LearnChem · Part IV — What makes a reaction go further, or faster?
An iron gate began as red rock in a mine. The rock is a compound. Inside it, the iron is not the shiny thing you know. This module asks what the rock did to the metal. Then it asks what it costs to undo that. Part IV has been heading somewhere. This module ends there, with one number read three ways.
Why was there a Stone Age, then a Bronze Age, then an Iron Age — in that order?
There is a second question inside the first. Why is iron only ever dug up as rust? By the end you can place a metal on the diagram and on the ladder. Together they say which method frees it, and why the easy ones came first.
Module 21 left an iron grille going brown on a balcony. Beside it stood a ladder of electrodes, each with its own push. Rust was a reaction that runs by itself. A battery was that same push, made to do work. This module turns the grille's story round. Underground, a reaction such as rusting ran to the end long ago. The rock in the mine is the finished product. Now someone wants the metal back. What does it cost to run that reaction backwards?
One note concerns the syllabus. CBSE removed the Class 12 unit on getting metals out of their ores. Only the Class 10 chapter on metals and non-metals is still taught. That chapter is in words, with no numbers. This module teaches the topic anyway. It uses the reactivity series from Class 10. It uses the redox language from Class 11 and Module 21. The board will not ask for it. It is here because the question is worth answering.
Look at two things made of metal. One is a gold chain. It has been in a drawer for fifty years and still shines. The other is an iron gate. It has stood outside for ten years and is brown all over. Module 21 explained the brown. Now go further back, to before either was made. Where does each metal come from? Iron never comes out of the ground as iron. It comes out as a red or black rock. That rock is called an ore. An ore is rock that holds a metal locked inside a compound. Gold has a different story, and it is the puzzle below.
Gold is one of the rarest metals in the crust. Yet people have taken it from riverbeds for thousands of years with no furnace at all. Why?
Answer from your gut. The reveal follows.
Gold is nearly always found as the metal itself. A few others sometimes are, such as silver and copper. Such a metal is called native. Native means found as the metal, not in a compound. Iron is almost never native. The exceptions are lumps that fell from space. On Earth it is always found combined, mostly with oxygen. Module 21 showed what combining with oxygen means. The iron atom gives electrons away. So an ore is a metal that has already given its electrons away. Gold in a river never gave anything away. That difference is the whole of this module.
This is bookkeeping, not a real per-atom charge. Gold sitting as the metal is 0. Iron locked in its ore is already +3.
The two numbers above are oxidation numbers, from Module 21. Gold as metal sits at 0. It has given nothing away. Iron in its ore sits at +3. Each atom has given three electrons away. Now look at the belief the gate tested. Gold is rare, and it is easy to get. Aluminium is the commonest metal in the crust. It was one of the hardest to free. Rarity and difficulty are different questions. Difficulty is about electrons, and the rest of this module says how.
Silver is often found native, like gold. Yet old silver jewellery goes black over the years. Does that break the native-metal idea?
So native occurrence is the exception. Every other metal starts in a compound. From here on, this module is about those. When it says ore, it means a compound.
— A gold chain never lost its electrons. An iron gate already has.
Iron ore is iron that gave its electrons to oxygen. Rust is the same thing. Module 14 gave you a referee for any change, ΔG. ΔG is the free energy change of a reaction. Written ΔG°, it is that change under standard conditions. A negative ΔG° means the change runs by itself. Iron turning to oxide has a very negative ΔG°. That is why it ran underground without help. Now the ore goes into a furnace. Out comes iron. Before the mechanism, say what you think does the work in there.
A furnace turns iron ore into iron. What is really doing that job?
Predict, then read what is really going on.
Getting a metal out is called smelting. Smelting is not melting. The metal gains its electrons back, and that is chemical reduction. Reduction here means exactly what it meant in Module 21, gaining electrons. Heat has two jobs. It makes the reaction fast, and it melts the metal so it runs clear of the waste. The waste is called slag. Now comes one limit on everything that follows. The diagram this module builds is drawn for oxides only. Zinc's main ore is not an oxide. It is zinc blende, a sulphide, ZnS. Lead's and copper's main ores are sulphides too. A sulphide ore is first roasted in air. Roasting swaps its sulphur for oxygen. Only then does the oxide diagram apply. The equation below is that roasting step.
