Metallurgical Thermodynamics: Solutions, Phase Equilibria, Ellingham Diagrams and Electrochemistry

Section 2 of the GATE Metallurgical Engineering (MT) paper, in one chapter and in the syllabus’s own order. It starts with the laws — the first law as energy conservation, the second law and entropy, enthalpy, the Gibbs and Helmholtz free energies, Maxwell’s relations and the chemical potential — and then applies them to the systems a metallurgist actually handles: solutions (ideal and regular, activity, Raoult’s and Henry’s laws, and Sieverts’ law for gases dissolving in liquid metal), the Gibbs phase rule, binary phase diagrams and the lever rule, free-energy versus composition curves, and the equilibrium constant with Ellingham and phase-stability (predominance) diagrams. The electrochemistry heading closes it: single electrode potentials, electrochemical cells, the Nernst equation and potential-pH (Pourbaix) diagrams. General chemical-engineering thermodynamics — equations of state, fugacity of gases, vapour-liquid equilibrium — is not what this section asks; everything here is at the depth of oxides, alloys, slags and corroding metals.

1. The laws, the free energies and Maxwell’s relations

The first law is energy conservation for a closed system: dU = δq − δw, with δw = P dV for expansion work only. At constant pressure the heat absorbed equals the change in enthalpy H = U + PV, so ΔH is what a calorimeter measures for a reaction at 1 atm; C_p = (∂H/∂T)_P and the temperature dependence of a reaction enthalpy is Kirchhoff’s law, ΔH(T₂) = ΔH(T₁) + ∫ΔC_p dT. Heats of reaction add by Hess’s law, which is how a metallurgist builds ΔH for an oxide reduction from two formation enthalpies. The second law introduces entropy, dS = δq_rev/T, and for an isolated system ΔS ≥ 0; the third law fixes S = 0 for a perfect crystal at 0 K, which is why absolute entropies are tabulated.

Two free energies turn the second law into a test the metallurgist can apply to a system rather than to the universe. The Helmholtz energy A = U − TS is minimum at equilibrium at constant T and V; the Gibbs energy G = H − TS is minimum at constant T and P, which is the condition of every furnace, ladle and heat-treatment, so ΔG < 0 is the criterion for a process to be spontaneous. The four fundamental equations dU = T dS − P dV, dH = T dS + V dP, dA = −S dT − P dV and dG = −S dT + V dP give Maxwell’s relations by equating mixed second derivatives: (∂T/∂V)_S = −(∂P/∂S)_V, (∂T/∂P)_S = (∂V/∂S)_P, (∂S/∂V)_T = (∂P/∂T)_V and (∂S/∂P)_T = −(∂V/∂T)_P. From dG, (∂G/∂T)_P = −S and (∂G/∂P)_T = V, and the Gibbs-Helmholtz equation [∂(G/T)/∂T]_P = −H/T² links a measured ΔG(T) to ΔH.

For an open system of several components, G depends on the amounts n_i too, and the chemical potential μ_i = (∂G/∂n_i) at constant T, P and the other n_j is the partial molar Gibbs energy. Matter flows from high μ to low μ, and at equilibrium the chemical potential of every component is the same in every phase in which it is present — the condition from which the phase rule, the common-tangent construction and every equilibrium constant follow. For a phase change of a pure substance, equal μ in both phases gives the Clausius-Clapeyron equation dP/dT = ΔH/(TΔV), and for a vapour ln P = −ΔH_v/(RT) + constant.

🧠 Reading Maxwell’s relations off the fundamental equations
Write dG = −S dT + V dP. The coefficients are (∂G/∂T)_P = −S and (∂G/∂P)_T = V; differentiate each by the other variable and equate: (∂S/∂P)_T = −(∂V/∂T)_P. The minus sign always comes from the −S or −P term in the equation used. This one relation is what shows that entropy falls with pressure for any material that expands on heating.

