Summary and Quick Revision

Card 1 — The three definitions, lined up on one example

Classical. Oxidation is the addition of oxygen or of a more electronegative element to a substance, or the removal of hydrogen or of a more electropositive element from it. Reduction is the same four clauses reversed.

Electronic. Oxidation is the loss of electrons by a species. Reduction is the gain of electrons by a species.

Oxidation number. Oxidation is an increase in oxidation number. Reduction is a decrease.

All three must agree on any reaction. Run them on magnesium burning in air:

$$\mathrm{2Mg(s) + O_2(g) \rightarrow 2MgO(s)}$$

Definition	Verdict on $\mathrm{Mg}$	Verdict on $\mathrm{O_2}$
Classical	oxygen added to it, so oxidised	electropositive Mg added to it, so reduced
Electronic	$\mathrm{Mg \rightarrow Mg^{2+} + 2e^-}$, lost, oxidised	$\mathrm{O_2 + 4e^- \rightarrow 2O^{2-}}$, gained, reduced
Oxidation number	$0 \rightarrow +2$, a rise, oxidised	$0 \rightarrow -2$, a fall, reduced

Three vocabularies, one verdict. If your three disagree, one has been applied wrongly, almost always the classical one stretched past its range.

Where each stops. The classical scheme says nothing about $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$, which holds no oxygen and no hydrogen; the electronic definition is exact only for ionic compounds, the shared pair in $\mathrm{HCl}$ merely shifting. The oxidation number fixes both gaps by pretence: give every bonding pair to the more electronegative atom, then count charges as if the transfer were complete.

Key Point: Oxidation and reduction always occur together, in the same reaction, at the same time. A half reaction is a bookkeeping device, not a reaction that happens by itself.

Card 2 — The table students most often invert

Key Point: The oxidising agent is the electron acceptor and is itself REDUCED. The reducing agent is the electron donor and is itself OXIDISED.

	Oxidation	Reduction
Electrons	lost	gained
Oxidation number	rises	falls
Classical test	oxygen or an electronegative element added; hydrogen or an electropositive element removed	oxygen or an electronegative element removed; hydrogen or an electropositive element added
In a half reaction, $\mathrm{e^-}$ sits on the	right	left
The species doing this is the	reducing agent	oxidising agent
At which electrode	anode	cathode

The word "oxidising" describes what the agent does to its partner, not what happens to it. Zinc is the reducing agent in $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$ and copper is in $\mathrm{Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag}$: the label belongs to the role, never to the element.

The strength rule alongside it. A metal high in the activity series is a strong reducing agent and its cation a weak oxidising agent; low in the series, the reverse. A metal and its own ion are never both strong. Fluorine is the strongest oxidising agent among the halogens, iodide the strongest reducing agent among the halide ions, and lithium the strongest reducing agent in aqueous solution, because of its very large hydration enthalpy, even though caesium has the lowest ionisation enthalpy.

[JEE/NEET] Answer "name the oxidising agent" by finding the species that gained electrons, never by matching the word in the stem to the word in an option.

Card 3 — The oxidation number rules, in priority order

Apply these top down. When two rules give different answers for the same atom, the one higher on the list wins.

Rule 1 — A free element is $0$. $\mathrm{Na}$, $\mathrm{Mg}$, and every elemental molecule: $\mathrm{H_2}$, $\mathrm{O_2}$, $\mathrm{O_3}$, $\mathrm{N_2}$, $\mathrm{P_4}$, $\mathrm{S_8}$, graphite.

Rule 2 — A monatomic ion equals its charge. $\mathrm{Na^+}$ is $+1$, $\mathrm{Fe^{3+}}$ is $+3$, $\mathrm{Cl^-}$ is $-1$.

Rule 3 — Fluorine is $-1$, always, in every compound.

Rule 4 — Group 1 metals are $+1$, group 2 metals are $+2$, in all their compounds. Aluminium is always $+3$.

Rule 5 — Hydrogen is $+1$, except where it is the more electronegative partner, and there it is $-1$: bonded to a metal, as in $\mathrm{LiH}$, $\mathrm{NaH}$, $\mathrm{CaH_2}$, $\mathrm{MgH_2}$, or to boron, as in $\mathrm{NaBH_4}$.

Rule 6 — Oxygen is $-2$, except in the four cases below.

Situation	Oxygen gets	Examples
Ordinary compounds	$-2$	$\mathrm{H_2O}$, $\mathrm{CO_2}$, $\mathrm{H_2SO_4}$, $\mathrm{KMnO_4}$
Peroxides, an O-O single bond	$-1$	$\mathrm{H_2O_2}$, $\mathrm{Na_2O_2}$, $\mathrm{BaO_2}$, $\mathrm{CaO_2}$
Superoxides, the $\mathrm{O_2^-}$ ion	$-\frac{1}{2}$	$\mathrm{KO_2}$, $\mathrm{RbO_2}$
Bonded to fluorine	positive	$\mathrm{OF_2}$ gives $+2$, $\mathrm{O_2F_2}$ gives $+1$

Rule 7 — The sum rule. The oxidation numbers in a neutral molecule add to $0$; in a polyatomic ion they add to the charge on the ion. $\mathrm{SO_4^{2-}}$ totals $-2$, $\mathrm{NH_4^+}$ totals $+1$.

