Electrochemical equilibrium
Week 6 · Onsite · 180 minutes
13 November 2026
Start with the chemical potential
Follow the same reaction throughout the derivation.
\mathrm{A_2(g)\rightleftharpoons 2A(g)}
The chemical potential of each species depends on its activity.
\mu_i=\mu_i^\circ+RT\ln a_i
\mu_i^\circ refers to the chosen standard state. The activity a_i is dimensionless. For a gas, use a_i=f_i/p^\circ.
Combine the potentials along the reaction
\mathrm{A_2(g)\rightleftharpoons 2A(g)}
At fixed temperature and pressure, a small forward reaction consumes one unit of A₂ and forms two units of A.
\mathrm dn_{\mathrm A}=2\,\mathrm d\xi,\qquad \mathrm dn_{\mathrm{A_2}}=-\mathrm d\xi
Substitute these amounts into \mathrm dG=\sum_i\mu_i\,\mathrm dn_i.
\mathrm dG=(2\mu_{\mathrm A}-\mu_{\mathrm{A_2}})\,\mathrm d\xi
The reaction Gibbs energy is the slope along the reaction extent \xi.
\Delta_rG=\left(\frac{\partial G}{\partial\xi}\right)_{T,P}=2\mu_{\mathrm A}-\mu_{\mathrm{A_2}}
The activity ratio is Q
Insert \mu_i=\mu_i^\circ+RT\ln a_i into the reaction Gibbs energy.
\Delta_rG=(2\mu_{\mathrm A}^\circ-\mu_{\mathrm{A_2}}^\circ)+RT(2\ln a_{\mathrm A}-\ln a_{\mathrm{A_2}})
Define \Delta_rG^\circ=2\mu_{\mathrm A}^\circ-\mu_{\mathrm{A_2}}^\circ and combine the logarithms.
\Delta_rG=\Delta_rG^\circ+RT\ln\left(\frac{a_{\mathrm A}^2}{a_{\mathrm{A_2}}}\right)
The ratio calculated from the mixture’s current activities is the reaction quotient.
Q=\frac{a_{\mathrm A}^2}{a_{\mathrm{A_2}}},\qquad \Delta_rG=\Delta_rG^\circ+RT\ln Q
At equilibrium, Q equals K
At reaction equilibrium, \Delta_rG=0. Call the equilibrium value of the activity ratio K.
0=\Delta_rG^\circ+RT\ln K,\qquad K=\exp\left(-\frac{\Delta_rG^\circ}{RT}\right)
For the written reaction and chosen standard states, the temperature determines K. Combine this relation with the expression for the current mixture.
\boxed{\Delta_rG=RT\ln\left(\frac{Q}{K}\right)}
Compare Q with K to predict the reaction direction. Which sign does \Delta_rG have when Q<K?
Q and K for the same reaction
For \mathrm{A_2(g)\rightleftharpoons 2A(g)}, use the dimensionless activity ratio a_{\mathrm A}^2/a_{\mathrm{A_2}}.
Reaction quotient, Q. Calculate the ratio using the activities in the mixture now.
Equilibrium constant, K. At equilibrium the ratio equals K. Its value is fixed at a given temperature for the written reaction and chosen standard states.
| If Q<K | If Q=K | If Q>K |
|---|---|---|
| The forward reaction forms A. | The reaction is at equilibrium. | The reverse reaction forms A₂. |
Does doubling the pressure always double Q?
A question from Week 5. How does the pressure argument change for a real gas?
Start with the ideal-gas reaction \mathrm{A_2(g)\rightleftharpoons 2A(g)}. At 350 K, K=3.953 for p^\circ=100 kPa. Compress the mixture from 100 to 200 kPa at fixed temperature. Consider the instant before the composition responds.
| Starting value | Predict Q immediately after compression | Which way will the reaction move? |
|---|---|---|
| Q=K=3.953 | Calculate on paper | Toward A or A₂? |
| Q=1.000 | Calculate on paper | Toward A or A₂? |
Take one minute. Compare your two answers with a neighbor.