2. Solutions: activity, ideal and regular solutions, Henry, Raoult and Sieverts

In a solution the chemical potential of component i is written μ_i = μ_i° + RT ln a_i, where μ_i° is its value in a chosen standard state and a_i is its activity; the activity coefficient γ_i = a_i/x_i measures departure from ideality. Raoult’s law, a_i = x_i, holds for the solvent as x_i → 1; Henry’s law, a_i = γ_i° x_i with a constant γ_i°, holds for a solute as x_i → 0. Molten iron and steel use a third standard state, the 1 wt% (Henrian) standard state, in which the activity of a dilute solute equals its weight percent — so the oxygen, sulphur or carbon activity in steel is written directly as [%O], [%S] or [%C] times an interaction-corrected coefficient. The Gibbs-Duhem equation, Σ x_i d ln a_i = 0, ties the components together: in a binary, if the solute obeys Henry’s law the solvent must obey Raoult’s.

An ideal solution has ΔH_mix = 0 and ΔS_mix = −R Σ x_i ln x_i, so ΔG_mix = RT Σ x_i ln x_i is always negative; for an equimolar binary ΔS_mix = R ln 2 = 5.76 J/(mol K). A regular solution keeps the ideal entropy but has ΔH_mix = Ω x_A x_B, with Ω the interaction parameter: Ω < 0 means unlike neighbours are preferred (ordering, compound formation, negative deviation), Ω > 0 means like neighbours are preferred (clustering, positive deviation). The activity coefficients follow directly: RT ln γ_A = Ω x_B² and RT ln γ_B = Ω x_A². When Ω > 0 the G-x curve develops two minima below the critical temperature T_c = Ω/(2R), and the alloy separates into two solutions of different composition — a miscibility gap.

Diatomic gases dissolve in liquid and solid metals as atoms: ½H₂(g) = [H], ½N₂(g) = [N]. The equilibrium constant is K = [%H]/√p_H₂, so the dissolved content is proportional to the square root of the gas partial pressure — Sieverts’ law, [%H] = K_H √p_H₂. It is the whole basis of vacuum degassing: lowering p_H₂ a hundredfold lowers the equilibrium hydrogen tenfold, not a hundredfold. Because the dissolution of hydrogen in iron is endothermic, K_H rises with temperature, which is why hydrogen picked up in liquid steel is rejected on solidification and cooling and causes porosity and flaking.

⚠️ Square root, not linear
Sieverts’ law applies to a gas that dissociates on dissolving. Halving the hydrogen pressure lowers the dissolved hydrogen to 1/√2 = 0.707 of its value, not to one half. Reading it as Henry’s law for a molecule is the standard wrong answer in degassing numericals.

3. The phase rule, binary phase diagrams, the lever rule and G-x curves

The Gibbs phase rule counts the intensive variables that can be changed independently without changing the number of phases: F = C − P + 2, with C components and P phases. Metallurgical phase diagrams are drawn at one atmosphere, and pressure has little effect on condensed phases, so the condensed phase rule F = C − P + 1 is used. In a binary at fixed pressure a single-phase field has F = 2 (temperature and composition both free), a two-phase field has F = 1 (fix the temperature and both phase compositions are fixed), and three phases coexist only at an invariant point, F = 0: the eutectic (L → α + β), the peritectic (L + α → β), the eutectoid (γ → α + β), the peritectoid and the monotectic (L₁ → α + L₂).

Inside a two-phase field a horizontal tie line at the temperature of interest ends on the two phase boundaries, which give the compositions of the two phases; the lever rule gives their amounts. For an alloy of overall composition C₀ between phases α (C_α) and β (C_β), the fraction of α is W_α = (C_β − C₀)/(C_β − C_α) and W_β = (C₀ − C_α)/(C_β − C_α) — each phase takes the arm of the lever on the far side. It is a mass balance and nothing more, so it works in weight fractions when compositions are in wt% and in mole fractions when they are in at%. On the iron-carbon diagram (Fe-Fe₃C) the numbers most used are the eutectoid at 727 °C and about 0.76 wt% C (texts quote 0.76 to 0.80), the maximum solubility of carbon in ferrite of about 0.022 wt% at 727 °C, cementite at 6.70 wt% C, the eutectic at 1147 °C and 4.3 wt% C and the maximum solubility in austenite of about 2.1 wt% C.

