🧪 Chemistry Exam 6 · Chapters 16–18
Chemistry · Exam 6

Exam 6 Study Guide

Exam 6 covers chapters 16, 17 and 18. Two of those chapters are about which way a change wants to go. The third one walks the main groups element by element. Chapter 16 gives you entropy and free energy. Chapter 17 puts free energy into a wire and gets a voltage out. Chapter 18 then shows that whole toolkit at work on real elements. Do the worked examples with a pencil first, then take the self-test at the end of each chapter.

What this guide holds

  • Chapter 16: spontaneous change, entropy, the second and third laws, ΔS°, Gibbs free energy, and the link from ΔG° to K.
  • Chapter 17: balancing half-reactions, galvanic cells, standard reduction potentials, the tie among E°, ΔG° and K, the Nernst equation, electrolysis and corrosion.
  • Chapter 18: main group trends, groups 1 and 2, boron through silicon, nitrogen and phosphorus, oxygen and sulfur, the halogens and the noble gases.
  • Every equation with its units, three worked examples per chapter, and the traps that cost the most points.
  • A 10-item self-test per chapter, and a one-page cram sheet at the end.

Chapter 16Thermodynamics

What to know

A change is spontaneous if it keeps going on its own once it starts. Nothing outside has to drive it. A change that needs steady outside work is nonspontaneous, and it is always the reverse of a spontaneous one.

Spontaneous says nothing about speed. Graphite is the favored form of carbon at room temperature, with ΔG° = −2.9 kJ/mol for diamond turning into it. No ring has ever gone grey. The atoms sit in strong bonds and cannot move. Thermodynamics names the destination. Kinetics sets the travel time.

Heat release helps, but it is not the whole rule. Ice melts in a warm room while soaking heat up. A cold pack does the same. So a second force is at work, and that force is how spread out the matter and the energy end up.

Entropy, written S, measures that spread. A state with many ways to arrange itself has high entropy. Each full arrangement is one microstate. Ludwig Boltzmann turned that counting into a formula, S = k ln W. The log matters here. Multiply the number of ways by ten and S rises by only k ln 10.

Four entropy patterns that answer most questions
  • Gas beats liquid beats solid. Water is 70.0 J/mol·K as a liquid and 188.8 J/mol·K as steam.
  • Warmer beats cooler for the same substance in the same phase.
  • Larger molecules beat smaller ones, since they hold more ways to vibrate.
  • A dissolved solid beats the same solid sitting in a pile.

The second law of thermodynamics gives the real test. A change is spontaneous when the total entropy of the universe goes up. Two ledgers are open at once, the system and the surroundings. Either one may show a loss. Only the sum has to grow.

The surroundings are enormous, so their temperature barely moves when they take in heat. That makes their entropy change easy to write as −ΔH ÷ T. Read the minus sign with care. Heat that leaves the system lands in the surroundings, so an exothermic change raises their entropy. And look at the T on the bottom. The same joule of heat buys more entropy in a cold room than in a hot one.

The third law of thermodynamics sets the zero point. A perfect crystal of a pure substance at 0 K has an entropy of exactly zero. Each particle sits in one fixed spot, so W = 1 and S = k ln 1 = 0. This is why tables can list a real, absolute standard molar entropy S° for every substance. Every listed value is positive, even for an element.

So ΔS° for a reaction is products minus reactants, each term multiplied by its coefficient. Elements do not drop out, because their S° is not zero. That is the big difference from an enthalpy table. Often the sign is all you need, so count the moles of gas on each side.

Gibbs free energy folds both ledgers into one number for the system alone. Start from the second law, swap in the heat term, then multiply through by −T. Multiplying by a negative number flips the direction of the sign, which is where the minus in the formula comes from.

The four sign cases of ΔG = ΔH − TΔS
  • ΔH negative and ΔS positive: spontaneous at every temperature. Both terms help.
  • ΔH positive and ΔS negative: never spontaneous. Both terms fight you.
  • ΔH negative and ΔS negative: spontaneous when cold, and it switches over somewhere.
  • ΔH positive and ΔS positive: spontaneous when hot, and it switches over somewhere.

Free energy also fixes the size of K. A negative ΔG° makes K larger than 1, so products win. A positive ΔG° makes K smaller than 1. A ΔG° of zero gives K = 1. At 298.15 K, every 5.71 kJ of ΔG° moves K by a factor of ten, and that figure is just RT ln 10.

Keep ΔG° and ΔG apart. The circle means the standard state, with each gas at 1 bar and each solute at 1 mol/L. The push you feel at some other mixture is ΔG = ΔG° + RT ln Q. Once Q climbs to K, the log term cancels ΔG° and ΔG drops to zero. That is equilibrium, seen from the free energy side.

In the two switching cases, one temperature marks the turn. Set ΔG to zero and solve. That value is the crossover temperature, and K = 1 there. The method assumes ΔH° and ΔS° hardly drift with temperature, which holds well enough over a few hundred kelvin.

Equations, units and symbols

EquationWhat each symbol meansUnits
S = k ln W W is the number of microstates. k is the Boltzmann constant, which is R ÷ NA. k = 1.381 × 10-23 J/K. W is a bare count. S in J/K.
ΔS = R ln(V2 ÷ V1) For one mole of gas changing volume at steady temperature. Doubling it gives R ln 2. R = 8.314 J/mol·K, so ΔS is in J/mol·K.
ΔSuniv = ΔSsys + ΔSsurr > 0 The second law. Positive means spontaneous, zero means equilibrium. J/K.
ΔSsurr = −ΔHsys ÷ T At steady pressure. Heat leaving the system warms the surroundings, so the sign flips. ΔH in J, T in K, result in J/K.
ΔS°rxn = ΣnS°(products) − ΣnS°(reactants) n is each coefficient. Elements stay in the sum, because S° for an element is not zero. S° in J/mol·K, so ΔS° comes out in J/K.
ΔG = ΔH − TΔS Negative means the forward change is spontaneous. Positive means the reverse is. Zero means equilibrium. Match the units first. ΔH in kJ needs ΔS in kJ/K.
ΔG° = ΣnΔG°f(products) − ΣnΔG°f(reactants) ΔG°f is zero for an element in its standard form. This route works at 298.15 K only. kJ/mol in the table, kJ in the answer.
ΔG° = −RT ln K Rearranged, K = e−ΔG°/RT. Since R is in joules, put ΔG° in joules too. R = 8.314 J/mol·K, T in K, ΔG° in J/mol.
ΔG = ΔG° + RT ln Q The push at a mixture that is not the standard state. ΔG = 0 once Q = K. J/mol.
T = ΔH° ÷ ΔS° The crossover temperature, found by setting ΔG to zero. K = 1 there. Same energy unit top and bottom. Answer in K.
5.71 kJ per power of ten in K Equal to RT ln 10 at 298.15 K. A handy size check. kJ/mol.
Standard molar entropy S°J/mol·KStandard molar entropy S°J/mol·K
H2(g)130.7H2O(l)70.0
O2(g)205.2H2O(g)188.8
N2(g)191.6NH3(g)192.8
Cl2(g)223.1HCl(g)186.9
C(graphite)5.7CO2(g)213.8
CaCO3(s)91.7CaO(s)38.1

