Corrected exercises: redox, oxidation numbers and nomenclature (202-SN1-RE)

$NaCl$ sodium chloride, $CaBr_{2}$ calcium bromide, $Al_{2}O_{3}$ aluminum oxide, $K_{2}S$ potassium sulfide. No numerical prefix appears: the subscripts follow from the charges, so they do not have to be said. Writing calcium dibromide is an error, not a harmless redundancy.

Iron, copper and most transition metals have several possible charges, so the charge of the metal must be given in Roman numerals in parentheses. $FeCl_{2}$ iron(II) chloride, $FeCl_{3}$ iron(III) chloride, $CuO$ copper(II) oxide, $Cu_{2}O$ copper(I) oxide. The charge is recovered by reading the anion, whose charge is fixed: in $Cu_{2}O$ the oxygen is at $-2$, so the two coppers share $+2$, that is $+1$ each.

Magnesium oxide: $Mg^{2+}$ and $O^{2-}$ give $MgO$. Aluminum sulfide: $Al^{3+}$ and $S^{2-}$ require $2$ aluminums and $3$ sulfurs, $Al_{2}S_{3}$. Lead(II) iodide: $Pb^{2+}$ and $I^{-}$, so $PbI_{2}$. Calcium nitride: $Ca^{2+}$ and $N^{3-}$, so $Ca_{3}N_{2}$. Crossing the charges gives the subscripts directly, provided you simplify afterward.

It is indeed neutral: two $Ca^{2+}$ contribute $+4$ and two $O^{2-}$ contribute $-4$. What is wrong lies elsewhere: the formula of an ionic compound is always the EMPIRICAL FORMULA, that is, the simplest ratio of the ions in the crystal lattice. There is no isolated $Ca_{2}O_{2}$ molecule, only a lattice in which there is one calcium for each oxygen. So we write $CaO$. This is exactly the step that crossing the charges forces you to take afterward, and that gets forgotten every second time.

Calcium, in group 2, loses two electrons to give $Ca^{2+}$; bromine, in group 17, gains one to give $Br^{-}$. The compound must be neutral overall: $1\times(+2)+2\times(-1)=0$. Two bromide ions are needed for each calcium, hence $CaBr_{2}$. $Ca_{2}Br$ would give $2\times(+2)+1\times(-1)=+3$, a nonzero charge: such a compound could not exist.

$Na_{2}SO_{4}$ sodium sulfate, $Ca(NO_{3})_{2}$ calcium nitrate, $NH_{4}Cl$ ammonium chloride, $K_{3}PO_{4}$ potassium phosphate. The ammonium ion $NH_{4}^{+}$ is the only common polyatomic ion that is a CATION: it is therefore named first, in the place of a metal.

$ClO^{-}$ hypochlorite, $ClO_{2}^{-}$ chlorite, $ClO_{3}^{-}$ chlorate, $ClO_{4}^{-}$ perchlorate. Oxidation numbers of chlorine: $x+(-2)=-1$ gives $+1$; $x+2(-2)=-1$ gives $+3$; $x+3(-2)=-1$ gives $+5$; $x+4(-2)=-1$ gives $+7$. The rule to remember uses two reference points: the reference ion takes the suffix -ate, one oxygen fewer gives -ite, two fewer gives hypo- plus -ite, one more gives per- plus -ate. Each oxygen removed lowers the oxidation number by two units.

$HCl$ hydrochloric acid, $HNO_{3}$ nitric acid, $H_{2}SO_{4}$ sulfuric acid, $H_{2}SO_{3}$ sulfurous acid. The rule: an anion ending in -ide gives an acid named hydro-...-ic, an anion ending in -ate gives an acid ending in -ic, an anion ending in -ite gives an acid ending in -ous. Sulfate gives sulfuric, sulfite gives sulfurous: one letter of difference in the name for one oxygen of difference in the formula, which makes this pair the most common trap on a test.

