Corrected exercises: redox and electrochemistry (202-SN2-RE)

In $KMnO_{4}$: $K$ is $+1$, each $O$ is $-2$, which makes $-8$ in total, and the molecule is neutral. So $+1+n(Mn)-8=0$ and $n(Mn)=+7$, the maximum possible value for manganese. In $Cr_{2}O_{7}^{2-}$: $2n(Cr)+7(-2)=-2$, so $2n(Cr)=+12$ and $n(Cr)=+6$.

In $H_{2}O_{2}$: the two $H$ are $+1$, the molecule is neutral, so $2(+1)+2n(O)=0$ and $n(O)=-1$. This is the PEROXIDE exception, where two oxygens are bonded to each other. In $NaH$: $Na$ is an alkali metal, it is $+1$, so $n(H)=-1$. This is the metal HYDRIDE exception, where hydrogen is bonded to an element LESS electronegative than itself.

In $S_{2}O_{3}^{2-}$: $2n(S)+3(-2)=-2$, so $2n(S)=+4$ and $n(S)=+2$. This value is an AVERAGE: in thiosulfate the two sulfurs are not equivalent, one is at $+5$ and the other at $-1$. Since the oxidation number is a counting convention, only the sum has meaning here. In $CH_{3}OH$: $n(C)+3(+1)+(-2)+(+1)=0$, so $n(C)=-2$.

Zinc goes from 0 to $+2$: it LOSES electrons, so it is OXIDIZED, and it is the REDUCING AGENT. Hydrogen goes from $+1$ to 0: it GAINS electrons, so it is REDUCED, and $H^{+}$ is the OXIDIZING AGENT. A classic vocabulary pitfall: the oxidizing agent is the one that is reduced, and the reducing agent is the one that is oxidized. Check of the balance: zinc gives up 2 electrons, two $H^{+}$ ions pick up 1 each, the count is right.

Reduction half-reaction: manganese goes from $+7$ to $+2$, so it gains 5 electrons. We write $MnO_{4}^{-}\rightarrow Mn^{2+}$, balance the oxygen with 4 $H_{2}O$ on the right, the hydrogen with 8 $H^{+}$ on the left, then the charge: $MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O$. Oxidation half-reaction: $Fe^{2+}\rightarrow Fe^{3+}+e^{-}$. To equalize the electrons, we multiply the second by 5. Overall: $MnO_{4}^{-}+8H^{+}+5Fe^{2+}\rightarrow Mn^{2+}+4H_{2}O+5Fe^{3+}$.

On the left: $(-1)+8(+1)+5(+2)=-1+8+10=+17$. On the right: $(+2)+4(0)+5(+3)=2+15=+17$. The charges are equal, and so are the atoms: 1 $Mn$, 4 $O$, 8 $H$, 5 $Fe$ on each side. This charge balance is the quickest check there is on a redox equation: do it every time.

Reduction: manganese goes from $+7$ to $+4$, which is 3 electrons. In basic solution we write directly $MnO_{4}^{-}+2H_{2}O+3e^{-}\rightarrow MnO_{2}+4OH^{-}$. Oxidation: iodine goes from $-1$ to $+5$, which is 6 electrons: $I^{-}+6OH^{-}\rightarrow IO_{3}^{-}+3H_{2}O+6e^{-}$. We multiply the first by 2 to get 6 electrons on both sides: $2MnO_{4}^{-}+4H_{2}O+6e^{-}\rightarrow 2MnO_{2}+8OH^{-}$. Adding, and cancelling the common $6OH^{-}$ and $3H_{2}O$: $2MnO_{4}^{-}+H_{2}O+I^{-}\rightarrow 2MnO_{2}+2OH^{-}+IO_{3}^{-}$. Charge check: on the left $-2-1=-3$, on the right $-2-1=-3$.

Because in basic solution the concentration of $H^{+}$ is negligible: at pH 13 it is $10^{-13}$ mol/L. Showing $H^{+}$ as a reactant would amount to claiming that a practically absent species takes part in stoichiometric amounts, which is chemically false even if the equation stays formally balanced. The technique is therefore to balance first in acidic solution if that is more convenient, then add to EACH side as many $OH^{-}$ as there are $H^{+}$, recombine $H^{+}+OH^{-}$ into $H_{2}O$, and finally cancel the excess water molecules.

The couple with the HIGHER potential undergoes reduction: it is $Cu^{2+}/Cu$. Cathode (reduction): $Cu^{2+}+2e^{-}\rightarrow Cu$. Anode (oxidation): $Zn\rightarrow Zn^{2+}+2e^{-}$. Overall equation, obtained by adding, the electrons cancelling: $Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$.