Worked examples

Example 16.1 · ΔS° from a table, twice

Find ΔS° for 2 H2(g) + O2(g) → 2 H2O(g). Then do it again for liquid water as the product, and say why the two differ.

Step 1 · add the products

Steam is 188.8 J/mol·K, and there are 2 moles. So 2 × 188.8 = 377.6 J/K.

Step 2 · add the reactants

2 × 130.7 = 261.4 J/K for the hydrogen, plus 205.2 J/K for the oxygen. Together that is 466.6 J/K.

Step 3 · subtract

ΔS° = 377.6 − 466.6 = −89.0 J/K.

Step 4 · now use liquid water

Liquid water is 70.0 J/mol·K, so the products come to 2 × 70.0 = 140.0 J/K. Then ΔS° = 140.0 − 466.6 = −326.6 J/K.

Answer and the sign check

Both values are negative, which the gas count predicted. Three moles of gas became two with steam, and zero with liquid water. The liquid case loses far more, because packing a gas into a liquid costs a huge amount of spread. Note that the elements stayed in the sum. Their S° values are not zero.

Example 16.2 · ΔG° at two temperatures, and the crossover

Ammonia breaks down: 2 NH3(g) → N2(g) + 3 H2(g), with ΔH° = +91.8 kJ and ΔS° = +198.1 J/K. Find ΔG° at 298.15 K and at 700 K, and find the crossover temperature.

Step 1 · fix the units

ΔH° is in kJ and ΔS° is in J/K, so they cannot be subtracted yet. Divide the entropy by 1000: ΔS° = 0.1981 kJ/K.

Step 2 · room temperature

ΔG° = 91.8 kJ − (298.15 K)(0.1981 kJ/K) = 91.8 − 59.1 = +32.7 kJ.

Step 3 · check that another way

From the table, ΔG°f for NH3(g) is −16.4 kJ/mol, and both elements are zero. So ΔG° = 0 − 2(−16.4) = +32.8 kJ. The two routes agree to the rounding.

Step 4 · now go hot

ΔG° = 91.8 kJ − (700 K)(0.1981 kJ/K) = 91.8 − 138.7 = −46.9 kJ.

Step 5 · the crossover

T = ΔH° ÷ ΔS° = (91.8 kJ) ÷ (0.1981 kJ/K) = 463 K, which is 190 °C.

Answer

Ammonia is stable in a cylinder at room temperature, where ΔG° is +32.7 kJ. Above 463 K the sign turns over, and by 700 K the breakdown is favored by 46.9 kJ. This is the hot-only sign case, with both ΔH° and ΔS° positive. Nothing about the molecules changed. Only the size of the TΔS term changed.

Example 16.3 · From ΔG° to K, at both temperatures

Use the same reaction, 2 NH3(g) → N2(g) + 3 H2(g). Find K at 298.15 K from ΔG° = +32.8 kJ, then at 700 K from ΔG° = −46.9 kJ.

Step 1 · put the energy in joules

R is 8.314 J/mol·K, so ΔG° = +32.8 kJ becomes +32,800 J/mol.

Step 2 · build the denominator

RT = (8.314 J/mol·K)(298.15 K) = 2478.8 J/mol.

Step 3 · solve for ln K

ln K = −ΔG° ÷ RT = −32,800 ÷ 2478.8 = −13.23. So K = e-13.23 = 1.8 × 10-6.

Step 4 · now at 700 K

RT = (8.314)(700) = 5819.8 J/mol. Then ln K = −(−46,900) ÷ 5819.8 = +8.06, so K = e8.06 = 3.2 × 103.

Answer and a cross-check

K rose from 1.8 × 10-6 to 3.2 × 103. The forward Haber reaction has K = 5.6 × 105 at 298.15 K, and 1 ÷ (5.6 × 105) = 1.8 × 10-6. Flipping a reaction inverts K, and the two numbers match.

The traps people fall into

Watch for these
  • Reading spontaneous as fast. Diamond turning into graphite is favored and still takes ages.
  • Subtracting J/K from kJ. Convert first. Forget this and the answer is off by a factor of a thousand.
  • Dropping the elements from a ΔS° sum. S° for an element is positive, not zero. Only ΔH°f and ΔG°f are zero for elements.
  • Losing the minus sign on ΔSsurr. An exothermic change gives the surroundings a positive entropy change.
  • Using Celsius. Every T in this chapter is in kelvin.
  • Using a ΔG°f sum away from 298.15 K. That temperature is baked into the listed values.
  • Mixing up ΔG° and ΔG. ΔG° is the standard state. ΔG is the push at the mixture you have.
  • Using R = 0.08206 in −RT ln K. That version of R is for gas law problems. Here use 8.314 J/mol·K.
  • Thinking a positive ΔS alone settles it. Only the total entropy of system plus surroundings decides.
  • Expecting S = 0 for any cold solid. The third law needs a perfect crystal of a pure substance at 0 K.