Compare the new quotient with K
At fixed mole fractions, the ideal-gas quotient for this reaction is proportional to pressure.
Q=\frac{y_{\mathrm A}^2}{y_{\mathrm{A_2}}}\frac{P}{p^\circ}
| Before compression | Immediately after | Response at 350 K |
|---|---|---|
| Q=3.953 | Q=7.906>K | More A₂ forms. |
| Q=1.000 | Q=2.000<K | More A forms. |
As the reaction proceeds, the mole fractions change. At the new equilibrium, Q=K=3.953 in both cases.
Does gas non-ideality change K?
At the same temperature, for the same written reaction and standard states, the thermodynamic K stays the same.
K=\exp\left(-\frac{\Delta_rG^\circ}{RT}\right)
Fugacity coefficients enter the current mixture’s activity ratio. For \mathrm{A_2(g)\rightleftharpoons 2A(g)},
Q=\frac{\phi_{\mathrm A}^2}{\phi_{\mathrm{A_2}}}\frac{y_{\mathrm A}^2}{y_{\mathrm{A_2}}}\frac{P}{p^\circ}
At equilibrium, this full ratio equals K. Non-ideality changes the equilibrium composition needed to satisfy that condition. We will now calculate the fugacities that enter it.
Fugacity expresses the chemical potential
For an ideal gas mixture, the partial pressure y_iP enters the chemical potential.
\mu_i^{\rm ideal}=\mu_i^\circ+RT\ln\left(\frac{y_iP}{p^\circ}\right)
For a real gas, define the component fugacity, f_i, through the same logarithmic form and standard state.
\mu_i=\mu_i^\circ+RT\ln\left(\frac{f_i}{p^\circ}\right)
Fugacity has pressure units. The activity a_i=f_i/p^\circ is dimensionless. In the ideal-gas limit, f_i=y_iP.
Compare fugacity with partial pressure
The fugacity coefficient, \phi_i, is the ratio of component fugacity to its partial pressure.
\phi_i=\frac{f_i}{y_iP},\qquad f_i=\phi_i y_iP
Subtract the ideal-mixture chemical potential at the same T,P,\mathbf y.
\mu_i-\mu_i^{\rm ideal}=RT\ln\phi_i
At \phi_i=1, f_i=y_iP and the chemical potential matches the ideal-mixture value. A value below one gives a negative departure in chemical potential; a value above one gives a positive departure.
Each component has its own \phi_i(T,P,\mathbf y).
Calculate phi from an equation of state
An EOS supplies the partial-molar properties used to calculate RT\ln\phi_i. Choose the EOS and its parameters for the stated T,P,\mathbf y.
For the low-density, pressure-form second-virial EOS, used in Lab 09,
Z=\frac{PV}{nRT}=1+\frac{B_{\rm mix}P}{RT},\qquad B_{\rm mix}=\sum_i\sum_j y_i y_j B_{ij}
The component fugacity coefficients follow from this mixture EOS.
\ln\phi_i=\left(2\sum_j y_jB_{ij}-B_{\rm mix}\right)\frac{P}{RT}
The coefficients B_{ij}(T) come from gas-property data or a correlation. The cross coefficient describes unlike pairs. Use consistent units and check the EOS validity range.
Work out both coefficients from the EOS
Use synthetic inputs at 350 K and 200 kPa. Let component 1 be A and component 2 be A₂, with y_1=0.80, y_2=0.20.
Take B_{11}=-1.00, B_{12}=B_{21}=-0.600, B_{22}=-0.200 L mol^{-1} and R=8.31446 L kPa mol^{-1} K^{-1}.