Anhydrous part: $63.55+32.07+4(16.00)=63.55+32.07+64.00=159.62$ g/mol. Five waters: $5\times 18.02=90.10$ g/mol. Total: $159.62+90.10=249.72$ g/mol. Percent water: $\dfrac{90.10}{249.72}\times 100=36.08\%$. The point of vocabulary matters: the dot in $CuSO_{4}\cdot 5\,H_{2}O$ is not a multiplication, it is an association, and the water is very much part of the molar mass.

Heating drives off the water of crystallization and leaves the anhydrous salt: $m=5.000\times\dfrac{159.62}{249.72}=3.196$ g. You can check with the complement: $5.000\times 0.3608=1.804$ g of water lost, and $5.000-1.804=3.196$ g. The solid goes from bright blue to white, and turns blue again if it is wetted: this is the classic laboratory test for the presence of water. A weighing that does not come down to $3.196$ g signals incomplete dehydration, not a calculation error.

In $ClO_{2}$, a NEUTRAL molecule, the sum is zero: $x+2(-2)=0$ gives $x=+4$. In the chlorite ion $ClO_{2}^{-}$, the sum is $-1$ and chlorine is at $+3$: same formula, one charge of difference, hence one oxidation number of difference. Copying the $+3$ from part b) is the error this question is looking for, and it comes from reading the formula without reading the superscript. In $Cl_{2}$, an element in its elemental form, chlorine is at $0$. In $NaClO$, sodium is at $+1$ and oxygen at $-2$, so $1+x-2=0$ and $x=+1$, as in the hypochlorite ion it contains. Chlorine ranges from $-1$, its minimum since it is only one electron short of completing its shell, to $+7$, its maximum since it has only seven valence electrons. $ClO_{4}^{-}$, at $+7$, can only GO DOWN: it can only be reduced, so it is an oxidizing agent and never a reducing agent. $Cl^{-}$, at $-1$, can only GO UP: it can only be oxidized, so it is a reducing agent. $Cl_{2}$ and $ClO_{2}$, in between, act as one or the other depending on their partner. A nuance to keep for later: a high oxidation number says what a species CAN do, not how strongly; it is the standard potential, seen in solution chemistry, that ranks oxidizing agents against one another.

$CO$ carbon monoxide, $CO_{2}$ carbon dioxide, $N_{2}O$ dinitrogen monoxide, $N_{2}O_{4}$ dinitrogen tetroxide, $PCl_{5}$ phosphorus pentachloride. A convention of usage: the prefix mono- is omitted on the FIRST element named, so we say carbon dioxide and not monocarbon dioxide, but it is kept on oxygen, hence monoxide and not oxide.

Sulfur hexafluoride: $SF_{6}$. Nitrogen trichloride: $NCl_{3}$. Dichlorine heptoxide: $Cl_{2}O_{7}$. Here the name gives the subscripts with no calculation at all, which is the whole point of molecular nomenclature: no charge is needed, so there is no rule to remember about the group of the element.

Because sodium chloride has only one possible formula. The charges $Na^{+}$ and $Cl^{-}$ force a one-to-one ratio, so stating the number would be redundant. Between two nonmetals, on the contrary, nothing forces a ratio: carbon and oxygen give $CO$ as well as $CO_{2}$, and nitrogen and oxygen give at least five compounds. Without a prefix, the name would not identify a compound. The practical rule on a test: a metal in the formula, no prefix; two nonmetals, prefixes.

Oxygen is at $-2$ throughout. $N_{2}O$: $2x+(-2)=0$ gives $x=+1$. $NO$: $x-2=0$ gives $+2$. $N_{2}O_{3}$: $2x-6=0$ gives $+3$. $NO_{2}$: $x-4=0$ gives $+4$. $N_{2}O_{5}$: $2x-10=0$ gives $+5$. Nitrogen therefore covers every state from $+1$ to $+5$, which explains the richness of its chemistry and the fact that nitric acid is both a strong acid and an oxidizing agent.