$E^{\circ}_{\text{cell}}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}=0.34-(-0.76)=1.10$ V. The value is positive, which confirms that the reaction is spontaneous in the direction written. An essential remark: although both half-reactions were written with 2 electrons, we do NOT multiply $E^{\circ}$ by 2. Potential is an intensive quantity, an energy PER charge, and it does not depend on the amount of substance.

By convention, the anode is written on the left and the cathode on the right, the double bar standing for the salt bridge: $Zn\left|Zn^{2+}(1.0\text{ mol/L})\right|\left|Cu^{2+}(1.0\text{ mol/L})\right|Cu$. In the wire, electrons go from the ANODE to the CATHODE, so from zinc to copper, as the arrow in the figure shows. The conventional current flows the other way.

In the left compartment, oxidation produces $Zn^{2+}$ ions: this compartment tends to become positive. In the right one, reduction consumes $Cu^{2+}$: it tends to become negative. To restore neutrality, the ANIONS of the salt bridge migrate toward the ANODE compartment, that is, to the left, and the cations toward the cathode. To remember: anions go to the anode, and the shared prefix makes it easy to remember.

The zinc electrode DISSOLVES, its mass decreases. The copper electrode becomes coated with deposited metallic copper, its mass INCREASES. In the right-hand compartment, the $Cu^{2+}$ ions, responsible for the characteristic blue colour, are consumed: the solution gradually FADES. These three observations are the simplest experimental check of the direction in which a cell operates, and they make up a classic laboratory question.

$\Delta G^{\circ}=-nFE^{\circ}=-2\times 96\,500\times 1.10=-2.12\times 10^{5}$ J/mol, that is $-212$ kJ/mol. The negative sign means that the reaction is SPONTANEOUS in the direction written, under standard conditions. The value is the maximum electrical work recoverable per mole of zinc consumed.

We can go through $\Delta G^{\circ}=-RT\ln K$, which gives $\ln K=\dfrac{nFE^{\circ}}{RT}=\dfrac{212\,300}{8.314\times 298}=85.7$, so $\log K=37.2$ and $K=1.6\times 10^{37}$. The shortcut from the course gives the same thing: $\log K=\dfrac{nE^{\circ}}{0.0592}=\dfrac{2\times 1.10}{0.0592}=37.2$.

A constant of $10^{37}$ means that the reaction goes to completion in the practical sense. Starting from concentrations of the order of 1 mol/L, a $Cu^{2+}$ concentration of the order of $10^{-37}$ mol/L would remain at equilibrium, far less than one ion per litre: the very notion of concentration stops making sense at this scale. A cell potential of a little over one volt therefore already corresponds to a complete reaction, which explains why no ordinary cell stops “halfway”.

$E^{\circ}=-0.25$ V is NEGATIVE, so $\Delta G^{\circ}=-2\times 96\,500\times(-0.25)=+48.3$ kJ/mol is positive: the reaction is NOT spontaneous in the direction written. It is, however, spontaneous in the REVERSE direction, for which $E^{\circ}=+0.25$ V. Its constant is $\log K=\dfrac{2\times(-0.25)}{0.0592}=-8.45$, so $K=3.6\times 10^{-9}$: very small, which means an equilibrium lying almost entirely on the side of the reactants. This reaction can nevertheless be forced by supplying electrical energy: that is exactly the principle of electrolysis.

Only two ions are present: at the cathode $Na^{+}+e^{-}\rightarrow Na(l)$, at the anode $2Cl^{-}\rightarrow Cl_{2}(g)+2e^{-}$. Metallic sodium is indeed produced, because no species easier to reduce is present: this is the Downs process, and it is the only industrial route to sodium.

Water. At the cathode, the species that is reduced is the one with the HIGHEST reduction potential, because it is the better oxidizing agent: $-0.83$ V for water against $-2.71$ V for $Na^{+}$, nearly two volts apart. Water is therefore reduced to $H_{2}$ long before sodium has the slightest chance. And even supposing a sodium atom did form, it would react immediately with water. Every aqueous electrolysis is thus bounded by water itself: under standard conditions, a metal with a potential below about $-0.83$ V is not deposited from an aqueous solution (hydrogen overpotential does allow zinc, chromium or manganese to be plated, however).

Cathode: $Cu^{2+}$ has the highest potential ($+0.34$ V against $-0.83$ V for water), so $Cu^{2+}+2e^{-}\rightarrow Cu(s)$, a deposit of metallic copper. Anode: the sulfate ion $SO_{4}^{2-}$ has its sulfur already at the maximum oxidation state $+VI$ and can no longer be oxidized; so it is water that is oxidized: $2H_{2}O\rightarrow O_{2}+4H^{+}+4e^{-}$. This is the principle of the electrolytic refining of copper.