Rules 1 to 6 fix what you know; rule 7 is the equation you solve for what you do not.

The three clashes the priority settles. Rule 3 beats rule 6 in $\mathrm{OF_2}$, forcing oxygen to $+2$; rule 4 beats rule 6 in $\mathrm{Na_2O_2}$, pushing oxygen to $-1$; rule 4 beats rule 5 in $\mathrm{NaH}$, driving hydrogen to $-1$.

The ceiling. A main-group element stops at its group number in groups 1 and 2, and at the group number minus $10$ from group 13 onwards: phosphorus at $+5$, sulphur at $+6$, chlorine at $+7$. A transition metal instead stops at the number of $d$ and $s$ electrons it can lose, which for chromium is six, giving the $+6$ of chromate and dichromate. An answer above the ceiling means a peroxide linkage was missed. Only fluorine can never go positive.

Card 4 — The twelve trap species, with the one-line reason

Species	Element	Value	The one-line reason
$\mathrm{H_2SO_5}$, Caro's acid	$\mathrm{S}$	$+6$	one peroxide linkage: three O at $-2$, two at $-1$. Blind gives $+8$
$\mathrm{H_2S_2O_8}$, Marshall's acid	$\mathrm{S}$	$+6$	one peroxide bridge in $\mathrm{HO-SO_2-O-O-SO_2-OH}$. Blind gives $+7$
$\mathrm{CrO_5}$	$\mathrm{Cr}$	$+6$	butterfly $\mathrm{CrO(O_2)_2}$: one O at $-2$, four at $-1$. Blind gives $+10$
$\mathrm{KO_2}$	$\mathrm{O}$	$-\frac{1}{2}$	superoxide ion $\mathrm{O_2^-}$, one extra electron over two identical oxygens
$\mathrm{OF_2}$	$\mathrm{O}$	$+2$	fluorine outranks oxygen and keeps $-1$, so oxygen absorbs the rest
$\mathrm{O_2F_2}$	$\mathrm{O}$	$+1$	same reason, two oxygens sharing the $+2$
$\mathrm{Fe_3O_4}$	$\mathrm{Fe}$	$+\frac{8}{3}$	average of one Fe(II) and two Fe(III); $\mathrm{FeO \cdot Fe_2O_3}$
$\mathrm{Na_2S_4O_6}$	$\mathrm{S}$	$+\frac{5}{2}$	average along $\mathrm{[\,O_3S-S-S-SO_3\,]^{2-}}$: really $+5, 0, 0, +5$
$\mathrm{C_3O_2}$	$\mathrm{C}$	$+\frac{4}{3}$	average along $\mathrm{O=C=C=C=O}$: really $+2, 0, +2$
$\mathrm{Br_3O_8}$	$\mathrm{Br}$	$+\frac{16}{3}$	average of two terminal Br at $+6$ and one middle Br at $+4$
$\mathrm{NaBH_4}$	$\mathrm{B}$	$+3$	hydrogen is bonded to boron and beats it for electronegativity, so each H is $-1$
$\mathrm{CaH_2}$	$\mathrm{H}$	$-1$	calcium is fixed at $+2$ by rule 4, so hydrogen is driven negative

Two habits clear the list. Check the ceiling first, since $+8$ for sulphur or $+10$ for chromium should stop you dead and the missing piece is always an O-O bond. Then, whenever you write a fraction, add the whole numbers in half a sentence. The superoxide is the exception: the two oxygens in $\mathrm{KO_2}$ are identical, so $-\frac{1}{2}$ is a genuine per-atom value, as is $+\frac{1}{2}$ in $\mathrm{O_2^+}$.

Three near-traps need no correction. $\mathrm{H_2S_2O_7}$ has an $\mathrm{S-O-S}$ bridge, so sulphur is $+6$ outright. $\mathrm{Pb_3O_4}$ averages $+\frac{8}{3}$ but splits $+2, +2, +4$, where $\mathrm{Fe_3O_4}$ and $\mathrm{Mn_3O_4}$ split $+2, +3, +3$. Ammonium nitrate averages a whole $+1$ describing neither nitrogen, $-3$ and $+5$.

Card 5 — Stock notation

Key Point (Definition): In Stock notation the oxidation number of an element is written as a Roman numeral in parentheses, placed immediately after the name or symbol of that element, with no space before the bracket. $\mathrm{FeO}$ is iron(II) oxide, written $\mathrm{Fe(II)O}$.

Roman numeral in capitals, never 2 or ii. No space before the bracket. No sign, though zero is written where it matters, as in nickel(0) tetracarbonyl. In a formula the numeral follows the symbol, $\mathrm{Sn(II)Cl_2}$. And the rule tested on its own: the numeral is the oxidation number of one atom, so $\mathrm{Fe_2O_3}$ is iron(III) oxide and never iron(VI) oxide, the subscript having already counted the atoms.