Lever rule on a 0.40 wt% C steel just below 727 °C (eutectoid 0.76, ferrite 0.022, cementite 6.70 wt% C)
QuantityTie line usedResult
Proeutectoid ferrite0.022 to 0.76 (just above 727 °C)(0.76 − 0.40)/(0.76 − 0.022) = 0.488
PearliteThe rest of the same tie line1 − 0.488 = 0.512
Total ferrite0.022 to 6.70(6.70 − 0.40)/(6.70 − 0.022) = 0.943
Total cementite0.022 to 6.701 − 0.943 = 0.057

Every phase diagram is a projection of free-energy versus composition curves, one curve per phase at each temperature. Where one curve lies lowest across a composition range, that phase alone is stable. Where the lowest total G is obtained by a mixture, the common tangent to two curves touches them at the equilibrium compositions of the two phases — which is exactly the statement that μ_A and μ_B are equal in both phases, since the tangent’s intercepts at x_B = 0 and x_B = 1 are μ_A and μ_B. As temperature falls the liquid curve rises relative to the solid curves, the tangent points sweep out the solidus and liquidus, and a single common tangent touching three curves at once is an invariant reaction. A curve with a central hump (Ω > 0) gives a miscibility gap and, inside its inflection points (d²G/dx² < 0), spinodal decomposition.

4. The equilibrium constant, Ellingham diagrams and predominance diagrams

For a reaction at constant T and P, ΔG = ΔG° + RT ln Q, where Q is the activity quotient; at equilibrium ΔG = 0 and Q becomes the equilibrium constant, ΔG° = −RT ln K. A pure solid or liquid in its standard state has activity 1, and a gas has activity equal to its partial pressure in atm, so for 2M(s) + O₂(g) = 2MO(s) with pure M and MO, K = 1/p_O₂ and the equilibrium oxygen partial pressure is p_O₂ = exp(ΔG°/RT). If the actual p_O₂ in the furnace atmosphere is above this value the metal oxidises; below it the oxide is reduced. Since ΔG° = ΔH° − TΔS°, and ΔH° and ΔS° change slowly with temperature, ΔG° is very nearly a straight line in T.

The Ellingham diagram plots ΔG° against T for the formation of each oxide, written per mole of O₂ so that every line can be compared directly. The intercept at 0 K is ΔH° and the slope is −ΔS°. Because a mole of gas is consumed and only condensed phases are formed, ΔS° is about −200 J/(mol K) for most metal oxides and the lines slope upward, nearly parallel. The slope increases at the melting point and much more at the boiling point of the metal, where the entropy lost on forming the oxide grows; it decreases if the oxide melts. Three carbon lines are the reason the diagram matters: C + O₂ = CO₂ is almost horizontal (ΔS° ≈ 0, one mole of gas for one), 2CO + O₂ = 2CO₂ slopes up, and 2C + O₂ = 2CO slopes DOWN (two moles of gas from one, ΔS° > 0). So carbon becomes a stronger reductant the hotter it is, and at a high enough temperature the CO line crosses below any oxide line — which is why iron, manganese and even chromium oxides can be smelted with carbon, while aluminium, magnesium and titanium oxides, whose lines lie far lower, cannot be reduced by carbon at practical temperatures.

  • Lower on the diagram = more stable oxide. A metal reduces the oxide of any metal whose line lies ABOVE its own at that temperature (the thermite reaction: aluminium reduces iron oxide).
  • The ΔG° of the reduction MO + M′ = M + M′O, per mole O₂, is the vertical gap between the two lines — the larger the gap, the more complete the reduction.
  • Nomographic scales around the edge read the equilibrium p_O₂, and the H₂/H₂O and CO/CO₂ ratios, by drawing a line from the O, H or C point on the left through the oxide line at the chosen temperature.
  • The same construction exists for sulphides and chlorides; the sulphide diagram explains why roasting precedes reduction for zinc and why copper matte smelting works.