OVERPOTENTIAL (or overvoltage). Standard potentials are thermodynamic quantities: they say what is favourable, not what is fast. The evolution of $O_{2}$ involves four electrons and the formation of an oxygen-oxygen bond, a slow kinetic process that in practice demands an extra voltage of several tenths of a volt. $Cl_{2}$, on the other hand, needs only two electrons and is released with a small overpotential. On a well-chosen anode, the real order is reversed and it is chlorine that comes off. Thermodynamics sets the race, kinetics picks the winner.

Cathode: $2H_{2}O+2e^{-}\rightarrow H_{2}+2OH^{-}$; anode: $2Cl^{-}\rightarrow Cl_{2}+2e^{-}$. Overall: $2NaCl+2H_{2}O\rightarrow Cl_{2}+H_{2}+2NaOH$, the $Na^{+}$ ions remaining spectators and ending up with the $OH^{-}$ formed. The three products sold are chlorine, hydrogen and caustic soda (sodium hydroxide). The process is power-hungry because an electrolysis does not merely steer a reaction: it FORCES a transformation whose $\Delta G$ is positive, and the electrical energy supplied must at the very least cover that $\Delta G$, plus all the overpotentials and ohmic losses. Here we pay in electricity for what the reaction refuses to give spontaneously.

$Q=\dfrac{\left[Zn^{2+}\right]}{\left[Cu^{2+}\right]}$. The solids $Zn$ and $Cu$ do not appear because their activity is 1: the concentration of a pure solid does not vary, its density being fixed. It is exactly the same convention as for a heterogeneous solubility equilibrium.

$Q=\dfrac{0.100}{0.00100}=100$, so $\log Q=2$. Then $E=E^{\circ}-\dfrac{0.0592}{n}\log Q=1.10-\dfrac{0.0592}{2}\times 2=1.10-0.0592=1.041$ V.

The voltage has DECREASED, from 1.10 V to 1.04 V. This could be predicted without calculation: compared with standard conditions, the concentration of the reactant $Cu^{2+}$ has been DIVIDED by 1000, whereas that of the product $Zn^{2+}$ has only been divided by 10. The ratio $Q$ has therefore increased by a factor of 100. The system is thus closer to its equilibrium than under standard conditions, it has less “driving force” left, and the measured voltage is lower. A cell that discharges sees its voltage fall continuously until it reaches zero.

Here the two couples are identical, so $E^{\circ}=0$: all the voltage comes from the concentration difference. The spontaneous reaction is the one that tends to EQUALIZE the two concentrations: copper dissolves on the dilute side and is deposited on the concentrated side. The ANODE is therefore the DILUTE compartment, at 0.0010 mol/L. With $Q=\dfrac{\left[Cu^{2+}\right]_{\text{dilute}}}{\left[Cu^{2+}\right]_{\text{concentrated}}}=\dfrac{0.0010}{1.0}=10^{-3}$: $E=0-\dfrac{0.0592}{2}\log\left(10^{-3}\right)=-0.0296\times(-3)=0.0888$ V, about 89 mV. Small but perfectly measurable: this is the principle of the pH electrode.

A dead cell has ZERO voltage: $E=0$. The Nernst equation then gives $0=E^{\circ}-\dfrac{0.0592}{n}\log Q$, so $\log Q=\dfrac{nE^{\circ}}{0.0592}=\log K$, that is, $Q=K$. A dead cell is therefore not an “empty” cell: it is a cell whose reaction has REACHED ITS EQUILIBRIUM. This is the most important conceptual link of the chapter, and it justifies after the fact the relation between $E^{\circ}$ and $K$ established in exercise 4.

The duration in seconds is $45.0\times 60=2700$ s. The charge is $q=It=2.50\times 2700=6750$ C. The amount of electrons is $n(e^{-})=\dfrac{q}{F}=\dfrac{6750}{96\,500}=6.99\times 10^{-2}$ mol.

The half-reaction is $Ag^{+}+e^{-}\rightarrow Ag$: one electron per atom, so $n(Ag)=n(e^{-})=6.99\times 10^{-2}$ mol. The mass deposited is $m=6.99\times 10^{-2}\times 107.87=7.55$ g.

The half-reaction is now $Cu^{2+}+2e^{-}\rightarrow Cu$: TWO electrons are needed per atom. The charge being the same, $n(Cu)=\dfrac{n(e^{-})}{2}=3.50\times 10^{-2}$ mol, that is $m=3.50\times 10^{-2}\times 63.55=2.22$ g. The same electricity therefore deposits half as many moles of copper as of silver: the charge of the ion, and not its molar mass, governs the split. This is the content of Faraday's law, and it is the point the problem tests.