B_{\rm mix}=0.80^2(-1.00)+2(0.80)(0.20)(-0.600)+0.20^2(-0.200)=-0.840\ {\rm L\,mol^{-1}}
| Component | \sum_j y_jB_{ij} (L mol^{-1}) | \ln\phi_i | \phi_i=\exp(\ln\phi_i) |
|---|---|---|---|
| A | -0.920 | (-1.000)(200)/(8.31446\times350)=-0.068727 | 0.93358 |
| A₂ | -0.520 | (-0.200)(200)/(8.31446\times350)=-0.013745 | 0.98635 |
Carry these coefficients into the activity calculation. Keep unrounded values until the final result.
A real gas uses fugacity in each activity
At a specified T,P,\mathbf y, a mixture equation of state gives each component’s fugacity coefficient \phi_i. With an ideal-gas standard state,
a_i=\frac{f_i}{p^\circ}=\frac{\phi_i y_iP}{p^\circ},\qquad Q=\frac{a_{\mathrm A}^2}{a_{\mathrm{A_2}}}.
For this dissociation, the multiplier on the ideal-gas estimate is \phi_{\mathrm A}^2/\phi_{\mathrm{A_2}}. The coefficients can change during compression and as the reaction proceeds. The same T, written reaction and standard states give the same K.
Try it. Continue with the same EOS calculation at 350 K and 200 kPa, with y_{\mathrm A}=0.80, y_{\mathrm{A_2}}=0.20. Use the calculated coefficients \phi_{\mathrm A}=0.93358 and \phi_{\mathrm{A_2}}=0.98635, with p^\circ=100 kPa. Calculate both fugacities, both activities and the reaction direction.
Work the real-gas example on paper
The pressure ratio is 200/100=2. At the specified composition,
| Quantity | Calculation | Value |
|---|---|---|
| f_{\mathrm A} | \phi_{\mathrm A}(0.80)(200) | 149.373 kPa |
| f_{\mathrm{A_2}} | \phi_{\mathrm{A_2}}(0.20)(200) | 39.454 kPa |
| a_{\mathrm A} | f_{\mathrm A}/(100\ {\rm kPa}) | 1.49373 |
| a_{\mathrm{A_2}} | f_{\mathrm{A_2}}/(100\ {\rm kPa}) | 0.394539 |
| Q | a_{\mathrm A}^2/a_{\mathrm{A_2}} | 5.6553 |
The ideal-gas estimate at this same state is 6.40. Both exceed K=3.953, so A₂ forms. To find the final extent for the real gas, update the composition and both fugacity coefficients until Q=K. Lab 12 solves the ideal-gas case.
The 180-minute class
| Class time | Activity |
|---|---|
| 0–20 min | Derive Q and K, then calculate fugacity and check the pressure question |
| 20–40 min | Revisit Weeks 2–5 in pairs with paper and a calculator |
| 40–65 min | Balance a cell reaction and calculate its potential by hand |
| 65–75 min | Demonstrate the concentration cell in Lab 13 |
| 75–85 min | Break |
| 85–155 min | Pair investigation and independent check |
| 155–180 min | Discuss results (15), exit ticket (5), feedback (3), closing (2) |
Week 2 · What did the VLE measurement tell us?
A VLE measurement gives P=100 kPa, x_1=0.40, y_1=0.60. Assume an ideal vapor, negligible Poynting correction and a pure-liquid reference state.
y_1P=x_1\gamma_1 P_1^{\rm sat},\qquad a_1=x_1\gamma_1.
- Use P_1^{\rm sat}=120 kPa. Find \gamma_1 and a_1.
- A second source gives 150 kPa at the same temperature. Predict which way your inferred \gamma_1 moves.
- Before fitting the data, which Antoine conventions would you check?
Week 3 · Would this feed split into two liquids?
At z_1=0.12, compare the curve and the supporting line at the same overall composition. Which state has lower Gibbs energy?
The homogeneous curve has positive local curvature here. What does the tangent tell you that curvature alone cannot?
A Gibbs–Duhem area test examines the VLE data. How could errors of opposite sign affect its signed integral?
Week 4 · Can the vessel supply that much gas?