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
Picture the electrons as a loan. The metal gave them away long ago. Whoever took them now holds the debt. Usually that is oxygen. In a sulphide it is sulphur. In a chloride it is chlorine. To get the metal back, someone must pay the debt. Carbon can pay it, by giving its own electrons to the oxygen. An electric current can pay it. Or another metal can take the debt on itself. Aluminium does that to iron oxide in the thermite reaction. The aluminium ends up owing, and the iron walks free. The picture breaks in two places. A loan is agreed to. An ore never agreed to anything, and the electrons did not choose to go. And a native metal never borrowed at all. Gold owes nothing, so nothing is paid.
So every price this module quotes is the price of an oxide. If the rock is a sulphide, roasting comes first, at its own cost. What is the price of an oxide? It is the ΔG° of making it from metal and oxygen. Chemists plot that ΔG° against temperature, one line per oxide. The plot is called an Ellingham diagram. The next section reads its lines.
A galvanised iron bucket wears a thin coat of zinc. That zinc began as zinc blende, a sulphide rock. It was roasted to zinc oxide first. Only then was the zinc taken back out of the oxide. So the coat that guards the bucket has been bound to sulphur, then to oxygen, and now to nothing. One more thing about zinc matters. Most of it is never smelted in a furnace at all. A furnace can free zinc, and for most plants it is simply not the cheaper way. In 2008 about 86 per cent was roasted, dissolved in acid, and pulled out by electric current. A furnace made about 10 per cent.
Look at the roasting equation above, 2ZnS + 3O2 → 2ZnO + 2SO2. Which element is oxidised in that step?
— Every oxide has a price in ΔG°, and most prices fall as it gets hotter.
Here is the Ellingham diagram in words. Along the bottom runs temperature. Up the side runs ΔG° for making an oxide, per mole of oxygen. Every line is written to use one mole of O2. That is what lets you compare iron with aluminium on one plot. Lower on the plot means a more stable oxide. Every metal-oxide line slopes upwards as it gets hotter, and steeper still once the metal boils. The oxide gets a little less stable with heat. One line does the opposite. It is carbon's line, for 2C + O2 → 2CO. It slopes down. The arithmetic below says why, from the table's own two points.
Every metal-oxide line rises as the furnace gets hotter. Carbon's own line falls instead. What is different about carbon's reaction?
Predict, then check the numbers.
Yes. Count the gas on each side. 2C + O2 → 2CO takes one mole of gas and makes two. Two moles of gas have more arrangements than one. So ΔS° for carbon's reaction is positive. Module 14 wrote ΔG° as ΔH° minus TΔS°. ΔS° is positive here. So the −TΔS° term grows more negative as T rises. So ΔG° falls, and the line slopes down. Every metal-oxide line does the reverse. It uses up a mole of O2 and makes a solid. Gas is lost. Eight oxide lines sit on the next bench. Carbon's CO line is the only one that gains gas. Nothing about carbon's character is at work here. It is bookkeeping about moles of gas. That one fact is carbon's whole power as a reducer. The arithmetic below puts a number on the slope.
Here is the arithmetic, worked in the open, from NIST-JANAF table C-093. At 298 K, ΔG° for 2C + O2 → 2CO is −274.3 kJ per mole of O2. At 2000 K the same line has reached −572.1 kJ per mole of O2. Subtract the first from the second, and −572.1 minus −274.3 is the change. Taking away a negative is adding, so that is −572.1 + 274.3, which is −297.8 kJ/mol. Subtract the temperatures too, and 2000 K minus 298 K is 1702 K. Divide the change in ΔG° by the change in temperature, so −297.8 divided by 1702. That slope comes to −0.175 kJ per mole per kelvin, and its sign is negative. Module 14 wrote ΔG as ΔH minus TΔS. ΔH is the heat taken in, and ΔS is the change in disorder, Module 14's count of arrangements. So the slope of ΔG° against T is −ΔS°, as long as both barely change with temperature. So ΔS° is −(−0.175), which is +0.175 kJ/(mol·K). Multiply 0.175 by 1000 to turn kilojoules into joules, and ΔS° is +175 J/(mol·K). The sign is positive, and the readout below repeats the slope and the entropy change.