Used for Fe, Cu, Sn, Pb, Mn, Cr, Au and Hg; not for Na, K, Mg, Ca, Al, Zn and F, which have one common state each.

Formula	Stock name
$\mathrm{FeO}$ / $\mathrm{Fe_2O_3}$	iron(II) oxide / iron(III) oxide
$\mathrm{Cu_2O}$ / $\mathrm{CuO}$	copper(I) oxide / copper(II) oxide
$\mathrm{SnCl_2}$ / $\mathrm{SnCl_4}$	tin(II) chloride / tin(IV) chloride
$\mathrm{MnO_2}$	manganese(IV) oxide
$\mathrm{KMnO_4}$	potassium manganate(VII), commonly potassium permanganate
$\mathrm{Hg_2Cl_2}$ / $\mathrm{HgCl_2}$	mercury(I) chloride / mercury(II) chloride
$\mathrm{AuCl_3}$	gold(III) chloride
$\mathrm{Fe_3O_4}$	$\mathrm{Fe(II)Fe_2(III)O_4}$

Name to formula. The numeral gives the charge on one cation: write the anion charge, combine to zero, reduce to lowest whole numbers, so lead(IV) oxide crosses to $\mathrm{Pb_2O_4}$ and reduces to $\mathrm{PbO_2}$. Mercury(I) chloride is the one place the recipe misleads, the $+1$ state being the bonded pair $\mathrm{Hg_2^{2+}}$, so the formula is $\mathrm{Hg_2Cl_2}$.

Card 6 — The four types, with the criterion and one example each

Two questions get asked of any equation. Is it redox at all? Assign oxidation numbers on both sides, and if even one changes, it is. What shape does it have? That is what the four classes name.

Class	General form	Redox when	Example
Combination	$\mathrm{A + B \rightarrow C}$	at least one reactant is a free element	$\mathrm{C + O_2 \rightarrow CO_2}$
Decomposition	$\mathrm{C \rightarrow A + B}$	at least one product is a free element	$\mathrm{2KClO_3 \rightarrow 2KCl + 3O_2}$
Displacement	$\mathrm{X + YZ \rightarrow XZ + Y}$	a free element takes the place of a combined one	$\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}$
Disproportionation	one substance in, two products of the same element	that element ends in both a higher and a lower state	$\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$

Metal displacement runs down the activity series, $\mathrm{K > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Au}$, a metal displacing any below it; the metallurgical versions are $\mathrm{V_2O_5 + 5Ca}$, $\mathrm{TiCl_4 + 2Mg}$ and the thermite $\mathrm{Cr_2O_3 + 2Al}$. Non-metal displacement takes hydrogen from cold water with the alkali metals and Ca, Sr and Ba, from hot water with magnesium, from steam with iron, and from dilute acid with any metal above hydrogen. Halogen displacement follows $\mathrm{F_2 > Cl_2 > Br_2 > I_2}$, with $E^\circ$ values $+2.87$, $+1.36$, $+1.09$ and $+0.54\ \mathrm{V}$, so $\mathrm{Cl_2 + 2Br^-}$, $\mathrm{Cl_2 + 2I^-}$ and $\mathrm{Br_2 + 2I^-}$ go and no reverse does. Oxygen displacement is the rare one, $\mathrm{2H_2O + 2F_2 \rightarrow 4HF + O_2}$.

One reaction can carry two honest labels: $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$ is a decomposition by shape and a disproportionation by oxidation number. The traps with an ordinary shape and no redox are $\mathrm{CaCO_3 \rightarrow CaO + CO_2}$, its reverse, $\mathrm{NH_3 + HCl \rightarrow NH_4Cl}$ and $\mathrm{SO_3 + H_2O \rightarrow H_2SO_4}$, with every neutralisation and precipitation.

[JEE/NEET] The commonest item here asks which of four equations is not redox. Run the oxidation numbers before you look at the shape.

Card 7 — Disproportionation, the fluorine exception, and the reverse

Key Point (Definition): In a disproportionation reaction, one element in a single oxidation state is simultaneously oxidised and reduced, part of it ending in a higher state and part in a lower one.

The two conditions, both worth writing out in an answer.

1. The element must start in an intermediate oxidation state, not its highest and not its lowest.
2. Both a higher and a lower oxidation state must be accessible to it.

So the substance needs an element with at least three states, sitting in a middle one.

The shortlist.

$$\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2} \qquad \text{O from } -1 \text{ to } -2 \text{ and to } 0$$

$$\mathrm{Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O} \qquad \text{cold dilute alkali}$$

$$\mathrm{3Cl_2 + 6OH^- \rightarrow 5Cl^- + ClO_3^- + 3H_2O} \qquad \text{hot concentrated alkali}$$

$$\mathrm{P_4 + 3OH^- + 3H_2O \rightarrow PH_3 + 3H_2PO_2^-} \qquad \mathrm{2Cu^+ \rightarrow Cu^{2+} + Cu}$$

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

Cannot disproportionate: anything in its highest state, so $\mathrm{ClO_4^-}$, $\mathrm{MnO_4^-}$, $\mathrm{SO_4^{2-}}$ and $\mathrm{NO_3^-}$ can only be reduced; anything in its lowest, so $\mathrm{Cl^-}$ and $\mathrm{H_2S}$ can only be oxidised.