A phase-stability, or predominance-area, diagram asks a different question: at a fixed temperature, which condensed phase is stable as two gas potentials vary together? For the M-S-O system the axes are log p_O₂ and log p_S₂ (or log p_SO₂), and the fields are M, MO, MS and MSO₄. Each boundary is one reaction at equilibrium — for example MS + ½O₂ = MO + ½S₂ gives a straight line whose slope comes from the powers of the pressures in K — so the diagram is built from a handful of ΔG° values and a ruler. It is the tool for designing a roast: an oxidising roast must sit in the MO field, a sulphating roast in the MSO₄ field. The analogous Kellogg-type diagrams for M-C-O and M-Cl-O systems do the same for reduction and chlorination.

🎯 Why the CO line slopes down
In 2C(s) + O₂(g) = 2CO(g) one mole of gas becomes two, so ΔS° is positive and the slope −ΔS° is negative. Every other common line consumes gas and slopes up. That single sign difference is the thermodynamic reason the blast furnace works and why carbothermic reduction is a high-temperature process.

5. Electrochemistry: electrode potentials, cells, the Nernst equation and Pourbaix diagrams

A single electrode potential cannot be measured alone; it is quoted against the standard hydrogen electrode (SHE, H⁺ at unit activity, H₂ at 1 atm, platinum), which is assigned 0 V. Standard reduction potentials at 25 °C place Cu²⁺/Cu at +0.34 V, Fe²⁺/Fe at −0.44 V and Zn²⁺/Zn at −0.76 V: the more negative the potential, the more readily the metal gives up electrons. In a galvanic cell the electrode with the lower potential is the anode (oxidation) and the one with the higher potential is the cathode (reduction); the emf is E = E_cathode − E_anode, and the free-energy change is ΔG = −nFE with F = 96 485 C/mol. The Daniell cell, Zn | Zn²⁺ ‖ Cu²⁺ | Cu, has E° = 0.34 − (−0.76) = 1.10 V. An electrolytic cell is the same arrangement driven backwards by an external supply, which is how metals are electrowon and electrorefined.

Away from unit activities the potential follows from ΔG = ΔG° + RT ln Q divided by −nF: the Nernst equation, E = E° − (RT/nF) ln Q. At 298 K, RT/F × ln 10 = 0.0592 V, so E = E° − (0.0592/n) log₁₀ Q. For a single electrode Mⁿ⁺ + ne⁻ = M, E = E° + (0.0592/n) log a_Mⁿ⁺: a tenfold dilution of the ions lowers a divalent electrode’s potential by 29.6 mV. A concentration cell — two identical electrodes in solutions of different activity — has E° = 0 and runs entirely on this term, which is also how a differential-aeration or concentration corrosion cell is set up in a crevice.

A potential-pH (Pourbaix) diagram maps, for a metal in water at 25 °C, which species is thermodynamically stable at each combination of electrode potential and pH, assuming a chosen ion activity (typically 10⁻⁶). Three kinds of boundary appear: horizontal lines for pure electron transfers (Fe²⁺ + 2e⁻ = Fe, independent of pH), vertical lines for pure acid-base reactions (Fe³⁺ + 3H₂O = Fe(OH)₃ + 3H⁺, independent of potential), and sloping lines for reactions involving both. Two dashed lines bound the stability of water itself: below E = 0 − 0.0592 pH hydrogen is evolved, and above E = 1.229 − 0.0592 pH oxygen is evolved. The fields are read as immunity (the metal is stable), corrosion (a soluble ion is stable) and passivation (a solid oxide or hydroxide is stable and may protect). The diagram says what can happen, never how fast; kinetics is the business of polarization, in the next chapter.