The readout gives the slope as −0.175 kJ per mole per kelvin. Read that as +175 J/(mol·K) of entropy. Now put iron beside it. Iron's line runs the other way. Its ΔS° is negative, so its line slopes up. The two lines lean towards each other as the furnace heats. The hotter the furnace, the cheaper carbon's oxide gets.
Carbon has a second line, for C + O2 → CO2. Count the gas on each side first. What should that line do as it gets hotter?
Now take the word crossing. Two lines cross where their prices are equal. Below the crossing, one oxide is the cheaper to make. Above it, the other is. Put carbon's CO line against iron's. Where they cross, carbon's oxide becomes the cheaper one. Above that temperature, carbon will take iron's oxygen. That crossing is a number, and the next bench reads it off.
— Carbon's line is the one line getting cheaper as it gets hotter.
Now put the lines on one plot and let them cross. Bench 1 draws eight oxide lines and both carbon lines. Every point comes from the NIST-JANAF tables, per mole of O2. Between the table's points the bench draws straight lines. That is interpolation, and it is the bench's one approximation. Where a metal's line crosses carbon's CO line, carbon wins above that temperature. So each oxide has its own crossing temperature. A low crossing means an easy furnace. Before you drive the bench, predict one pair. Iron and magnesium are both on it.
Which oxide's line crosses carbon's line at the lower temperature, iron's or magnesium's?
Predict, then drive the bench.
Bench 1 is the diagram itself. Pick an oxide and slide the temperature. The readout gives both lines' ΔG° at that point and names the crossing. Iron's line crosses carbon's near 1044 K. Magnesium's crosses far higher, above 2100 K, which is about 1840 °C. What may the page claim from a crossing? Only that above that temperature carbon's oxide is the cheaper one on paper. Nothing on this bench says what happens to the metal next. The next section is about that gap.
Tick the box to add zinc's line. It is worked out from other tables, not read from JANAF, so it is marked as derived. Two oxides are missing on purpose, and the note under the bench says which.
Manganese oxide and silver oxide are left off this bench, on or off the box. Neither has a number this course could check well enough to plot. That is a stated gap, not a hidden one.
Read the crossings from the lowest up. Copper's line crosses lowest of all. Iron's crosses near 1044 K. A Delhi University handout, built from other tables, puts it near 1000 K. So the bench is not marking its own homework. Chromium's crosses just above 1500 K. Silicon's and titanium's cross higher, above 1900 K and above 2000 K. Magnesium's crosses higher still. Now tick zinc. Its derived line crosses carbon's at about 942 °C, near 1216 K. Zinc boils at 907 °C. So the zinc leaves a furnace as a gas, not a liquid. Now turn to the history. Copper was smelted long before iron. Iron came before anything that crosses higher. So the crossings give the order of the ages, for every metal that stays put once it is freed. Zinc shows why that clause is needed. Its crossing sits only about 170 K above iron's, yet zinc was smelted long after iron. It does not stay put, and the next section says what that costs. One more honest caveat sits in the box below.
Stone, then Bronze, then Iron is a museum order. Christian Jürgensen Thomsen arranged the Danish National Museum's finds that way between 1816 and 1825. Later it became a timeline for everyone, and it does not fit everyone. In much of sub-Saharan Africa people went from stone to iron, with no bronze age at all. In the Americas people smelted copper and made alloys, and never smelted iron before 1492. So the diagram explains why the order, where the order happened. It does not say history had to run that way. One more distinction matters. Native copper hammered cold is not smelting. Only copper taken from ore, around 5000 BC in Serbia, is reduction. Now come the dates nearest home. Iron in the middle Ganga valley is well dated by Tewari's radiocarbon work, to about 1800 BC. Tamil Nadu's own claims are older still. The state archaeology department dates Sivagalai charcoal to about 3345 to 2953 BC. That claim is contested. Rice grains from the same site date to about 1248 to 1155 BC, two thousand years later. At Mayiladumparai, archaeologist Disha Ahluwalia has pointed out one more gap. The dated charcoal there is not shown beside iron in the same layer. The Serbian and Ganga dates also rest on single teams. Nothing found at those sites contradicts them, and that is the difference. Until Sivagalai's gap is explained, the claim is a claim. It is reported here as one.