For 5.00 g of copper: $n(Cu)=\dfrac{5.00}{63.55}=7.87\times 10^{-2}$ mol, so $n(e^{-})=2\times 7.87\times 10^{-2}=0.157$ mol and $q=0.157\times 96\,500=1.52\times 10^{4}$ C. The time is $t=\dfrac{q}{I}=\dfrac{15\,190}{2.50}=6074$ s, about 101 minutes, a little over an hour and a half. Consistency check: this is more than twice the initial duration, as expected since we want more than double the mass deposited in c).

Half-reactions: $MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O$ and $Fe^{2+}\rightarrow Fe^{3+}+e^{-}$. To equalize the electrons we multiply the second by 5: $MnO_{4}^{-}+5Fe^{2+}+8H^{+}\rightarrow Mn^{2+}+5Fe^{3+}+4H_{2}O$. Charge check: $-1+10+8=+17$ on the left, $+2+15=+17$ on the right.

$E^{\circ}=1.51-0.77=0.74$ V, with $n=5$ electrons exchanged. $\log K=\frac{(5)(0.74)}{0.0592}=62.5$, so $K\approx 10^{62}$. A constant of this order means that the reaction is complete to a degree beyond any measurement: this is the very condition of a titration, in which each drop added is assumed to react entirely before the next.

$n(MnO_{4}^{-})=(0.02450)(0.0200)=4.900\times 10^{-4}$ mol. The 1 to 5 stoichiometry gives $n(Fe^{2+})=5(4.900\times 10^{-4})=2.450\times 10^{-3}$ mol, that is $m=(2.450\times 10^{-3})(55.85)=0.1368$ g. The ore therefore contains $13.68$% iron.

Two irons per $Fe_{2}O_{3}$: $n(Fe_{2}O_{3})=\frac{2.450\times 10^{-3}}{2}=1.225\times 10^{-3}$ mol, that is $m=(1.225\times 10^{-3})(159.69)=0.1956$ g, so $19.56$% $Fe_{2}O_{3}$. Same sample, same measurement: only the unit of account changes, which is why an ore grade must always state the form in which it is expressed.

Permanganate is its OWN indicator: the $MnO_{4}^{-}$ ion is an intense purple, whereas $Mn^{2+}$ is practically colourless. As long as $Fe^{2+}$ remains, each drop added is decolourized instantly. One drop past the equivalence point has nothing left to reduce: the purple persists and colours the whole solution pale pink. This endpoint is remarkably sharp because it depends not on a pH range but on the disappearance of the limiting reactant, and a trace of excess permanganate is enough to be visible.

Oxidation: $Fe\rightarrow Fe^{2+}+2e^{-}$; reduction: $O_{2}+2H_{2}O+4e^{-}\rightarrow 4OH^{-}$. $E^{\circ}=0.40-(-0.44)=0.84$ V, clearly positive: the corrosion of iron in moist air is spontaneous, and the $Fe^{2+}$ formed is then oxidized to hydrated $Fe^{3+}$, rust. That is precisely why the problem is universal and costly: thermodynamics is on the side of rust.

Iron-zinc: zinc has the lowest potential, so it is the one that is oxidized, and $E^{\circ}=-0.44-(-0.76)=0.32$ V. Iron-tin: this time it is IRON that has the lowest potential, so it is the one that is oxidized, with $E^{\circ}=-0.14-(-0.44)=0.30$ V. In both cases the reaction is spontaneous; only the identity of the sacrificed metal changes.

As long as the coating is intact, both protect equally, as a simple physical barrier. The scratch changes everything, because it creates a short-circuited cell between the two metals, in the presence of the electrolyte that water provides. With zinc, it is the coating that becomes the anode and dissolves: the iron, now the cathode, cannot be oxidized as long as zinc remains, and it is protected PRECISELY where it is exposed. With tin, the orientation is reversed: iron is the anode, and it corrodes faster than it would have alone, because it is now coupled to a large cathodic area of tin. An intact tin can keeps for years; scratched and wet, it is perforated in a few weeks.

One year is $(365)(24)(3600)=3.154\times 10^{7}$ s, so $Q=(0.50)(3.154\times 10^{7})=1.577\times 10^{7}$ C. Moles of electrons: $\frac{1.577\times 10^{7}}{96\,485}=163.4$ mol. Zinc gives up 2 electrons, hence $\frac{163.4}{2}=81.7$ mol of zinc, that is $m=(81.7)(65.38)=5.34\times 10^{3}$ g, about 5.3 kg per year. This sizing fixes the size of the anodes and how often they are replaced.

Because the principle is to force the structure to play the role of CATHODE: a cathode receives electrons and undergoes reduction, so it cannot be oxidized, that is, corroded. The sacrificial anode supplies these electrons by being consumed in its place. When an electrical source is available, impressed current is used instead: a generator injects the electrons into the structure and the anode, inert, is hardly consumed at all. Same principle, but we pay in electricity instead of paying in metal.