⚠️ The sign convention that decides the anode
Tables list REDUCTION potentials. In a cell, the electrode with the more negative reduction potential runs as the anode and corrodes. Subtracting in the wrong order gives a negative emf for a cell that actually runs, and ΔG = −nFE then says the spontaneous cell is impossible.

Key takeaways

  • At constant T and P the criterion is ΔG < 0; Maxwell’s relations come from equating mixed second derivatives of U, H, A and G, and the chemical potential is equal in every phase at equilibrium.
  • Ideal solution: ΔH_mix = 0, ΔS_mix = −RΣx ln x. Regular solution: ΔH_mix = Ωx_Ax_B, RT ln γ_A = Ωx_B², miscibility gap below T_c = Ω/(2R). Sieverts: dissolved gas ∝ √p.
  • Condensed phase rule F = C − P + 1; three phases in a binary is an invariant point. The lever rule is a mass balance: each phase takes the far arm of the tie line.
  • ΔG° = −RT ln K; for 2M + O₂ = 2MO, p_O₂(eq) = exp(ΔG°/RT). On the Ellingham diagram slope = −ΔS°, lower is more stable, and only the 2C + O₂ = 2CO line slopes down.
  • E = E_cathode − E_anode, ΔG = −nFE, Nernst E = E° − (0.0592/n) log Q at 298 K; a Pourbaix diagram shows immunity, corrosion and passivation fields between the water lines, and says nothing about rate.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which of the following is a correct Maxwell relation?

    1. (∂S/∂P)_T = −(∂V/∂T)_P
    2. (∂S/∂P)_T = (∂V/∂T)_P
    3. (∂S/∂V)_T = −(∂P/∂T)_V
    4. (∂T/∂V)_S = (∂P/∂S)_V
    Show answer

    Answer: A — (∂S/∂P)_T = −(∂V/∂T)_P

    From dG = −S dT + V dP, (∂G/∂T)_P = −S and (∂G/∂P)_T = V; equating the cross derivatives gives (∂S/∂P)_T = −(∂V/∂T)_P. From dA = −S dT − P dV the V-relation is (∂S/∂V)_T = +(∂P/∂T)_V, and from dU the S-V relation carries a minus sign, so the other three have a sign wrong.
  2. For a ternary alloy system at constant pressure, how many degrees of freedom remain when three phases are in equilibrium?

    Numerical answer — type the value.

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    Answer: 1

    Condensed phase rule at fixed pressure: F = C − P + 1 = 3 − 3 + 1 = 1, so a three-phase equilibrium in a ternary is univariant (a line, not a point). Using the full rule F = C − P + 2 gives 2, which forgets that pressure has already been fixed.
  3. A plain-carbon steel with 0.40 wt% C is slowly cooled to just below 727 °C. Taking the eutectoid composition as 0.76 wt% C and the carbon content of ferrite as 0.022 wt%, what is the weight fraction of proeutectoid ferrite (to two decimal places)?

    Numerical answer — type the value.

    Show answer

    Answer: 0.49

    Lever rule on the tie line just above the eutectoid, from ferrite (0.022) to austenite (0.76): W_α,pro = (0.76 − 0.40)/(0.76 − 0.022) = 0.36/0.738 = 0.488. Check: pearlite = (0.40 − 0.022)/0.738 = 0.512 and the two sum to 1. Using the ferrite-to-cementite tie line (to 6.70) gives 0.943, which is the TOTAL ferrite including that inside pearlite.
  4. What is the ideal molar entropy of mixing of an equimolar binary solution, in J/(mol K) (to two decimal places)? Take R = 8.314 J/(mol K).

    Numerical answer — type the value.