Tin's ore is cassiterite, SnO2. Tin bronze was in use by about 4650 BC, long before smelted iron. Where should tin's line sit on the diagram?
— The crossings order the ages, for every metal that stays put once it is freed.
Bench 1 is a real calculation from real tables. Every crossing on it is right, as far as ΔG° goes. Take magnesium. Its line crosses carbon's high up, as you saw. Above that crossing, carbon's oxide is the cheaper one. So a hot enough furnace should hand you magnesium metal. Commit to what comes out before you read what a real furnace gives.
Above about 2120 K, which is about 1850 °C, the diagram says carbon can take the oxygen off magnesium oxide. Run that furnace for real. What happens to the magnesium you make?
Commit before you read what a real furnace does.
At the crossing, the products are magnesium gas and carbon monoxide gas. Magnesium boils at 1363 K, far below the crossing. Both products must be cooled to collect the metal. Cooling reverses the sign. Mg(g) + CO(g) → MgO(s) + C(s) runs as soon as the gas mixture cools, and it runs fast. The metal comes out coated in its own oxide again. Brooks and colleagues called stopping this the major technical challenge, in 2006.
Two crossing figures exist, and both are right. Brooks gives 1764 °C, about 2037 K. That is where the product gases together reach one atmosphere, the pressure of ordinary air. The strict reading takes ΔG° = 0 from the JANAF points. It gives about 2110 to 2120 K, which is 1840 to 1850 °C. Standard state means each gas at one atmosphere on its own, not shared. The first asks when gas starts to form. The second asks when every species reaches standard state. So the crossing is real, and it lies between about 1750 and 1850 °C. Real magnesium comes two other ways. The Pidgeon process uses silicon to take the oxygen, under vacuum. Or molten magnesium chloride is split by electric current. Neither uses carbon.
Zinc is the milder case of the same trap. Its crossing sits above zinc's boiling point, so zinc also leaves as a vapour. Cooled slowly in the furnace gas, it turns back to oxide. A real zinc furnace catches the vapour in a spray of molten lead, fast. That is engineering, and the diagram shows none of it.
Here is the arithmetic, worked in the open, for titanium's crossing. The bench's grid gives TiO2's line and carbon's CO line at 2000 K and at 2100 K. At 2000 K, TiO2's line minus carbon's line is −13.4 kJ per mole of O2. The sign is negative, so titanium's oxide is still the cheaper one there. At 2100 K the same subtraction gives +21.2 kJ per mole of O2. The sign has turned positive, so carbon's oxide is now the cheaper one. The crossing sits between the two, where the gap is zero. The gap moves from −13.4 to +21.2 over those 100 K. That change is +21.2 − (−13.4), which is +34.6 kJ per mole of O2. The gap needs +13.4 of that +34.6 to reach zero. Divide 13.4 by 34.6, and the fraction is 0.387. Multiply 0.387 by 100 K, which is 38.7 K, and add it to 2000 K. The crossing comes to about 2039 K, a little above 1765 °C. A common figure is 1650 °C, from a course handout that gives no source for it. This checked figure is more than 100 °C higher, and it was not tuned to match anything.
That crossing is moot. Carbon heated with titanium oxide does not give titanium. It gives titanium carbide, TiC, at any temperature this bench can reach. TiC is more stable than titanium metal beside carbon. The JANAF tables put the margin at about 158 to 172 kJ/mol, across 1100 to 2000 K. Any titanium that forms reacts on into the carbide. So a real plant does not use carbon on the oxide. It uses the Kroll process. The oxide is first turned into titanium chloride, TiCl4. Then molten magnesium takes the chlorine, as TiCl4 + 2Mg → Ti + 2MgCl2. Here is the loop worth noticing. Titanium is made with magnesium. The magnesium chloride left over is split back into magnesium and chlorine by electric current. Both go round again.