The fluorine exception. Fluorine never disproportionates, having no positive oxidation state: it is $0$ in $\mathrm{F_2}$ and $-1$ in every compound. With cold dilute alkali it gives $\mathrm{2F_2 + 2OH^- \rightarrow 2F^- + OF_2 + H_2O}$, and fluorine is $-1$ in both products; the element oxidised is oxygen, from $-2$ to $+2$.

Key Point (Definition): In a comproportionation reaction, two species holding the same element in a higher and a lower oxidation state react to give one product in which that element is in an intermediate state. It is disproportionation run backwards.

$$\mathrm{5Cl^- + ClO_3^- + 6H^+ \rightarrow 3Cl_2 + 3H_2O} \qquad \mathrm{IO_3^- + 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O}$$

$$\mathrm{2H_2S + SO_2 \rightarrow 3S + 2H_2O} \qquad \mathrm{2Fe^{3+} + Fe \rightarrow 3Fe^{2+}}$$

[JEE Main] Two states in and one out is comproportionation; one state in and two out is disproportionation. Read the direction before you name it.

Card 8 — Balancing by the oxidation number method

The whole method rests on one identity: total increase in oxidation number = total decrease in oxidation number.

Step 1. Write the skeletal equation with correct formulae for every reactant and product.

Step 2. Assign oxidation numbers to every atom and mark the ones that change. One element goes up and one goes down; in a disproportionation the same element does both.

Step 3. Compute the total increase and the total decrease per formula unit: change per atom multiplied by the number of atoms in the formula unit.

Step 4. Equalise the two totals by multiplying the oxidised and reduced species, the lowest common multiple giving the smallest coefficients.

Step 5. Balance every other atom except H and O.

Step 6. Balance oxygen with $\mathrm{H_2O}$, then hydrogen with $\mathrm{H^+}$. Oxygen first, always, because water carries hydrogen with it.

Step 7. Convert to basic medium if the question says basic: add as many $\mathrm{OH^-}$ to both sides as there are $\mathrm{H^+}$, combine each pair into one $\mathrm{H_2O}$, then cancel water now on both sides.

Step 8. Verify. Count every element on both sides, then total the charges on both sides. Both must match.

Step 3 decides the answer. In $\mathrm{Cr_2O_7^{2-} \rightarrow 2Cr^{3+}}$ each chromium falls $3$ over two atoms, so the decrease is $6$, not $3$; in $\mathrm{C_2O_4^{2-} \rightarrow 2CO_2}$ the increase is $2$, not $1$. Iron gives $1$ against a dichromate demand of $6$, fixing the $1:6$ ratio; permanganate takes $5$ against an oxalate supply of $2$, whose lowest common multiple of $10$ fixes the $2:5$ ratio behind every permanganate titration.

If the charge check fails by $k$, the $\mathrm{H^+}$ count is almost always wrong by $k$, $\mathrm{H^+}$ being the only species added on charge grounds. Use this method for molecular equations and whenever the oxidation numbers are visible at a glance.

Card 9 — Balancing by the half-reaction (ion-electron) method

Step 1. Write the skeletal ionic equation, spectators left out, marking which species is oxidised and which reduced.

Step 2. Split into two half reactions, one line each, carrying only the species holding the atom that changed.

Step 3. Balance each half in this order: (a) every atom except H and O; (b) oxygen with $\mathrm{H_2O}$; (c) hydrogen with $\mathrm{H^+}$; (d) charge with $\mathrm{e^-}$ on the more positive side.

Step 4. Equalise the electrons by the lowest common multiple, then add the halves. Electrons must cancel completely, and so must any $\mathrm{H^+}$ or $\mathrm{H_2O}$ common to both sides.

Step 5. Convert to basic medium if required.

Step 6. Verify atoms, then charge.

Which side the electrons go on. Oxidation is loss, so $\mathrm{e^-}$ sits on the right, $\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$; reduction is gain, so it sits on the left, $\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$. The fallback that never fails is the charge rule: electrons go on the more positive side.

How the acidic and basic routes differ. Not at all in the first four steps. In acid you balance oxygen with $\mathrm{H_2O}$ and hydrogen with $\mathrm{H^+}$ and stop. In base you do that, then add to both sides as many $\mathrm{OH^-}$ as there are surviving $\mathrm{H^+}$, collapse each pair into one $\mathrm{H_2O}$, cancel duplicated water and divide out any common factor.

Key Point: A basic-medium answer must contain no $\mathrm{H^+}$, and an acidic-medium answer no $\mathrm{OH^-}$. Adding hydroxide to one side only is not an identity operation, and the charge check catches it at once.

Disproportionation. The same species is the reactant of both halves: $\mathrm{Cl_2 + 2e^- \rightarrow 2Cl^-}$ alongside $\mathrm{Cl_2 + 6H_2O \rightarrow 2ClO_3^- + 12H^+ + 10e^-}$. After equalising, the two $\mathrm{Cl_2}$ terms land on the same side and add rather than cancel.