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    Answer: 5.76

    ΔS_mix = −R(x_A ln x_A + x_B ln x_B) = −8.314 × 2 × 0.5 ln 0.5 = 8.314 × 0.6931 = 5.763 J/(mol K). Equivalently R ln 2. Forgetting the minus sign gives a negative entropy of mixing, which would make ideal mixing non-spontaneous; using log₁₀ gives 2.50.
  5. A binary A-B liquid behaves as a regular solution with interaction parameter Ω = −20 kJ/mol. What is the activity of A at x_A = 0.5 and 1000 K (to three decimal places)? Take R = 8.314 J/(mol K).

    Numerical answer — type the value.

    Show answer

    Answer: 0.274

    ln γ_A = Ωx_B²/(RT) = −20 000 × 0.25/(8.314 × 1000) = −0.6014, so γ_A = 0.548 and a_A = γ_A x_A = 0.548 × 0.5 = 0.274. Cross-check through ΔḠ_A = RT ln a_A = RT ln 0.5 + Ωx_B² = −5763 − 5000 = −10 763 J/mol, and exp(−10 763/8314) = 0.274. Using x_A² in place of x_B² gives the same number here only because x_A = x_B; a negative Ω must give an activity BELOW the ideal 0.5.
  6. A regular solution has Ω = +25 kJ/mol. Above what temperature, in K, does its miscibility gap disappear (to the nearest whole number)? Take R = 8.314 J/(mol K).

    Numerical answer — type the value.

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    Answer: 1503

    The gap closes at the critical point, where d²ΔG_mix/dx² = 0 first holds at x = 0.5: −2Ω + RT/(x(1 − x)) = 0 gives T_c = Ω/(2R) = 25 000/(2 × 8.314) = 1503 K. Check: at 1503 K, RT/(x(1 − x)) = 8.314 × 1503 × 4 = 50 000 = 2Ω. Forgetting the factor 2 gives 3007 K.
  7. Liquid steel in equilibrium with hydrogen at 1 atm dissolves 4 ppm hydrogen at a given temperature. To what hydrogen partial pressure, in atm, must a vacuum degasser lower the gas to bring the equilibrium content to 1 ppm at the same temperature?

    Numerical answer — type the value.

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    Answer: 0.0625

    Sieverts’ law: [H] = K√p_H₂. With [H] = 4 ppm at 1 atm, K = 4 ppm/atm^½. For 1 ppm, √p = 1/4, so p = 1/16 = 0.0625 atm. Ratio check: the content falls fourfold, so the pressure falls by 4² = 16. A linear (Henry’s law) reading gives 0.25 atm, which leaves 2 ppm in the steel.
  8. For 2M(s) + O₂(g) = 2MO(s), ΔG° = −500 kJ per mole of O₂ at 1500 K, with M and MO pure. What is log₁₀ of the equilibrium oxygen partial pressure (in atm) at 1500 K (to two decimal places)? Take R = 8.314 J/(mol K).

    Numerical answer — type the value.

    Show answer

    Answer: -17.41

    With unit activities K = 1/p_O₂, so ΔG° = −RT ln K = RT ln p_O₂ and ln p_O₂ = ΔG°/RT = −500 000/(8.314 × 1500) = −40.09. Then log₁₀ p_O₂ = −40.09/2.3026 = −17.41, i.e. p_O₂ ≈ 3.9 × 10⁻¹⁸ atm. Directly: log p_O₂ = ΔG°/(2.303RT) = −500 000/28 721 = −17.41. Writing K = p_O₂ flips the sign to +17.41, an absurd pressure.
  9. For MO(s) + CO(g) = M(s) + CO₂(g), ΔG° = +5.0 kJ at 1200 K, with M and MO pure. What is the minimum ratio p_CO/p_CO₂ in the gas needed to reduce MO at 1200 K (to two decimal places)? Take R = 8.314 J/(mol K).

    Numerical answer — type the value.

    Show answer

    Answer: 1.65

    K = p_CO₂/p_CO = exp(−ΔG°/RT) = exp(−5000/(8.314 × 1200)) = exp(−0.5011) = 0.606. Reduction needs Q < K, i.e. p_CO₂/p_CO < 0.606, so p_CO/p_CO₂ > 1/0.606 = 1.65. Check directly: exp(+0.5011) = 1.650. Reporting 0.606 gives the equilibrium CO₂/CO ratio, the inverse of what was asked; a positive ΔG° does not forbid reduction, it only demands a CO-rich gas.
  10. Which of the following statements about the Ellingham diagram for oxides are correct?