Self-test · Chapter 16

1. Predict the sign of ΔS° for 2 SO2(g) + O2(g) → 2 SO3(g), without a table.
Negative. Count the gas moles: 3 on the left and 2 on the right. Fewer moles of gas among the products means less spread, so ΔS° comes out negative. Gas moles decide the sign almost every time.
2. Find ΔS° for H2(g) + Cl2(g) → 2 HCl(g).
ΔS° = +20.0 J/K. Products: 2 × 186.9 = 373.8 J/K. Reactants: 130.7 + 223.1 = 353.8 J/K. Subtract: 373.8 − 353.8 = +20.0 J/K. Two moles of gas became two, so the value is small, and the phases had to decide it.
3. One mole of ice melts at 273.15 K, taking in 6.01 kJ. Find ΔS for the ice itself.
ΔS = +22.0 J/K. Put the heat in joules: 6.01 kJ = 6010 J. Then ΔS = 6010 J ÷ 273.15 K = +22.0 J/K. Positive, because a liquid is far more spread out than a crystal.
4. A reaction gives off 100.0 kJ at 298.15 K. Find ΔSsurr.
ΔSsurr = +335 J/K. Giving off heat means ΔHsys = −100.0 kJ, or −100,000 J. Then ΔSsurr = −(−100,000 J) ÷ 298.15 K = +335 J/K. The surroundings gained the heat, so they gained spread.
5. Water boils with ΔH° = +44.0 kJ and ΔS° = +118.8 J/K. Find the crossover temperature.
T = 370 K. T = ΔH° ÷ ΔS° = 44,000 J ÷ 118.8 J/K = 370 K. The real boiling point is 373 K, so two table entries landed within three degrees. Below this temperature ΔG° is positive and the liquid holds.
6. A reaction has ΔG° = −30.0 kJ at 298.15 K. Find K.
K = 1.8 × 105. RT = (8.314)(298.15) = 2478.8 J/mol. Then ln K = 30,000 ÷ 2478.8 = 12.10, so K = e12.10 = 1.8 × 105. A negative ΔG° always gives a K above 1.
7. Show that 5.71 kJ of ΔG° is worth one power of ten in K at 298.15 K.
It is RT ln 10. RT = (8.314 J/mol·K)(298.15 K) = 2478.8 J/mol, and ln 10 = 2.303. Multiply: (2478.8)(2.303) = 5708 J/mol, which is 5.71 kJ/mol. So a modest −30 kJ already buys about five powers of ten.
8. Which of the four sign cases is spontaneous only when hot, and why?
ΔH positive with ΔS positive. The TΔS term grows with temperature while ΔH holds still. Once TΔS outgrows ΔH, the difference turns negative. Roasting limestone is the classic case, with a crossover near 1120 K.
9. One mole of gas doubles its volume at steady temperature. Find ΔS.
ΔS = +5.76 J/mol·K. Each particle gets twice as many spots, so W doubles once per particle, giving a ratio of 2 raised to Avogadro's number. The log turns that into NAk ln 2, which is R ln 2. So ΔS = (8.314 J/mol·K)(0.693) = +5.76 J/mol·K.
10. Roasting limestone has ΔH° = +179.2 kJ and ΔS° = +160.2 J/K. Find ΔG° at 298.15 K and at 1200 K.
+131.4 kJ, then −13.0 kJ. First convert: ΔS° = 0.1602 kJ/K. At 298.15 K, ΔG° = 179.2 − (298.15)(0.1602) = 179.2 − 47.8 = +131.4 kJ. At 1200 K, ΔG° = 179.2 − (1200)(0.1602) = 179.2 − 192.2 = −13.0 kJ. A lime kiln works because of that second line.

Chapter 17Electrochemistry

What to know

A redox reaction moves electrons from one species to another. Oxidation is the loss of electrons and reduction is the gain. Neither one happens alone. The reducing agent is the one that gets oxidized. The oxidizing agent is the one that gets reduced. Track the move with oxidation numbers. Up means the atom lost electrons. Down means it gained them.

Big redox equations fight back if you balance them by eye. So split them. A half-reaction shows one side of the trade with the electrons written in. Balance each half alone, then scale them so the electrons cancel.

The recipe in acid, in order
  • Write one half-reaction for each atom whose oxidation number changed.
  • Balance every atom except oxygen and hydrogen.
  • Add H2O to the side that needs oxygen.
  • Add H+ to the side that needs hydrogen.
  • Add electrons until the total charge matches on both sides.
  • Multiply each half so the electron counts agree, then add and cancel.
  • For a basic solution, balance in acid first, then add OH- to both sides, one for every H+ you used.

A galvanic cell keeps the two halves apart so the electrons have to travel through a wire. Each beaker plus its electrode is a half-cell. The anode is where oxidation happens. The cathode is where reduction happens. Two short phrases lock it in. An ox: anode, oxidation. Red cat: reduction, cathode. Electrons always leave the anode and travel to the cathode.

Without help, charge piles up and the current stops within seconds. A salt bridge fixes that. It passes ions between the beakers, so neither one builds up charge. In cell notation the anode goes on the left, a single bar is a phase boundary, and a double bar is the bridge. A half-cell with no solid of its own needs an inert electrode such as platinum or graphite.

No single half-cell can be measured on its own, because a voltmeter needs two leads. So chemists picked one as the zero mark. The standard hydrogen electrode is platinum in 1 M H+ with H2 at 1 bar and 25 °C, defined as exactly 0.000 V. Pair any half-cell with it and the reading is that half-cell's standard reduction potential, E°. Every table value is written as a reduction.

High on the table means a strong oxidizing agent. Low on the table means a strong reducing agent. Any species high on the table will oxidize anything lower down. To get the cell potential, take the cathode value minus the anode value, both straight from the table. Do not flip a sign first. The subtraction already handles that.

One fact with three faces
  • A positive E°cell means a negative ΔG° and a K above 1. The cell runs on its own.
  • A zero E°cell means ΔG° = 0 and K = 1.
  • A negative E°cell means a positive ΔG° and a K below 1. The reaction needs a push.
  • E° never scales with coefficients, because potential is energy per unit of charge.

The Nernst equation corrects the potential for the concentrations you really have. Read its minus sign. Pile up products and Q rises, so the potential drops. Pile up reactants and the potential climbs. That is a battery's whole life story in one line. When Q finally reaches K, the log term cancels E° and the cell reads zero. A dead battery is a cell that reached equilibrium.