An oxidation number is not a real charge: it is bookkeeping. We act as if every covalent bond were entirely ionic, assigning the two electrons of the bond to the more electronegative atom. In $N_{2}O$, the nitrogen-oxygen bonds are polarized toward oxygen, so the electrons are assigned to oxygen, which leaves $+1$ per nitrogen on average. This fiction makes no physical claim: its only function is to track, from one equation to the next, who gave up electrons and who received them. That is the thread of the whole set. An immediate consequence: an oxidation number can be fractional, which no real charge could be.

$H_{2}O$: hydrogen is at $+1$, so $2(+1)+x=0$ and oxygen is at $-2$. $NH_{3}$: $x+3(+1)=0$, nitrogen is at $-3$. $CH_{4}$: $x+4(+1)=0$, carbon is at $-4$, its lowest state. $SO_{2}$: $x+2(-2)=0$, sulfur is at $+4$.

In $H_{2}O_{2}$, the hydrogens impose $2(+1)+2x=0$, so each oxygen is at $-1$ and not at $-2$: this is the definition of a peroxide, two oxygens bonded to each other, neither of which can then pull the electron away from its twin. In $OF_{2}$, fluorine is the only element more electronegative than oxygen: it keeps its $-1$, so $x+2(-1)=0$ and oxygen rises to $+2$. These are the only two exceptions to know, and they come up regularly because they are handled by applying the rules IN ORDER.

In $NaH$, sodium is a group 1 metal: it can only give up its electron, so it is at $+1$. The sum being zero, hydrogen is at $-1$. This is consistent with electronegativity, that of hydrogen (2.20) exceeding that of sodium (0.93): hydrogen is here the more electronegative element of the pair, so it takes the electrons of the bond. A hydride is the only case where hydrogen is negative, and it is also what makes it a violent reducing agent.

$Cr_{2}O_{7}^{2-}$: the sum equals the charge of the ion, that is $-2$. So $2x+7(-2)=-2$, hence $2x=12$ and $x=+6$. $MnO_{4}^{-}$: $x+4(-2)=-1$, hence $x=+7$. Manganese at $+7$ is at its maximum state, that of group 7: it can only go down, which makes permanganate a very powerful oxidizing agent, and the same reading applies to dichromate at $+6$.

$6x+12(+1)+6(-2)=0$ gives $6x=0$, so $x=0$. What this result does NOT mean: that each carbon in glucose is at zero. The oxidation number calculated this way is an AVERAGE over the six carbons, and in reality they range from $-1$ to $+1$ depending on their neighbours in the molecule. The average remains perfectly useful for the electron balance of a combustion, where only the total matters, but it describes no particular atom. It is the same warning as for propane, where we will find a fractional number.

(1) synthesis, two reactants give a single product. (2) decomposition, one reactant gives two products. (3) single displacement, an element drives another element out of its compound. (4) double displacement, two compounds exchange their partners. (5) combustion, a compound reacts with oxygen gas to give the oxides of its elements.

(1) yes: magnesium goes from $0$ to $+2$. (2) NO: calcium stays at $+2$, carbon at $+4$ and oxygen at $-2$ on both sides; nothing moves, it is a rearrangement, not a transfer. (3) yes: zinc goes from $0$ to $+2$ and copper from $+2$ to $0$. (4) NO: silver $+1$, sodium $+1$, chlorine $-1$, nitrate unchanged on both sides. (5) yes: carbon goes from $-4$ to $+4$ and oxygen from $0$ to $-2$. Remember the method rather than the classification: you do not GUESS whether a reaction is a redox reaction, you calculate the oxidation numbers on both sides and see whether any of them moved.