Prefer this method when the product is unfamiliar, when an average is fractional, when the medium is basic, when the electrons transferred are wanted, and whenever you need the two halves for a cell.

Card 10 — n-factor and equivalent mass for every common reagent

Key Point (Definition): The n-factor of a species is the number of electrons gained or lost per formula unit of that species in that particular reaction. It is a property of the reaction, not of the bottle.

Species	Change in the reaction	n	Molar mass	Equivalent mass
$\mathrm{KMnO_4}$, acidic	$\mathrm{MnO_4^- \rightarrow Mn^{2+}}$	$5$	$158$	$158/5 = 31.6$
$\mathrm{KMnO_4}$, neutral or weakly basic	$\mathrm{MnO_4^- \rightarrow MnO_2}$	$3$	$158$	$158/3 = 52.67$
$\mathrm{KMnO_4}$, strongly alkaline	$\mathrm{MnO_4^- \rightarrow MnO_4^{2-}}$	$1$	$158$	$158$
$\mathrm{K_2Cr_2O_7}$, acidic	$\mathrm{Cr_2O_7^{2-} \rightarrow 2Cr^{3+}}$	$6$	$294$	$294/6 = 49$
$\mathrm{H_2C_2O_4 \cdot 2H_2O}$	$\mathrm{C_2O_4^{2-} \rightarrow 2CO_2}$	$2$	$126$	$126/2 = 63$
$\mathrm{FeSO_4}$	$\mathrm{Fe^{2+} \rightarrow Fe^{3+}}$	$1$	$152$	$152$
Mohr salt	$\mathrm{Fe^{2+} \rightarrow Fe^{3+}}$	$1$	$392$	$392$
$\mathrm{Na_2S_2O_3 \cdot 5H_2O}$ with iodine	$\mathrm{2S_2O_3^{2-} \rightarrow S_4O_6^{2-}}$	$1$	$248$	$248$
$\mathrm{I_2}$	$\mathrm{I_2 \rightarrow 2I^-}$	$2$	$254$	$254/2 = 127$
$\mathrm{H_2O_2}$ as oxidant	$\mathrm{H_2O_2 \rightarrow 2H_2O}$	$2$	$34$	$34/2 = 17$
$\mathrm{H_2O_2}$ as reductant	$\mathrm{H_2O_2 \rightarrow O_2}$	$2$	$34$	$34/2 = 17$
$\mathrm{Fe_3O_4}$ oxidised to $\mathrm{Fe^{3+}}$	only the one Fe(II) is available	$1$	$232$	$232$

$$\text{Equivalent mass} = \frac{\text{Molar mass}}{n\text{-factor}} \qquad \text{Normality} = \text{Molarity} \times n\text{-factor} \qquad N_1V_1 = N_2V_2$$

The mole-ratio form is $n_1M_1V_1 = n_2M_2V_2$, and the two routes cannot disagree, normality being only molarity carrying its n-factor around with it. One bottle of $0.1\ \mathrm{M}$ $\mathrm{KMnO_4}$ is $0.5\ \mathrm{N}$ in acid, $0.3\ \mathrm{N}$ in neutral or weakly basic solution and $0.1\ \mathrm{N}$ in strong alkali: molarity never changes, normality changes with the reaction. For hydrogen peroxide, volume strength $= 5.6 \times$ normality $= 11.2 \times$ molarity, so an $11.2$ volume solution is $2\ \mathrm{N}$, $1\ \mathrm{M}$ and $34\ \mathrm{g\ L^{-1}}$.

Card 11 — The three titrations, on one page

	Permanganate	Dichromate	Iodometric
Reaction	$\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$	$\mathrm{Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O}$	$\mathrm{2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 + I_2}$, then $\mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}}$
Mole ratio	$\mathrm{MnO_4^- : C_2O_4^{2-}} = 2:5$; against iron(II) it is $1:5$	$\mathrm{Cr_2O_7^{2-} : Fe^{2+}} = 1:6$	$\mathrm{I_2 : S_2O_3^{2-}} = 1:2$; moles of Cu equal moles of thiosulphate
Indicator	none, self-indicating	diphenylamine, an internal redox indicator, added to the flask and oxidised there; the genuine external indicator it replaced was potassium ferricyanide, spotted on a tile	starch, added only near the end point
End point	first permanent pale pink tinge	green turning deep blue-violet	the blue colour disappearing
Medium	dilute sulphuric acid only; warm to about $333\ \mathrm{K}$ for oxalate, cold for iron(II)	acid; $\mathrm{H_2SO_4}$ or $\mathrm{HCl}$ both allowed	weakly acidic, cool, titrated without delay
Standard	$\mathrm{KMnO_4}$ is a secondary standard	$\mathrm{K_2Cr_2O_7}$ is a primary standard	thiosulphate standardised against dichromate

Why the acid matters. Permanganate at $+1.51\ \mathrm{V}$ sits above the chlorine couple at $+1.36\ \mathrm{V}$, so it oxidises chloride as well as the sample and the result comes out high. Nitric acid is an oxidant itself and attacks the reductant before the titration begins, so its result comes out low. Dichromate at $+1.33\ \mathrm{V}$ sits just below the chlorine couple, cannot oxidise chloride, and tolerates hydrochloric acid.