A concentration cell is the strange case. Both half-cells use the same couple at different concentrations, so E°cell is exactly zero. All of its voltage comes from the Nernst term. The dilute half-cell is always the anode, and the cell dies when the two concentrations meet.

Electrolysis runs a cell backward on purpose, using an outside supply to force a reaction with a negative E°cell. The job names do not change, so oxidation is still at the anode. What flips is the sign on the terminals. In water the solvent often reacts instead of the ion you expected. Real cells also need extra voltage above the calculated value, and that extra is the overpotential.

Faraday's law of electrolysis treats electrons as a reagent you can measure out with an ammeter and a clock. Current times time gives charge. Divide by F to get moles of electrons. Divide by n to get moles of product.

Corrosion is a galvanic cell you did not build. One patch of a damp nail acts as the anode and iron dissolves. Another patch acts as the cathode, where oxygen from the air is reduced. Those two half-cells are worth 0.84 V, so rusting is spontaneous. You can block the water, pick a metal that makes a tight oxide skin, or bolt on a sacrificial anode of zinc or magnesium.

Equations, units and symbols

EquationWhat each symbol meansUnits
E°cell = E°cathode − E°anode Both values come from the table as reductions. A positive answer means the cell runs by itself. Volts. 1 V = 1 J/C.
ΔG° = −nFE°cell n is moles of electrons in the balanced overall reaction. F is the Faraday constant. F = 96,485 C/mol. Answer in J/mol.
E°cell = (RT ÷ nF) ln K The general link between potential and the equilibrium constant. R = 8.314 J/mol·K, T in K, answer in volts.
E°cell = (0.0592 V ÷ n) log K The same link at 25 °C, in base-ten logs. Every 0.0592 V per electron is one power of ten in K. Volts. Good at 25 °C only.
Ecell = E°cell − (0.0592 V ÷ n) log Q The Nernst equation at 25 °C. Build Q from the overall cell reaction, with solids and pure liquids as 1. Volts.
q = It Charge equals current times time. Convert minutes and hours to seconds first. I in amperes (C/s), t in s, q in C.
moles of e- = It ÷ F Faraday's law. The charge passed, divided by the charge on one mole of electrons. mol.
moles of product = (moles of e-) ÷ n n is the electrons per formula unit in that half-reaction. Silver takes 1, copper 2, aluminum 3. mol, then times molar mass for grams.
Half-reaction, as a reductionE° (V)Half-reaction, as a reductionE° (V)
F2 + 2e- → 2F-+2.872H+ + 2e- → H20.00
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O+1.51Pb2+ + 2e- → Pb−0.13
Cl2 + 2e- → 2Cl-+1.36Ni2+ + 2e- → Ni−0.25
O2 + 4H+ + 4e- → 2H2O+1.23Fe2+ + 2e- → Fe−0.44
Br2 + 2e- → 2Br-+1.07Zn2+ + 2e- → Zn−0.76
Ag+ + e- → Ag+0.80Al3+ + 3e- → Al−1.66
Fe3+ + e- → Fe2++0.77Mg2+ + 2e- → Mg−2.37
I2 + 2e- → 2I-+0.54Li+ + e- → Li−3.04
O2 + 2H2O + 4e- → 4OH-+0.40Cu2+ + 2e- → Cu+0.34

Worked examples

Example 17.1 · Balance in acid, and read off n

Dichromate oxidizes iron(II) in acid. Balance Cr2O72- + Fe2+ → Cr3+ + Fe3+.

Step 1 · find what changed

Chromium goes from +6 in the dichromate down to +3. Iron goes from +2 up to +3. So chromium is reduced and iron is oxidized.

Step 2 · build the chromium half

Start with Cr2O72- → 2 Cr3+. Seven oxygens need 7 H2O on the right, and those waters need 14 H+ on the left. Charge is then −2 + 14 = +12 on the left and +6 on the right, so add 6 electrons to the left.

Step 3 · the iron half

Fe2+ → Fe3+ + e-. One electron leaves, so multiply this half by 6 to match.

Step 4 · add and cancel

Cr2O72- + 6 Fe2+ + 14 H+ → 2 Cr3+ + 6 Fe3+ + 7 H2O.

Step 5 · check it twice

Atoms: 2 Cr, 6 Fe, 7 O and 14 H on each side. Charge on the left: −2 + 12 + 14 = +24. Charge on the right: 6 + 18 = +24. Balanced.

Answer

Six electrons move, so n = 6. That is the n you will use in every later equation for this reaction. Guessing it from a coefficient is the fastest way to a wrong answer.

Example 17.2 · From the table to E°, ΔG° and K

Build a cell from magnesium and silver at standard conditions. Find E°cell, n, ΔG° and K at 25 °C.

Step 1 · pull both values

Ag+ + e- → Ag is +0.80 V. Mg2+ + 2e- → Mg is −2.37 V.

Step 2 · the higher value is the cathode

Silver is higher, so silver ion is reduced and magnesium is the anode. E°cell = (+0.80) − (−2.37) = +3.17 V.

Step 3 · balance and count electrons

Mg + 2 Ag+ → Mg2+ + 2 Ag moves two electrons, so n = 2. Doubling the silver half did not change its +0.80 V.

Step 4 · free energy

ΔG° = −nFE°cell = −(2)(96,485 C/mol)(3.17 J/C) = −611,700 J/mol, which rounds to −612 kJ/mol.

Step 5 · the equilibrium constant

log K = nE°cell ÷ 0.0592 V = (2)(3.17) ÷ 0.0592 = 6.34 ÷ 0.0592 = 107. So K = 10107.

Answer

+3.17 V, n = 2, ΔG° = −612 kJ/mol and K near 10107. A K that size means the reaction runs to completion. Drop magnesium into silver nitrate and almost no silver ion survives.

Example 17.3 · Nernst, away from standard conditions

A zinc and copper cell has [Zn2+] = 0.0100 M and [Cu2+] = 1.50 M. Find Ecell at 25 °C.

Step 1 · the standard cell first

E°cell = 0.34 V − (−0.76 V) = 1.10 V. The overall reaction is Zn(s) + Cu2+ → Zn2+ + Cu(s), so n = 2.