Zinc is OXIDIZED, it loses two electrons: it is the REDUCING AGENT. The copper ion $Cu^{2+}$ is REDUCED, it gains two electrons: it is the OXIDIZING AGENT. The formula that prevents a mix-up: the oxidizing agent is the one that gets reduced, because it oxidizes the other. Sulfate plays no part, it is a spectator ion.

Each magnesium goes from $0$ to $+2$: it loses $2$ electrons. The equation contains $2\,Mg$, so $2\times 2=4$ electrons are exchanged in total. Check from the other side: oxygen gas supplies $2$ oxygens that each go from $0$ to $-2$, that is $2\times 2=4$ electrons gained. The two counts must always come out equal, and it is this equality that will be used to balance the next exercise.

Because a double displacement merely exchanges ionic partners: each ion keeps its charge and only changes neighbour. No electron leaves one species for another, so no oxidation number can change. The driving force of these reactions says the same thing: what makes them proceed is the formation of a precipitate, a gas or water, that is, the removal of ions from solution, never a transfer of electrons. Practical consequence: faced with a precipitation or an acid-base neutralization, you can say with no calculation that there is no redox reaction.

(6) NO: calcium $+2$, oxygen $-2$, hydrogen $+1$, before and after. Yet it is a synthesis, two reactants and a single product, like reaction (1). The difference lies here: a synthesis between two ELEMENTS is always a redox reaction, since an element starts at $0$ and ends up bound in a compound, but a synthesis between two COMPOUNDS, here a basic oxide being hydrated, may well not be one. (7) YES: sodium goes from $0$ to $+1$, it is oxidized, it is the REDUCING AGENT; the hydrogen of water goes from $+1$ to $0$ in $H_{2}$, so water is the oxidizing agent. Two sodiums give up $2$ electrons, two hydrogens gain $2$: the balance closes. This is the behaviour of the whole of group 1, which gives up its single electron to almost any partner, water included. (8) NO, and the trap is double. Nitrogen appears at two values, $-3$ in $NH_{4}^{+}$ and $+5$ in $NO_{3}^{-}$, but each stays the same on both sides: two different oxidation numbers do not make an oxidation number that changes. And a dissolution separates ions that already existed in the crystal, it creates none. The lesson of the three: only calculation settles it, and the reaction type is enough only for single displacement, combustion and double displacement.

Metallic copper is an element in its elemental form, so its oxidation number is $0$. In $Cu(NO_{3})_{2}$, each nitrate carries $-1$, so copper is at $+2$. Change: from $0$ to $+2$, an oxidation, and it LOSES $2$ electrons per atom.

In $HNO_{3}$: $(+1)+x+3(-2)=0$ gives $x=+5$. In $NO$: $x+(-2)=0$ gives $x=+2$. Change: from $+5$ to $+2$, a reduction, and nitrogen GAINS $3$ electrons per atom.

The electron balance must close: what is lost is exactly what is gained. So we look for the least common multiple of $2$ and $3$, which is $6$. It takes $3$ coppers to lose $6$ electrons and $2$ reduced nitrogens to gain $6$. The coefficients are therefore $3$ in front of $Cu$ and $2$ in front of $NO$.

$3\,Cu+8\,HNO_{3}\rightarrow 3\,Cu(NO_{3})_{2}+2\,NO+4\,H_{2}O$. Check: copper, $3$ on the left and $3$ on the right. Nitrogen, $8$ on the left, and on the right $3\times 2=6$ in the nitrates plus $2$ in the monoxide, that is $8$. Hydrogen, $8$ on the left, $4\times 2=8$ on the right. Oxygen, $8\times 3=24$ on the left, and on the right $3\times 6=18$ in the nitrates plus $2$ in the monoxide plus $4$ in the water, that is $24$. The coefficient $8$ of the acid and the $4$ of the water are deduced afterward, from hydrogen and then oxygen: the electron balance gives only the first two, and that is normal.