Permanganate must be standardised on the day of use, against oxalic acid dihydrate of equivalent mass $63$ or sodium oxalate, since commercial crystals carry $\mathrm{MnO_2}$ and the solutions decompose slowly. Dichromate needs no standardising.

Iodimetry against iodometry. Iodimetry titrates a reductant directly against standard iodine, iodine in the burette; iodometry liberates iodine from iodide with an oxidant, then titrates it with thiosulphate. Starch goes in only at pale straw yellow, since iodine added early binds inside the amylose helix, the blue fades late and the reading is high. Dichromate, iodine and thiosulphate react in the ratio $1:3:6$.

Card 12 — Electrode processes

Cell notation. The anode goes on the left, the cathode on the right. A single bar is a phase boundary, a double bar the salt bridge, a comma a separation within one phase.

$$\mathrm{Zn(s) \mid Zn^{2+}(aq,\ 1\ M) \parallel Cu^{2+}(aq,\ 1\ M) \mid Cu(s)}$$

When neither form of a couple can serve as its own electrode, an inert conductor is written at the far end: $\mathrm{Pt(s) \mid Fe^{2+}(aq),\ Fe^{3+}(aq) \parallel Ag^+(aq) \mid Ag(s)}$.

Key Point (Definition): The anode is the electrode at which oxidation occurs; the cathode is the electrode at which reduction occurs. This holds in every cell of every kind.

	Galvanic (spontaneous)	Electrolytic (driven)
Anode	oxidation, negative	oxidation, positive
Cathode	reduction, positive	reduction, negative
Energy	chemical to electrical	electrical to chemical

The reaction labels never move; only the sign flips. An ox for anode-oxidation, red cat for reduction-cathode. The salt bridge completes the circuit, maintains neutrality in both solutions and keeps them from mixing, its anions going to the anode compartment and its cations to the cathode compartment.

The standard hydrogen electrode, $\mathrm{Pt(s) \mid H_2(g,\ 1\ bar) \mid H^+(aq,\ 1\ M)}$ at $298\ \mathrm{K}$ on platinised platinum, runs $\mathrm{2H^+(aq) + 2e^- \rightleftharpoons H_2(g)}$ and is assigned exactly $0.00\ \mathrm{V}$ by convention, not by measurement. Every other value came from a cell built against it.

$$E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}}$$

with both values taken as reduction potentials straight from the table, unmultiplied. Positive means spontaneous as written; negative means the reverse is.

Key Point: $E^\circ$ is intensive. $\mathrm{2Ag^+ + 2e^- \rightarrow 2Ag}$ still has $E^\circ = +0.80\ \mathrm{V}$, not $+1.60\ \mathrm{V}$, because potential is energy per unit charge and doubling the reaction doubles both. Reversing a half reaction flips the sign; multiplying it never does.

In electrolysis an external source drives a reaction with a negative $E^\circ_{\mathrm{cell}}$. Molten $\mathrm{NaCl}$ gives sodium at the cathode and chlorine at the anode; brine gives hydrogen instead, water at $-0.83\ \mathrm{V}$ being far easier to reduce than $\mathrm{Na^+}$ at $-2.71\ \mathrm{V}$, while chlorine still comes off at the anode because oxygen carries a large overvoltage.

Card 13 — The electrochemical series, and the three things to read off it

Standard reduction potentials at $298\ \mathrm{K}$, most positive first.

Reduction half reaction	$E^\circ$ / V
$\mathrm{F_2 + 2e^- \rightarrow 2F^-}$	$+2.87$
$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$	$+1.51$
$\mathrm{Au^{3+} + 3e^- \rightarrow Au}$	$+1.40$
$\mathrm{Cl_2 + 2e^- \rightarrow 2Cl^-}$	$+1.36$
$\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$	$+1.33$
$\mathrm{O_2 + 4H^+ + 4e^- \rightarrow 2H_2O}$	$+1.23$
$\mathrm{MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O}$	$+1.23$
$\mathrm{Br_2 + 2e^- \rightarrow 2Br^-}$	$+1.09$
$\mathrm{Ag^+ + e^- \rightarrow Ag}$	$+0.80$
$\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$	$+0.77$
$\mathrm{I_2 + 2e^- \rightarrow 2I^-}$	$+0.54$
$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$	$+0.34$
$\mathrm{AgCl + e^- \rightarrow Ag + Cl^-}$	$+0.22$
$\mathrm{Cu^{2+} + e^- \rightarrow Cu^+}$	$+0.16$
$\mathrm{Sn^{4+} + 2e^- \rightarrow Sn^{2+}}$	$+0.15$
$\mathrm{AgBr + e^- \rightarrow Ag + Br^-}$	$+0.10$
$\mathrm{2H^+ + 2e^- \rightarrow H_2}$	$0.00$
$\mathrm{Pb^{2+} + 2e^- \rightarrow Pb}$	$-0.13$
$\mathrm{Sn^{2+} + 2e^- \rightarrow Sn}$	$-0.14$
$\mathrm{Ni^{2+} + 2e^- \rightarrow Ni}$	$-0.25$
$\mathrm{Fe^{2+} + 2e^- \rightarrow Fe}$	$-0.44$
$\mathrm{Cr^{3+} + 3e^- \rightarrow Cr}$	$-0.74$
$\mathrm{Zn^{2+} + 2e^- \rightarrow Zn}$	$-0.76$
$\mathrm{2H_2O + 2e^- \rightarrow H_2 + 2OH^-}$	$-0.83$
$\mathrm{Al^{3+} + 3e^- \rightarrow Al}$	$-1.66$
$\mathrm{Mg^{2+} + 2e^- \rightarrow Mg}$	$-2.36$
$\mathrm{Na^+ + e^- \rightarrow Na}$	$-2.71$
$\mathrm{Ca^{2+} + 2e^- \rightarrow Ca}$	$-2.87$
$\mathrm{K^+ + e^- \rightarrow K}$	$-2.93$
$\mathrm{Li^+ + e^- \rightarrow Li}$	$-3.05$