The Pidgeon process makes magnesium with silicon under a vacuum, not in open air. Why does a vacuum help?
— A crossing prices the oxygen, and says nothing about what you can collect afterwards.
Your Class 10 book sorts metals on a ladder. The most reactive sit at the top. Reactive here means quick to give electrons away. A metal high on the ladder holds its oxide very tightly. That oxide's line sits very low on Bench 1. Some lines sit so low that carbon's line never reaches them at any furnace temperature. For those, heat is not enough. Something else must pay for the electrons. Two ladder metals are named below. Predict which one carbon cannot help.
Lead and calcium both sit on the ladder, one below iron and one above. Which one can carbon never free from its oxide, anywhere Bench 1 can show?
Predict, then check the ladder.
Bench 2 is the same ladder your Class 10 book prints, NCERT Table 3.2. It has 13 entries, in the book's own order, and nothing is added. Hydrogen is one of them, in brackets between lead and copper. That rung is the hinge to Module 21. Module 21's electrode ladder set hydrogen at 0 V by definition. It sits in the same place here. A metal higher up can take oxygen from the oxide of one lower down. Pick two metals. The bench says which sits higher, and which method frees each one.
Your book groups the ladder into three bands. The top band is freed by electrolysis. Electrolysis means using an electric current to force a reaction the other way. The middle band is freed by carbon. The bottom band is found native, or freed by heat alone. What may the page claim from this bench? Only the order the book prints, and the band the book gives each metal. It measures nothing.
Calcium sits near the top of the ladder, and its oxide line never meets carbon's anywhere Bench 1 can show. Aluminium sits one rung below calcium, and its line never meets carbon's there either. No real furnace reaches further, so carbon has never paid either debt. Electricity does. The Hall–Héroult cell does it for aluminium. The oxide is dissolved in a molten salt bath near 960 °C. A current pushes electrons back onto the aluminium at one electrode. Oxygen is released at the other electrode, the anode, which is made of carbon. An anode is the electrode where electrons are taken away. Module 21 gave you the tool for the price. A ΔG° can be read as a voltage. The arithmetic below reads aluminium's.
Here is the arithmetic, worked in the open, at 1500 K. That is a JANAF grid point, not the bath, which is cooler. Tier 2 below works the same sum at the bath's own 1233 K. Making aluminium oxide at 1500 K has ΔG° = −797.75 kJ per mole of O2, from table Al-096. Undoing it costs the same amount with the sign reversed, +797.75 kJ per mole of O2. Per mole of O2, four electrons move, because O2 + 4e− → 2O2−. Module 21 gave the Faraday constant F as 96,485.332 coulombs per mole of electrons. Four moles of electrons carry 4 × 96,485.332 C, which is 385,941.328 C. One coulomb times one volt is one joule, so each volt here is worth 385,941.328 J per mole of O2. Convert the cost to joules, and 797.75 kJ is 797,750 J per mole of O2. Divide 797,750 J by 385,941.328 J per volt, and the minimum voltage is 2.0670 V. That is with a bare anode that gives off oxygen gas. A real cell has a carbon anode, and the oxygen burns the carbon to CO2 as it forms. Making CO2 at 1500 K has ΔG° = −396.3 kJ per mole of O2, from table C-095. So the net cost falls to 797.75 − 396.3, which is +401.45 kJ per mole of O2. Divide 401,450 J by 385,941.328 J per volt, and the minimum falls to 1.04 V. Real cells run at about 4.2 to 4.5 V, about four times that minimum. The gap is not an error. It is overvoltage and ohmic loss. Both are extra push that a real electrode and a real bath demand. Overvoltage is the extra voltage an electrode needs before its reaction runs at a useful rate. Ohmic loss is the voltage spent pushing current through the bath's own resistance. The readout below repeats the bare-anode figure.