Only two. Of the $8$ nitric acid molecules, $2$ see their nitrogen go down from $+5$ to $+2$ to become nitrogen monoxide: these are the ones that play the role of OXIDIZING AGENT. The other $6$ pass through the reaction without changing oxidation number and end up unchanged in the $3\,Cu(NO_{3})_{2}$: they serve only as a supply of nitrate ions to accompany the copper, in other words as an acid in the ordinary sense. This double role is the reason the equation resists inspection, and it is also what the oxidation number method makes visible in a single line.

Bleach: chlorine starts at $0$ in $Cl_{2}$ and ends at $-1$ in $NaCl$ and at $+1$ in $NaClO$. Of the two atoms of a single chlorine molecule, one gains an electron and the other loses one: chlorine gas is both the oxidizing agent and the reducing agent. This is a DISPROPORTIONATION, an element at an intermediate state that splits toward a lower and a higher state; $1$ electron is exchanged per equation. Hydride: hydrogen is at $-1$ in $NaH$, a metal hydride, and at $+1$ in water, and both end at $0$ in $H_{2}$. The hydride gives up an electron, so it is the reducing agent; a hydrogen of the water picks it up, so it is the oxidizing agent; and the two end up in the same molecule of hydrogen gas. This is the reverse of a disproportionation, a COMPROPORTIONATION, also called synproportionation: two states that converge toward an intermediate state; $1$ electron is exchanged per equation. Sodium, oxygen and the other hydrogen of the water do not move. The trap, in both cases, is to conclude too quickly that there is no redox reaction because only one element changes: it is indeed the same element, but carried by different atoms, and each has its own balance.

$C_{3}H_{8}+5\,O_{2}\rightarrow 3\,CO_{2}+4\,H_{2}O$. Method: first place carbon, $3$ carbon dioxides, then hydrogen, $4$ waters, and count oxygen last, $3\times 2+4=10$ atoms, that is $5$ molecules of oxygen gas.

$3x+8(+1)=0$ gives $x=-\dfrac{8}{3}\approx -2.67$: a fractional number, therefore an average, which is normal since the carbons of propane do not all have the same neighbours. In $CO_{2}$ carbon is at $+4$. Change per carbon: $+4-\left(-\dfrac{8}{3}\right)=\dfrac{20}{3}$, and for the three carbons $3\times\dfrac{20}{3}=20$ electrons lost. Check from the other side: $5\,O_{2}$ supply $10$ oxygen atoms that each go from $0$ to $-2$, that is $10\times 2=20$ electrons gained. The two counts match, and hydrogen stays at $+1$ from start to finish: it takes no part in the transfer.

$2\,C_{3}H_{8}+7\,O_{2}\rightarrow 6\,CO+8\,H_{2}O$. Check: carbon $6$ and $6$, hydrogen $16$ and $16$, oxygen $14$ on the left against $6+8=14$ on the right. The propane had to be doubled to avoid a fractional coefficient in front of oxygen gas, which is the standard manoeuvre whenever a combustion produces the monoxide.

For $1.00$ mol of propane, $5.00$ mol of oxygen gas is needed, that is $V(O_{2})=5.00\times 24.5=122.5$ L. Since air is only $21.0\%$ oxygen, $V(air)=\dfrac{122.5}{0.210}=583$ L. Remember the order of magnitude: nearly $600$ litres of air are needed for a single mole of propane, which explains why a poorly ventilated appliance slips so easily into incomplete combustion.

In incomplete combustion, carbon rises only to $+2$ instead of $+4$: it has given up only $\dfrac{14}{3}$ electrons per atom instead of $\dfrac{20}{3}$, that is $14$ instead of $20$ for three carbons. The oxidation is UNFINISHED, so part of the chemical energy remains locked in the carbon monoxide instead of being released: that is the loss of efficiency. And the same incompleteness creates the danger, since carbon monoxide, colourless and odourless, binds to hemoglobin about two hundred times better than oxygen does. Lack of air is thus the single cause of both problems, which is why the technical answer is always the same: ventilate.