Step 2 · write Q

Both metals are pure solids, so they enter as 1 and drop out. Q = [Zn2+] ÷ [Cu2+] = 0.0100 ÷ 1.50 = 6.67 × 10-3.

Step 3 · take the log

log(6.67 × 10-3) = −2.18.

Step 4 · put it in Nernst

Ecell = 1.10 V − (0.0592 V ÷ 2)(−2.18) = 1.10 V + (0.0296 V)(2.18) = 1.10 + 0.0645 = 1.16 V.

Answer and the sense check

1.16 V, which is above the standard 1.10 V. You loaded the cell with reactant and starved it of product, so the potential had to climb. Flip the two concentrations and the same math gives 1.04 V instead. A concentration cell is the limit of this idea, with E°cell = 0 and the dilute side as the anode.

The traps people fall into

Watch for these
  • Flipping the anode's sign before subtracting. E°cell = cathode minus anode handles the flip for you.
  • Scaling E° with the coefficients. Doubling a half-reaction doubles n and leaves E° alone.
  • Guessing n. Get it from the half-reactions after you match the electrons.
  • Mixing kJ and J in −nFE°. F is in coulombs per mole, so the answer lands in joules.
  • Using R = 0.08206 with ln K. Use 8.314 J/mol·K here.
  • Building Q from one half-reaction. Q comes from the overall cell reaction.
  • Leaving time in minutes. q = It needs seconds, so multiply minutes by 60.
  • Forgetting to divide by n in Faraday's law. Three electrons per aluminum atom means one third as many moles.
  • Expecting sodium metal from brine. Water is far easier to reduce, so you get hydrogen at the cathode instead.
  • Calling a dead battery empty. It reached Q = K, so its potential is zero even though material is left.

Self-test · Chapter 17

1. Balance the permanganate half-reaction in acid: MnO4- → Mn2+.
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. Four waters carry the four oxygens, and eight H+ feed those waters. Charge then reads −1 + 8 − 5 = +2 on the left and +2 on the right. Manganese went from +7 to +2, which is the five electrons.
2. Convert O2 + 4H+ + 4e- → 2H2O into its basic form.
O2 + 2H2O + 4e- → 4OH-. Add 4 OH- to both sides. On the left each H+ and OH- pair becomes water, giving 4 waters there against 2 on the right. Cancel two from each side and 2 waters remain on the left.
3. For the cell Ni(s) │ Ni2+(1 M) ││ Cu2+(1 M) │ Cu(s), name the electrodes and find E°cell and n.
Nickel is the anode, copper the cathode, E°cell = +0.59 V, n = 2. The left side of the notation is always the anode. E°cell = (+0.34) − (−0.25) = +0.59 V. The overall reaction Ni + Cu2+ → Ni2+ + Cu moves two electrons.
4. Will chlorine gas oxidize bromide ion? Give the potential.
Yes, by +0.29 V. Cl2 sits at +1.36 V and Br2 at +1.07 V. Chlorine is higher, so it takes the electrons: E°cell = 1.36 − 1.07 = +0.29 V. Try it the other way and you get −0.29 V, so bromine cannot touch chloride.
5. Find ΔG° for the nickel and copper cell above.
ΔG° = −114 kJ/mol. ΔG° = −nFE° = −(2)(96,485 C/mol)(0.59 J/C). First 2 × 96,485 = 192,970, then times 0.59 gives 113,850 J. So ΔG° = −1.14 × 105 J/mol, or −114 kJ/mol.
6. Find K for that same cell at 25 °C.
K = 8.5 × 1019. log K = nE° ÷ 0.0592 = (2)(0.59) ÷ 0.0592 = 1.18 ÷ 0.0592 = 19.93. Then K = 1019.93 = 8.5 × 1019. Half a volt looks small and hides twenty powers of ten.
7. Find Ecell for Cu │ Cu2+(0.010 M) ││ Cu2+(1.0 M) │ Cu, and name the anode.
+0.0592 V, and the dilute side is the anode. Both electrodes are copper, so E°cell = 0. The reaction just moves ions, with Q = 0.010 ÷ 1.0 = 0.010 and log Q = −2. Then Ecell = 0 − (0.0592 ÷ 2)(−2) = +0.0592 V. Copper dissolves on the dilute side to raise its own concentration.
8. A current of 2.00 A runs through an aluminum cell for 1.00 hour. What mass of aluminum forms? Aluminum is 26.98 g/mol.
0.671 g. Time in seconds: 1.00 h × 3600 s/h = 3600 s. Charge: q = (2.00 C/s)(3600 s) = 7200 C. Moles of electrons: 7200 ÷ 96,485 = 0.0746 mol. Al3+ + 3e- → Al, so moles of aluminum = 0.0746 ÷ 3 = 0.0249 mol. Mass: (0.0249)(26.98) = 0.671 g.
9. Electrolysis of molten NaCl gives sodium metal. Electrolysis of salt water does not. Why?
Water is far easier to reduce than Na+. In the melt there is nothing else to reduce, so sodium plates out at the cathode and chlorine leaves at the anode. In water the cathode gives hydrogen gas instead, the anode still gives chlorine, and sodium hydroxide builds up in the solution. One process makes chlorine, lye and hydrogen at once.
10. Show that rusting is spontaneous, and name one way to stop it.
E°cell = +0.84 V, and a sacrificial anode stops it. Iron dissolves at the anode, Fe → Fe2+ + 2e-, with the table value −0.44 V. Oxygen is reduced at the cathode, O2 + 2H2O + 4e- → 4OH-, at +0.40 V. So E°cell = 0.40 − (−0.44) = +0.84 V. Bolt on zinc or magnesium and that metal oxidizes instead, since both sit lower on the table.

Chapter 18Representative Metals, Metalloids and Nonmetals

What to know

The main group elements fill groups 1, 2 and 13 through 18. They are also called the representative elements. What ties a group together is its outer shell, and that electron count sets the charges and the bond count. Two trends explain nearly all of this chapter. One is size. The other is how tightly the nucleus grips the outer electrons, measured as effective nuclear charge, Zeff.

Walk across a period and each step adds one proton and one outer electron. The new electron joins a shell that is already there, so it screens badly. Zeff climbs, atoms shrink, and ionization energy and electronegativity both rise. So the left side hands electrons out and the right side takes them in. That shift is a loss of metallic character.