One: strength. A large positive $E^\circ$ marks a strong oxidising agent, fluorine at $+2.87\ \mathrm{V}$ heading the table; a large negative $E^\circ$ marks a strong reducing agent, lithium at $-3.05\ \mathrm{V}$ being the strongest in aqueous solution.

Two: the reactivity series, free. The metal couples carry the activity series quoted in every book, $\mathrm{K > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Au}$; that quoted list leaves out lithium, which its $-3.05\ \mathrm{V}$ puts above potassium, and chromium, which its $-0.74\ \mathrm{V}$ puts between zinc and iron. Among the non-metals the halogens rank $\mathrm{F_2 > Cl_2 > Br_2 > I_2}$ as oxidants, the halide ions the other way round, $\mathrm{I^- > Br^- > Cl^- > F^-}$.

Three: whether a reaction goes. Make the species reduced the cathode, subtract, read the sign. Daniell $0.34 - (-0.76) = +1.10\ \mathrm{V}$; zinc with $\mathrm{Ag^+}$ $0.80 - (-0.76) = +1.56\ \mathrm{V}$; copper with $\mathrm{Ag^+}$ $0.80 - 0.34 = +0.46\ \mathrm{V}$; iron with $\mathrm{Cu^{2+}}$ $0.34 - (-0.44) = +0.78\ \mathrm{V}$; silver with $\mathrm{Cu^{2+}}$ $0.34 - 0.80 = -0.46\ \mathrm{V}$, so a silver spoon is safe in copper sulphate. A metal liberates hydrogen from dilute acid only if its own $E^\circ$ is negative.

Card 14 — The fifteen mistakes that cost marks

1. Swapping the two agents: the oxidising agent is the species reduced, the reducing agent the species oxidised.
2. Calling a whole reaction "an oxidation" instead of naming which species is oxidised and which reduced.
3. Applying rule 6 before rules 3, 4 and 5, so $\mathrm{OF_2}$, $\mathrm{Na_2O_2}$ and $\mathrm{NaH}$ all come out wrong.
4. Missing a peroxide linkage, giving $+8$ for $\mathrm{H_2SO_5}$, $+7$ for $\mathrm{H_2S_2O_8}$ or $+10$ for $\mathrm{CrO_5}$ instead of $+6$ every time.
5. Ignoring the subscript on the unknown element, so $\mathrm{K_2Cr_2O_7}$ gives $+12$ and $\mathrm{H_4P_2O_7}$ gives $+10$.
6. Setting the sum to zero for a polyatomic ion, when $\mathrm{NH_4^+}$ sums to $+1$ and $\mathrm{Cr_2O_7^{2-}}$ to $-2$.
7. Quoting a fractional average as the state of a real atom, when no iron in $\mathrm{Fe_3O_4}$ is at $+\frac{8}{3}$.
8. Adding the Roman numerals of all the atoms, so $\mathrm{Fe_2O_3}$ becomes iron(VI) oxide instead of iron(III) oxide.
9. Labelling a combination or decomposition redox without checking, when $\mathrm{CaCO_3 \rightarrow CaO + CO_2}$ changes no oxidation number.
10. Claiming a species in its highest state disproportionates, or calling the fluorine and alkali reaction a disproportionation when it is oxygen that is oxidised.
11. Balancing hydrogen before oxygen, or adding electrons before the atoms are done, so every later count is stale.
12. Leaving $\mathrm{H^+}$ in a basic-medium answer, or adding $\mathrm{OH^-}$ to one side only.
13. Treating normality as a property of the bottle, when the same $0.1\ \mathrm{M}$ permanganate is $0.5\ \mathrm{N}$, $0.3\ \mathrm{N}$ or $0.1\ \mathrm{N}$ according to the medium.
14. Dividing Mohr salt by six, using $90$ for oxalic acid dihydrate when the crystals weigh $126$, or forgetting to multiply an aliquot back up to the full volume.
15. Multiplying $E^\circ$ by the number of electrons, adding the two potentials instead of subtracting, or subtracting anode minus cathode.