Two processes in this section run at extreme temperatures, and both are hazards, not demonstrations. The thermite reaction, aluminium with iron oxide, can reach more than 3000 °C. Wang, Munir and Maximov reported that in 1993. Aluminium powder catches fire on its own in air. That is GHS hazard statement H250. In contact with water it releases flammable gas, H261. Aluminium smelting runs a molten fluoride bath at 940 to 980 °C, and a fluoride bath can release hydrogen fluoride. Magnesium's chloride bath releases chlorine at the anode. Chlorine is an oxidiser, H270, and toxic if breathed in, H331. NIOSH's immediately-dangerous limits are 10 ppm for chlorine and 30 ppm for hydrogen fluoride. Nothing here is to be tried. Periodic Videos is this page's demonstration source of record. Look there, not at a bench.
Wang, Munir & Maximov (1993), J. Mater. Sci. 28, 3693–3708; ILO/WHO ICSC 0988 (aluminium powder, H250/H261); ERCO Worldwide chlorine SDS (H270/H331); NIOSH IDLH, chlorine CAS 7782-50-5 and hydrogen fluoride CAS 7664-39-3.
Strong oxidisers are the other hazard class this section touches. An oxidiser takes electrons from other substances. A strong one does it fast enough to feed a fire. Oxygen gas is the anode product of electrolytic extraction. It is classed as an Oxidizing Gas, Category 1, hazard statement H270, may cause or intensify fire. Potassium permanganate is a named example of the solid class. It is an Oxidizing Solid, Category 2, hazard statement H272, may intensify fire. Category 1 solids and liquids carry H271, may cause fire or explosion. Demonstrations of oxidisers are cited video only, from Periodic Videos.
Airgas Oxygen SDS (H270, Oxidizing Gases Cat. 1); Geneseo College potassium permanganate safety document (H272, Oxidizing Solids Cat. 2).
Copper sits below hydrogen on this ladder. A copper coin sits in dilute hydrochloric acid. What happens?
— Where carbon cannot pay, an electric current or a metal higher up the ladder does.
Three numbers have run through Part IV. Module 14 gave ΔG°, the referee. Modules 17 and 18 gave K, the equilibrium constant. K is the ratio of products to reactants when a reaction has settled. Module 21 gave E, the push of a cell in volts. Under standard conditions it is written E°. They arrived in different chapters, in different units. This section says they were one fact all along. Start where a crossing on Bench 1 lives. At the crossing, ΔG° is exactly zero. Predict what K is there.
At the exact temperature where a reaction's ΔG° is zero, what is its equilibrium constant K?
Predict before the identity settles it.
Here is the identity in one line. ΔG° = −nFE° = −RT ln K. n is the moles of electrons and F is the Faraday constant. R is the gas constant and T is the temperature. Read it left to right. Divide ΔG° by the charge that moves, nF, and you have E° with its sign flipped. That is the last section's coulomb times volt is a joule, read backwards. K is how far a reaction can run before its push is spent. ln K is the logarithm of that ratio, and RT turns the logarithm into an energy. So all three are one energy in three units.
Set ΔG° to zero and both other terms go to zero. E° is 0 V. ln K is 0, so K is exactly 1. All three happen together, at one temperature, exactly. That temperature is an Ellingham crossing. So a crossing is the triangle's zero. That is why a plant cares about the crossing. For any oxide line written per mole of O2, n is 4. So 4F is 385,941.328 joules per volt, and it converts every line on Bench 1 into volts.
Bench 3 runs that identity. Pick a case and an offset from its reference temperature. One calculation writes all three numbers at once. The Daniell cell is Module 21's zinc and copper cell, with n = 2. Its reference is 298.15 K. The aluminium oxide case is the Hall–Héroult cell from the last section, with n = 4. Its reference is 1500 K. Tick the carbon-anode box and watch the cost fall. What may the page claim from this bench? That the three numbers move together, for the case shown, and nothing more. Only the Daniell corner can be checked against numbers measured another way. That check is the arithmetic below.