Go down a group and each step adds a whole new shell. The outer electron sits farther out with more core electrons in the way, so the grip loosens. Metallic character grows down every main group. Group 14 shows it best, running from nonmetal carbon through metalloid silicon down to metallic tin and lead.

You can place an element by how its oxide treats water. A metal oxide is basic and makes a base. A nonmetal oxide is acidic and makes an acid. Between them sits the amphoteric oxide, which reacts with acid and with base. Aluminum oxide is the textbook case, and that one fact puts aluminum right on the metal border.

Groups 1 and 2, side by side
  • Group 1 has one outer s electron and forms M+. Group 2 has two and forms M2+.
  • Every alkali metal gives hydrogen gas plus a hydroxide in water. Only the speed changes, from a steady fizz for lithium to a cracked flask for cesium.
  • With oxygen, lithium gives a plain oxide, sodium a peroxide with an O–O bond, and potassium and heavier a superoxide.
  • Lithium alone attacks N2 at room temperature, giving Li3N.
  • Group 2 holds on harder. Beryllium does not react with water, magnesium needs steam, and calcium works away in cold water.
  • Beryllium oxide and hydroxide are amphoteric. The rest of group 2 gives plain bases.
  • The lime cycle runs limestone to quicklime to slaked lime and back to limestone.

Boron is the only member of group 13 that is not a metal. It has three valence electrons, one short of a filled set of four bonds. So boron compounds hunt for electron pairs and act as Lewis acids. Diborane solves its shortage with bridging hydrogens in three-center two-electron bonds. Boric acid does not hand over its own proton. It takes a hydroxide from water instead, which leaves an H+ behind.

Aluminum is the most plentiful metal in the crust and is never found free. Winning it takes electricity, in the Hall–Héroult process, with alumina dissolved in molten cryolite near 960 °C. The metal then protects itself with a thin tight film of Al2O3. Aluminum is amphoteric too, dissolving in strong acid and in strong base and giving hydrogen both times.

Carbon forms more allotropes than any other element. In diamond every carbon bonds to four others at 109.5°, so the whole crystal is one giant molecule that cannot slide and holds no free electron. In graphite each carbon bonds to three others in a flat sheet, and the fourth electron roams above and below. So graphite conducts along its sheets and slides apart in a pencil. Silicon sits below carbon, yet CO2 is a gas and SiO2 is rock. Carbon is small enough for strong double bonds. Silicon is not, so it makes four single bonds to oxygen and the solid never ends.

Nitrogen is 78% of the air and almost nothing reacts with it. The N≡N bond is worth 941 kJ/mol and only 110 pm long. Breaking it is the whole problem of nitrogen fixation. The Haber process solves it near 450 °C and 200 atm over iron, which is a deal struck between yield and speed. The Ostwald process then walks ammonia up to nitric acid through NO and NO2. Nitric acid attacks metals as an oxidizer, not just as an acid, and the product depends on how concentrated it is.

Phosphorus comes in three solids. White phosphorus is built from strained P4 tetrahedra and catches fire in air near 34 °C. Red phosphorus is a chain polymer safe enough for match strikers, and black phosphorus is layered like graphite. The acids hold the most tested idea in the chapter. Count the O–H groups, not the hydrogens. A hydrogen bonded straight to phosphorus never leaves in water.

Oxygen reacts far more readily than nitrogen, because its double bond is only 498 kJ/mol. Its allotrope ozone is bent, costs energy to make, shields the ground from above and irritates lungs at ground level. Oxygen is almost always −2, but a peroxide holds O22- at −1 each and a superoxide holds O2- at −½ each. Adding electrons fills antibonding levels, so the bond order drops and the O–O distance grows.

Sulfur is too big for strong double bonds, so instead of S2 it builds eight-atom crown rings. Heat them and the rings snap open into chains, which is why molten sulfur thickens instead of thinning. Sulfuric acid comes from the contact process: burn sulfur, oxidize SO2 over a vanadium oxide catalyst, absorb SO3 into strong acid, then add water. The third step avoids the acid mist that water alone would throw.

The halogens each have seven outer electrons and each needs one more, so each is a hungry oxidizer. Their appetite falls from fluorine at +2.87 V down to iodine at +0.54 V, and that order predicts every displacement reaction. Three of the four hydrogen halides are strong acids. Hydrofluoric acid is the odd one out, because the H–F bond is worth 569 kJ/mol. Weak does not mean safe, since HF is also the only common acid that eats glass. For the chlorine oxyacids, each extra oxygen makes the acid stronger.

The noble gases have filled outer shells and the highest ionization energies in every period. Still, the heavier ones react. Xenon needs only 1170 kJ/mol to lose an electron, which is about what an oxygen molecule needs. Fluorine is the partner, and the amount you use sets the product. XeF2 is linear and XeF4 is square planar, both shaped by extra lone pairs.

Equations, units and symbols

EquationWhat each symbol meansUnits
Zeff ≈ Z − (core electrons) Z is the proton count. Treating filled inner shells as perfect screens gives a ceiling, not an exact value. A charge in units of e, so a bare number with a plus sign.
2 M(s) + 2 H2O(l) → 2 MOH(aq) + H2(g) M is any group 1 metal. The mole ratio of metal to hydrogen gas is 2 to 1. Mole ratios from the coefficients.
M(s) + 2 H2O(l) → M(OH)2(aq) + H2(g) M is calcium, strontium or barium. One mole of metal gives one mole of hydrogen. Mole ratios from the coefficients.
V = nRT ÷ P Turns the moles of a gas product into a volume. Use it after any gas-forming stoichiometry. R = 0.08206 L·atm/mol·K, T in K, P in atm, V in L.
moles of e- = It ÷ F, then moles of metal = (moles of e-) ÷ n For the Hall–Héroult cell, n = 3, since Al3+ + 3e- → Al. F = 96,485 C/mol. I in A, t in s.
mass of product = moles × molar mass The last step of any mass-to-mass route, such as sulfur to sulfuric acid. g, from mol times g/mol.
Ka1 = x2 ÷ (c0 − x) For a polyprotic acid such as H3PO4, where x is the hydronium made. Ka1 does nearly all the work. [ ] in mol/L. Ka1 has no units.
E°cell = E°cathode − E°anode Used here to rank the halogens and to judge whether one halogen will displace another from its salt. Volts.
Number to knowValueNumber to knowValue
N≡N bond energy941 kJ/molO=O bond energy498 kJ/mol
F–F bond energy159 kJ/molCl–Cl bond energy242 kJ/mol
H–F bond energy569 kJ/molKa of HF6.8 × 10-4
H3PO4 acid constants7.5 × 10-3, 6.2 × 10-8, 4.2 × 10-13Ka of HClO2.9 × 10-8
Halogen E° for X2/X-F +2.87, Cl +1.36, Br +1.07, I +0.54 VKa of HClO21.1 × 10-2