Three more cost as much: assuming a reaction needs oxygen to be redox; expecting dihydrogen from copper and nitric acid, when the gas is an oxide of nitrogen; and saying the iodometric end point is the appearance of blue, when the blue disappears.

Card 15 — The last sixty seconds

Definitions. Oxidation: loss of electrons, oxidation number rises. Reduction: gain, oxidation number falls. Oxidising agent is reduced; reducing agent is oxidised. A half reaction is bookkeeping only.

Rules, in priority. Free element $0$. Monatomic ion equals its charge. F always $-1$. Group 1 $+1$, group 2 $+2$, Al $+3$. H is $+1$ except $-1$ when bonded to a metal or to boron. O is $-2$ except $-1$ in peroxides, $-\frac{1}{2}$ in superoxides, $+2$ in $\mathrm{OF_2}$ and $+1$ in $\mathrm{O_2F_2}$. Sum $= 0$ for a molecule, $=$ charge for an ion.

Fractions and traps. $\mathrm{Fe_3O_4}$ $+\frac{8}{3}$, really $+2, +3, +3$; $\mathrm{Na_2S_4O_6}$ $+\frac{5}{2}$, really $+5, 0, 0, +5$; $\mathrm{C_3O_2}$ $+\frac{4}{3}$; $\mathrm{Br_3O_8}$ $+\frac{16}{3}$; $\mathrm{KO_2}$ $-\frac{1}{2}$, and that one is real. The peroxo trio $\mathrm{H_2SO_5}$, $\mathrm{H_2S_2O_8}$ and $\mathrm{CrO_5}$ all give $+6$.

Four types. Combination is redox if a reactant is a free element, decomposition if a product is; displacement runs from the more reactive element to the less; disproportionation needs an intermediate state with a higher and a lower both accessible, and fluorine never qualifies.

Balancing. Total increase $=$ total decrease, change per atom times number of atoms. Oxygen with $\mathrm{H_2O}$, hydrogen with $\mathrm{H^+}$, charge with $\mathrm{e^-}$, in that order; for base add $\mathrm{OH^-}$ to both sides afterwards. Verify atoms, then charge.

Arithmetic. Equivalent mass $=$ molar mass $/\ n$; normality $=$ molarity $\times\ n$; $N_1V_1 = N_2V_2$, or $n_1M_1V_1 = n_2M_2V_2$. Volume strength $= 5.6 \times N = 11.2 \times M$.

n-factors. $\mathrm{KMnO_4}$ $5$, $3$, $1$ with equivalent masses $31.6$, $52.67$, $158$; $\mathrm{K_2Cr_2O_7}$ $6$, equivalent mass $49$; oxalic acid dihydrate $2$, equivalent mass $63$; $\mathrm{FeSO_4}$ and Mohr salt $1$, equivalent masses $152$ and $392$; thiosulphate $1$; $\mathrm{I_2}$ $2$; $\mathrm{H_2O_2}$ $2$ either way, equivalent mass $17$.

Ratios. $\mathrm{MnO_4^- : C_2O_4^{2-}} = 2:5$; $\mathrm{MnO_4^- : Fe^{2+}} = 1:5$; $\mathrm{Cr_2O_7^{2-} : Fe^{2+}} = 1:6$; $\mathrm{I_2 : S_2O_3^{2-}} = 1:2$; $\mathrm{Cr_2O_7^{2-} : I_2 : S_2O_3^{2-}} = 1:3:6$; moles of Cu $=$ moles of thiosulphate.

Molar masses. $\mathrm{KMnO_4}$ $158$, $\mathrm{K_2Cr_2O_7}$ $294$, $\mathrm{FeSO_4}$ $152$, Mohr salt $392$, $\mathrm{H_2C_2O_4 \cdot 2H_2O}$ $126$, $\mathrm{Na_2C_2O_4}$ $134$, $\mathrm{Na_2S_2O_3 \cdot 5H_2O}$ $248$, $\mathrm{I_2}$ $254$, $\mathrm{H_2O_2}$ $34$.

Cells. Anode left, oxidation, negative in a galvanic cell and positive in electrolysis. $E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}}$, positive means spontaneous, $E^\circ$ is intensive, SHE is $0.00\ \mathrm{V}$ by definition. Daniell $+1.10\ \mathrm{V}$, $\mathrm{Zn}$ with $\mathrm{Ag^+}$ $+1.56\ \mathrm{V}$, $\mathrm{Cu}$ with $\mathrm{Ag^+}$ $+0.46\ \mathrm{V}$, $\mathrm{Fe}$ with $\mathrm{Cu^{2+}}$ $+0.78\ \mathrm{V}$. Series: $\mathrm{K > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Au}$; $\mathrm{F_2 > Cl_2 > Br_2 > I_2}$; $\mathrm{I^- > Br^- > Cl^- > F^-}$.