot be oxidised (S is already at $$+6$$). Water is oxidised instead:

$$\mathrm{2\,H_2O\,(l) \rightarrow O_2\,(g) + 4\,H^+\,(aq) + 4\,e^-}$$

The overall change is therefore the electrolysis of water — $$\mathrm{H_2}$$ at the cathode and $$\mathrm{O_2}$$ at the anode in a $$2 : 1$$ volume ratio — while the $$\mathrm{H_2SO_4}$$ merely carries the current and becomes slightly more concentrated.

The solution contains $$\mathrm{Cu^{2+}}$$, $$\mathrm{Cl^-}$$ and $$\mathrm{H_2O}$$; the electrodes are inert.

At the cathode (reduction): $$\mathrm{Cu^{2+}}$$ ($$E^\circ(\mathrm{Cu^{2+}/Cu}) = +0.34\ \mathrm{V}$$) is reduced in preference to water, so copper is deposited:

$$\mathrm{Cu^{2+}\,(aq) + 2\,e^- \rightarrow Cu\,(s)}$$

At the anode (oxidation): although the oxidation of water to $$\mathrm{O_2}$$ has a slightly lower standard potential, the high overvoltage of oxygen on platinum (together with the appreciable concentration of $$\mathrm{Cl^-}$$) means that chloride is oxidised to chlorine gas:

$$\mathrm{2\,Cl^-\,(aq) \rightarrow Cl_2\,(g) + 2\,e^-}$$

So copper is deposited at the cathode and chlorine gas is liberated at the anode.

In this cell zinc is oxidised and silver ion is reduced, so the galvanic cell is represented as

$$\mathrm{Zn\,(s)\,|\,Zn^{2+}\,(aq)\,||\,Ag^+\,(aq)\,|\,Ag\,(s)}$$

The zinc electrode is the anode: the half-reaction $$\mathrm{Zn \rightarrow Zn^{2+} + 2\,e^-}$$ occurs there, leaving electrons behind on the metal. In a galvanic (voltaic) cell the anode is the source of electrons and is therefore the negatively charged electrode. (The silver electrode, the cathode, is positively charged.)

The current is carried by different species in different parts of the circuit:

• In the external circuit (the metallic wire joining the electrodes): the current is carried by electrons, which flow from the zinc (anode, $$-$$) to the silver (cathode, $$+$$).
• Inside the cell — within the two electrolyte solutions and the salt bridge: the current is carried by ions. Cations ($$\mathrm{Zn^{2+}}$$, $$\mathrm{Ag^+}$$, and cations from the salt bridge) migrate towards the cathode, while anions (e.g. $$\mathrm{NO_3^-}$$ from the salt bridge) migrate towards the anode, completing the circuit.

At the anode (zinc electrode) — oxidation:

$$\mathrm{Zn\,(s) \rightarrow Zn^{2+}\,(aq) + 2\,e^-}$$

At the cathode (silver electrode) — reduction:

$$\mathrm{Ag^+\,(aq) + e^- \rightarrow Ag\,(s)}$$

To match the two electrons released by each zinc atom, the cathode half-reaction is multiplied by $$2$$:

$$\mathrm{2\,Ag^+\,(aq) + 2\,e^- \rightarrow 2\,Ag\,(s)}$$

Adding the two half-reactions reproduces the overall cell reaction:

$$\mathrm{Zn\,(s) + 2\,Ag^+\,(aq) \rightarrow Zn^{2+}\,(aq) + 2\,Ag\,(s)}$$