Worked examples

Example 18.1 · Potassium meets water

Drop 4.60 g of potassium into excess water. How much hydrogen gas forms at 0 °C and 1.00 atm, and how much potassium hydroxide do you end up with? Potassium is 39.10 g/mol and KOH is 56.11 g/mol.

Step 1 · write the equation

2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g).

Step 2 · moles of metal

4.60 g ÷ 39.10 g/mol = 0.1176 mol K.

Step 3 · use the ratio

Two moles of metal give one mole of hydrogen. So 0.1176 ÷ 2 = 0.0588 mol H2.

Step 4 · turn moles into a volume

V = nRT ÷ P = (0.0588 mol)(0.08206 L·atm/mol·K)(273.15 K) ÷ 1.00 atm = 1.32 L.

Step 5 · the base

Potassium and hydroxide come out one to one, so 0.1176 mol KOH. Mass = (0.1176)(56.11) = 6.60 g.

Answer and the check

1.32 L of hydrogen and 6.60 g of KOH. Check the mass: 4.60 g of metal became 6.60 g of base, so it picked up 2.00 g of oxygen and hydrogen from the water. A heavier product is what has to happen. Sodium and lithium give the same equation, just at a different speed.

Example 18.2 · Aluminum out of a smelting cell

A Hall–Héroult cell passes 5.00 × 105 C of charge. What mass of aluminum forms? Use F = 96,485 C/mol and 26.98 g/mol.

Step 1 · moles of electrons

(5.00 × 105 C) ÷ (96,485 C/mol) = 5.18 mol of electrons.

Step 2 · the half-reaction sets n

Al3+ + 3 e- → Al. Each atom costs three electrons, so n = 3.

Step 3 · moles of metal

5.18 mol ÷ 3 = 1.73 mol Al.

Step 4 · mass

(1.73 mol)(26.98 g/mol) = 46.6 g.

Answer and why it matters

46.6 g, from half a million coulombs. Run the same charge through a silver bath, where n = 1, and you would get 3 times as many moles. That is why an aluminum smelter is built next to a power station, and why recycling the metal is so much cheaper than making it.

Example 18.3 · Phosphoric acid, and counting O–H groups

Find the pH of 0.250 M H3PO4 using Ka1 = 7.5 × 10-3. Then say how many protons H3PO3 and H3PO2 give up.

Step 1 · first step only

The later steps are 105 and 1010 times weaker, so they add almost no hydronium. Let x be the hydronium from the first step.

Step 2 · test the shortcut

c0 ÷ Ka1 = 0.250 ÷ (7.5 × 10-3) = 33. That is far under 400, so the shortcut is not allowed.

Step 3 · see how far off it would be

Dropping the x gives x2 = (7.5 × 10-3)(0.250) = 1.875 × 10-3, so x = 0.0433 M. That is 17% of 0.250, which the five percent rule rejects.

Step 4 · solve the quadratic

x2 ÷ (0.250 − x) = 7.5 × 10-3 rearranges to x2 + (7.5 × 10-3)x − 1.875 × 10-3 = 0. Then b2 − 4ac = 5.63 × 10-5 + 7.50 × 10-3 = 7.556 × 10-3, whose square root is 0.0869. So x = (−0.0075 + 0.0869) ÷ 2 = 0.0397 M.

Step 5 · take the log

pH = −log(0.0397) = 1.40. The shortcut would have given 1.36, so it was off by 0.04 units.

Answer and the counting rule

pH 1.40. Now the rule: count the O–H groups. H3PO4 has three, so it is triprotic. H3PO3 has one hydrogen bonded straight to phosphorus, so only two can leave and it is diprotic. H3PO2 has two P–H bonds, so it gives up just one proton.

The traps people fall into

Watch for these
  • Counting hydrogens instead of O–H groups. H3PO3 is diprotic and H3PO2 is monoprotic.
  • Giving every alkali metal the same oxide. Lithium gives the oxide, sodium the peroxide, potassium and heavier the superoxide.
  • Letting any group 1 metal attack nitrogen. Only lithium does it at room temperature.
  • Expecting beryllium to behave like calcium. Beryllium does not react with water, and its oxide and hydroxide are amphoteric.
  • Reading the F–F bond as the strongest. At 159 kJ/mol it is weaker than Cl–Cl, because lone pairs crowd on a tiny atom.
  • Calling HF safe because it is weak. It soaks through skin, attacks bone, and etches glass.
  • Ranking oxyacids backward. More oxygens pull charge away from the leftover ion, so HClO4 beats HClO by a huge margin.
  • Putting the small halogen in the middle of an interhalogen. The larger, less electronegative one goes in the center.
  • Treating nitric acid as just an acid. It dissolves copper because nitrate does the oxidizing. Concentrated gives NO2, dilute gives NO.
  • Adding water to SO3 in the contact process. That throws an acid mist, so the gas goes into strong acid first.