Now the third case, and it is the one that matters. Set the case to iron. It is 2FeO + 2C → 2Fe + 2CO, the blast furnace's own reaction. Both lines are NIST-JANAF's own, Fe-018 and C-093. Its reference temperature is not chosen. The bench computes it, as the crossing of those two lines. That is the crossing Bench 1 drew, near 1044 K, which is about 771 °C. At that temperature the books exactly balance. Nothing is made yet. It is where carbon starts to win, and a real furnace runs hundreds of degrees above it. Leave the offset at zero and read all three. ΔG° sits at 0.00, E° at 0.0000 V and log K at 0.00, so K is exactly 1. Now slide 200 K down. ΔG° reads +61.04 kJ per mole of O2, E° turns negative and K drops below one. The oxide still holds. Slide 200 K up instead. ΔG° reads −60.00, E° turns positive and K climbs above one. Carbon wins, and iron comes out. All three move together, every time. That whole window sits inside FeO's own tabulated range, so nothing here is extrapolated. That is one temperature in two pictures. On Bench 1 two lines cross. Here three readouts hit zero together. That is why a plant cares about the crossing. It is the floor a furnace must stay above.
Here is the same triangle for any oxide line at 1200 K, with n = 4. E° is −ΔG° divided by 385,941.328 J per volt, with ΔG° in joules. log10K is −ΔG° divided by RT ln 10. At 1200 K that divisor is 8.314 × 1200 × 2.3026, which is 22,974 J/mol. So one factor of ten in K costs 22.974 kJ/mol at this temperature. The table below runs four values of ΔG° through both divisions. Watch the second row, where ΔG° is zero, E° is zero and K is one.
Here is the arithmetic, worked in the open, on the Daniell cell at 298.15 K. Module 21 gave its standard voltage, E° = 1.10 V, with n = 2 electrons per reaction. So ΔG° = −nFE° = −(2)(96,485.332 C/mol)(1.10 V), three factors and a minus sign. Multiply 2 by 96,485.332 by 1.10, and the product is 212,268 J/mol. Put the minus sign back, and ΔG° is −212,268 J/mol, which is −212.268 kJ/mol. Now the same number from a different laboratory tradition, calorimetry, which never touches a voltmeter. Tabulated ΔfG° gives Zn2+ as −147.06 kJ/mol and Cu2+ as +65.49 kJ/mol. ΔG° is products minus reactants, so −147.06 − (+65.49), which is −212.55 kJ/mol. The two routes differ by 212.550 − 212.268, which is 0.282 kJ/mol. Divide 0.282 by 212.55 and multiply by 100, and the difference is 0.13 per cent. Heat measurements and voltage measurements agree to about one part in a thousand. Now take the third corner. ΔG° = −RT ln K, so ln K is −ΔG° divided by RT. ΔG° is −212,268 J/mol, and minus a negative is positive, so ln K is +212,268 divided by 8.314 × 298.15. That denominator is 2478.8, and 212,268 divided by 2478.8 is 85.63. Divide 85.63 by 2.3026 to turn ln into log10, and log10K is 37.19. Ten to that power is about 1.5 × 1037. Nobody can weigh out a ratio of ten to the thirty-seven. This corner exists only through the identity. The readout below repeats the first and third numbers.
At 1200 K a reaction has ΔG° = −80 kJ/mol O2. That sits between two rows the table showed, −40 and −120. About what is E°?
Look at the −40 row again. A reduction 40 kJ per mole of O2 downhill is worth 0.104 V across four electrons. That is a tenth of a volt. A torch battery is 1.5 V, fourteen times more. So a blast furnace, on that reading, runs on a fourteenth of a torch battery. And it makes a million tonnes of iron. Nothing is wrong here. What a plant buys is ΔG° times tonnage, not voltage. Repeat that tenth of a volt for every atom in a million tonnes. That is the whole of the iron industry. E° and ΔG° are one statement in two units. This row is the cleanest place in the course to see it.
— ΔG°, E° and K are the same fact, read off in three different units.