    1. The slope of each line equals −ΔS° of the oxidation reaction per mole of O₂
    2. The line for 2C + O₂ = 2CO has a negative slope
    3. A metal can reduce the oxide of any metal whose line lies below its own
    4. The slope of a metal’s line increases sharply at the boiling point of the metal
    Show answer

    Answer: A — The slope of each line equals −ΔS° of the oxidation reaction per mole of O₂; B — The line for 2C + O₂ = 2CO has a negative slope; D — The slope of a metal’s line increases sharply at the boiling point of the metal

    ΔG° = ΔH° − TΔS°, so the slope is −ΔS° (A). 2C + O₂ = 2CO makes two moles of gas from one, ΔS° > 0 and the slope is negative (B). At the metal’s boiling point the metal reactant becomes a gas, so forming the condensed oxide loses much more entropy and the slope jumps (D). C is reversed: a metal reduces oxides whose lines lie ABOVE its own; the lower line is the more stable oxide, which the higher metal cannot reduce.
  11. A Daniell cell Zn | Zn²⁺ (1 M) ‖ Cu²⁺ (0.01 M) | Cu operates at 298 K. Taking E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = −0.76 V and activities equal to molarities, what is the cell emf, in V (to two decimal places)?

    Numerical answer — type the value.

    Show answer

    Answer: 1.04

    E° = 0.34 − (−0.76) = 1.10 V. For Zn + Cu²⁺ = Zn²⁺ + Cu, Q = [Zn²⁺]/[Cu²⁺] = 1/0.01 = 100 and n = 2, so E = 1.10 − (0.0592/2) log 100 = 1.10 − 0.0592 = 1.041 V. Electrode by electrode: E_Cu = 0.34 + 0.0296 log 0.01 = 0.281 V, E_Zn = −0.76 V, difference 1.041 V. Putting Q upside down gives 1.16 V.
  12. Which of the following statements about potential-pH (Pourbaix) diagrams are correct?

    1. A boundary for a reaction involving only electron transfer, with no H⁺ or OH⁻, is horizontal
    2. Below the line E = −0.0592 pH (at 25 °C) water is reduced and hydrogen is evolved
    3. A metal lying in a passivation field is guaranteed not to corrode at any measurable rate
    4. The diagram predicts the rate at which a metal corrodes in its corrosion field
    Show answer

    Answer: A — A boundary for a reaction involving only electron transfer, with no H⁺ or OH⁻, is horizontal; B — Below the line E = −0.0592 pH (at 25 °C) water is reduced and hydrogen is evolved

    A reaction with no H⁺ or OH⁻ has a potential independent of pH, so its boundary is horizontal (A). The hydrogen line E = 0 − 0.0592 pH bounds water’s stability from below (B). C overstates the diagram: passivation means a solid film is thermodynamically stable, not that it is protective — porous or chloride-attacked films still corrode. D is the diagram’s central limitation: it is thermodynamics only and says nothing about kinetics.
  13. On an Ellingham diagram, the line for the formation of an oxide shows a distinct INCREASE in slope at a certain temperature. The most likely cause is:

    1. The oxide melts at that temperature
    2. The metal boils at that temperature
    3. The oxide becomes more stable than the metal
    4. The reaction becomes exothermic
    Show answer

    Answer: B — The metal boils at that temperature

    The slope is −ΔS°. When the reactant metal boils, the reaction now consumes metal vapour as well as O₂ to form a condensed oxide, so ΔS° becomes much more negative and the slope rises sharply (it rises less at the metal’s melting point). Melting of the OXIDE adds entropy to the product and DECREASES the slope. Stability and exothermicity are not what a slope change reports.