Self-test · Chapter 18

1. Estimate Zeff for an outer electron in sulfur, and compare it with sodium.
About +6 for sulfur against +1 for sodium. Sulfur has 16 protons and 10 core electrons, so Zeff is about 16 − 10 = +6. Sodium has 11 protons and the same 10 core electrons, giving 11 − 10 = +1. Six times the pull is why sulfur is much smaller and far harder to ionize.
2. Sort Na2O, Al2O3 and SO3 as basic, amphoteric or acidic, and write the water reaction for the first one.
Basic, amphoteric, acidic, and Na2O + H2O → 2 NaOH. Oxides shift from basic to amphoteric to acidic as you move left to right across a period. Sulfur trioxide with water gives H2SO4, a strong acid. Aluminum oxide shrugs off water and reacts with both acid and base.
3. Write the balanced reaction of calcium with water.
Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g). Calcium forms a +2 ion, so it needs two hydroxides. Check the hydrogens: 4 on the left, and 2 in the hydroxide plus 2 in the gas on the right. Magnesium needs steam for this, and beryllium will not do it at all.
4. Which group 1 metals give a superoxide with plenty of oxygen, and what is the formula?
Potassium and the heavier ones, giving KO2. The ion is O2-, so each oxygen sits at −½. Lithium gives Li2O and sodium gives Na2O2. Potassium superoxide fills emergency breathing canisters, because it gives oxygen back.
5. A kiln roasts 1.00 kg of CaCO3. How much CaO comes out? CaCO3 is 100.09 g/mol and CaO is 56.08 g/mol.
560 g. The equation is CaCO3(s) → CaO(s) + CO2(g), a 1 to 1 ratio. Moles: 1000 g ÷ 100.09 g/mol = 9.99 mol. Mass: (9.99 mol)(56.08 g/mol) = 560 g. The missing 440 g left as carbon dioxide, which is 9.99 mol × 44.01 g/mol.
6. Graphite conducts electricity and diamond does not. Explain both.
The fourth electron is the difference. In diamond each carbon makes four bonds, so every electron is locked into a bond and nothing can carry current. In graphite each carbon makes only three bonds in a flat sheet, and the fourth electron roams above and below. Those loose electrons carry current along the sheets, and the weak forces between sheets let them slide.
7. The Haber process runs near 450 °C and 200 atm. Why not colder, and why so high a pressure?
Cold helps the yield but kills the rate, and pressure shifts the balance. The forward reaction gives off heat, so cooling raises K. It also slows the reaction to a crawl, so the plant settles on a middle temperature. Four moles of gas become two, so squeezing shifts it toward ammonia. Pulling the liquid ammonia out keeps Q below K as well.
8. Rank HClO, HClO2 and HClO4 from weakest to strongest, and say why.
HClO, then HClO2, then HClO4. Hypochlorous acid has Ka = 2.9 × 10-8 and chlorous acid has 1.1 × 10-2, so the second one is about 400,000 times stronger. Perchloric acid is one of the strongest acids known. Each extra oxygen pulls electron density away and spreads the leftover charge, so the ion holds together without the proton.
9. Bromine water is poured on potassium iodide. Does it react? Give E°cell and ΔG°.
Yes, at +0.53 V and −102 kJ. Bromine is at +1.07 V and iodine at +0.54 V, so bromine takes the electrons: E°cell = 1.07 − 0.54 = +0.53 V. Then ΔG° = −nFE° = −(2)(96,485)(0.53) = −1.02 × 105 J = −102 kJ. The solution turns brown as iodine appears. Iodine on bromide gives −0.53 V, so nothing happens.
10. How much sulfuric acid comes from 1.00 metric ton of sulfur, if every step runs perfectly? Sulfur is 32.06 g/mol and H2SO4 is 98.08 g/mol.
3.06 metric tons. Each step carries one sulfur atom forward, so the ratio is 1 to 1 all the way. Moles: 1.00 × 106 g ÷ 32.06 g/mol = 3.12 × 104 mol. Mass: (3.12 × 104)(98.08) = 3.06 × 106 g. Check it the short way: 98.08 ÷ 32.06 = 3.06, so the acid weighs about three times the sulfur.

Cram sheet

One page for the last pass before the exam. Every value below is for 25 °C unless the line says otherwise.

Constants worth knowing cold

  • R = 8.314 J/mol·K for free energy work, and R = 0.08206 L·atm/mol·K for gas volumes.
  • k = 1.381 × 10-23 J/K, which is R ÷ NA. NA = 6.022 × 1023.
  • F = 96,485 C/mol, and 1 V = 1 J/C.
  • 0 °C = 273.15 K, and standard conditions here mean 298.15 K.
  • RT at 298.15 K = 2478.8 J/mol. RT ln 10 = 5.71 kJ/mol, which is one power of ten in K.
  • 0.0592 V per electron is one power of ten in K, at 25 °C only.
  • The hydrogen electrode is exactly 0.000 V, by agreement.

Every equation on one screen

Use it forEquation
Counting arrangementsS = k ln W, and ΔS = R ln(V2 ÷ V1) for one mole of gas
The real testΔSuniv = ΔSsys + ΔSsurr > 0
The surroundingsΔSsurr = −ΔHsys ÷ T
Entropy of reactionΔS° = ΣnS°(products) − ΣnS°(reactants), in J/K
Direction, system onlyΔG = ΔH − TΔS, with matched units
From a table at 298.15 KΔG° = ΣnΔG°f(products) − ΣnΔG°f(reactants)
Free energy and KΔG° = −RT ln K, and ΔG = ΔG° + RT ln Q
Where the sign flipsT = ΔH° ÷ ΔS°, where K = 1
Cell potentialE°cell = E°cathode − E°anode, and E° never scales
Potential to free energyΔG° = −nFE°cell
Potential to KE°cell = (0.0592 V ÷ n) log K at 25 °C
Real conditionsEcell = E°cell − (0.0592 V ÷ n) log Q
Plating and smeltingmoles of e- = It ÷ F, then divide by n for moles of product
ScreeningZeff ≈ Z − (core electrons)
Gas from a reactionV = nRT ÷ P

Facts that show up every year

  • Spontaneous is about direction, never about speed.
  • S° for an element is positive. ΔH°f and ΔG°f for an element are zero.
  • An ox and red cat: oxidation at the anode, reduction at the cathode, in every cell.
  • Positive E°cell, negative ΔG° and K above 1 are the same fact three times.
  • A dead battery has reached Q = K.
  • Group 1 in water always gives hydrogen plus a hydroxide. Only the speed changes.
  • Al2O3 and BeO are amphoteric. That places both elements on the metal border.
  • Count O–H groups to get the proton count of a phosphorus acid.
  • Halogen oxidizing power runs F > Cl > Br > I, and that order predicts every displacement.

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