A reaction has ΔG° = {dg} kJ/mol O2 at T = {t} K. Find log10K. Module 14's referee is ΔG°, and Module 12 gave its unit, the kilojoule per mole. Module 17 built K. Use ΔG° = −RT ln K with R = 8.314 J/(mol·K), and put ΔG° into joules first. Divide ln K by 2.3026 for log10K. Watch the sign, because a negative ΔG° gives a positive log K. Four of the eight oxides on Bench 1 are Module 11's transition metals. This formula gives K for every one of them.
1.
2. Aluminium oxide's real Hall–Héroult bath runs at about 1233 K, with a carbon anode. What is the minimum decomposition voltage there? JANAF gives the oxide line as −907.63 kJ/mol O2 at 1000 K and −797.75 at 1500 K. Interpolate between them in a straight line. Take the CO2 line at 1233 K as −396.09 kJ/mol O2.
Here is the worked solution, and the interpolation is this page's own, not a JANAF number. The bath at 1233 K sits 233 K above the 1000 K grid point. The grid points are 500 K apart, so the fraction is 233 divided by 500, which is 0.466. The line rises from −907.63 to −797.75 between them, a rise of 109.88 kJ/mol O2. Multiply 0.466 by 109.88, which is 51.20, and add it to −907.63. That gives ΔG° for making the oxide at 1233 K as −856.43 kJ/mol O2. Undoing it costs +856.43 kJ/mol O2, with the sign reversed. The carbon anode turns the oxygen into CO2, whose line at 1233 K is −396.09 kJ/mol O2. Subtract, and the net cost is 856.43 − 396.09, which is 460.34 kJ/mol O2. Convert to joules, 460,340 J, and divide by 385,941.328 J per volt. The minimum is 1.19 V, and the readout below rounds the same way. Without the carbon anode, 856,430 J divided by 385,941.328 J per volt gives 2.22 V. The commonly quoted theoretical figure for the real process is about 1.2 V. This page's 1.19 V is a cross-check against that figure, not a citation of it. Real cells run at 4.2 to 4.5 V, and the difference is overvoltage and ohmic loss.
In ore, a metal has given its electrons away. Getting it out pays them back.
Sources for this module, in the order the page uses them. Every line on Bench 1 is from the NIST-JANAF Thermochemical Tables, fourth edition, Chase 1998, at janaf.nist.gov. The tables used are C-093, C-095, Fe-018, Mg-008, Al-096, Ca-027, Cr-014, Cu-019, O-037 and O-043. Zinc's derived line uses CODATA key values with JANAF Zn-004 and Zn-005 for the phase changes. The iron crossing's independent check is a Shivaji College, Delhi University, metallurgy handout. The magnesium crossing and its reversion are from Brooks and colleagues, JOM 58, 2006, and Yang and colleagues, Journal of Magnesium and Alloys 2, 2014. Titanium carbide is from JANAF C-107 and from Lv, Tian and Hu, Processes 12, 2024. The Kroll process is from Okabe and Takeda, 2020. Zinc production shares are from Sohn and Olivas-Martinez, Treatise on Process Metallurgy, 2014, and are 2008 figures. Bench 2 is NCERT Science, Class X, Chapter 3, Table 3.2, page 45, reached through the Internet Archive's mirror because ncert.nic.in refused every request. The Daniell cell's 1.10 V is Module 21's, and the ion ΔfG° values are from a standard thermodynamic data table. F = 96,485.332 C/mol and R = 8.314 J/(mol·K) are CODATA values. The antiquity dates are from Radivojević and colleagues, 2010, 2013 and 2021, Erb-Satullo, 2019, and Tewari, Antiquity 77, 2003. The Tamil Nadu dates are the Tamil Nadu State Department of Archaeology's reports of 2020 to 2022, with Disha Ahluwalia's critique, and are printed as contested. The syllabus note was checked against the CBSE 2025–26 Class XII syllabus. Safety statements are from the documents named under each box. Periodic Videos is the demonstration source of record. Two secondary figures found during research failed against these sources and appear nowhere on this page. Verified 27 September 2026.