At 300 K, a 1.00 L vessel contains 0.00200 mol of volatile A. Its equilibrium sublimation pressure is 10.0 kPa. Use R=8.314 kPa L mol^{-1} K^{-1} and assume an ideal gas.
- Calculate the gas inventory required for P_A=10.0 kPa. Will any bulk solid remain?
- Calculate the actual P_A. What happens to P_A if you add an inert gas at fixed T,V?
- In a separate adsorption vessel, an adsorbent of mass m_s holds q(P_A,T) mol per kg. Which term must you add to the gas inventory to close the balance?
Week 5 · Change pressure, then change temperature
Lab 12 used 1 mol of A₂ initially, zero A, no inert, and p^\circ=100 kPa. Use the same ideal-gas reaction in all three cases.
| Case | T (K) | P (kPa) | K | Equilibrium extent (mol) |
|---|---|---|---|---|
| 12A | 350 | 100 | 3.953 | 0.7050 |
| 12B | 350 | 400 | 3.953 | 0.4451 |
| 12C | 400 | 100 | 33.861 | 0.9457 |
Which pair isolates pressure? Which isolates temperature? For 12C assume a constant positive reaction enthalpy of 50 kJ mol^{-1}. Explain the change in K.
With a partner, use the extent in 12B to check 2n_{\mathrm{A_2}}+n_{\mathrm A} against the feed.
Check your four predictions
| Week | Compare with your calculation or prediction |
|---|---|
| 2 · VLE | With P_1^{\rm sat}=120 kPa, \gamma_1=1.25 and a_1=0.50. With P_1^{\rm sat}=150 kPa they become 1.00 and 0.40. Check Antoine units, log base and temperature range. |
| 3 · Stability | At z_1=0.12 the two-liquid line lies lower. Positive local curvature describes small fluctuations. Opposite errors can cancel in the signed area test. |
| 4 · Inventory | Saturation requires 0.00401 mol. All 0.00200 mol enters the gas, giving P_A=4.99 kPa. Adding ideal inert at fixed T,V raises total pressure while P_A stays at 4.99 kPa. For adsorption add m_s q mol. |
| 5 · Reaction | 12A–12B isolate pressure. 12A–12C isolate temperature. Heating raises K for this endothermic reaction. In 12B, 2(0.5549)+0.8902=2.0000 mol of A atoms. |
The question for today
What determines the sign of a concentration-cell potential?
Week 5 gave us \Delta_rG=RT\ln(Q/K). A reversible cell turns that reaction driving force into a potential difference, E=-\Delta_rG/(nF).
Predict on paper. For the reaction as written, what sign should E have when Q<K? What voltage would you measure at complete cell equilibrium, Q=K?
Two reduction half-reactions
Enter both half-reactions as reductions. Reverse the left reaction and add the right after matching electron counts.
E_{cell}^\circ=E_{right}^\circ-E_{left}^\circ
Both potentials need the same reference electrode and temperature. Multiplying a half-reaction does not multiply its potential.
Nernst equation from reaction thermodynamics
Combine ΔrG=ΔrG°+RTlnQ with ΔrG=−nFE: E=E^\circ-\frac{RT}{nF}\ln Q
At 298.15 K, E=E^\circ-\frac{0.05915935\ \mathrm V}{n}\log_{10}Q
Changing T requires valid E° data at the new T.
A concentration-cell calculation
Use the same \mathrm{M^{2+}/M} couple on both sides at 298.15 K. For oxidation on the left and reduction on the right, Q=a_{\mathrm L}/a_{\mathrm R}. Take a_{\mathrm L}=0.01, a_{\mathrm R}=1, n=2, and neglect the junction potential. Q=0.01,\qquad E=+0.05915935\ \mathrm V
E°=0, K=1, but E is nonzero. Reversing the concentrations changes the sign of E. Equal activities give E=0.
ΔrG≈−11.416 kJ per mol of the written reaction.
An independent check
Use dimensionless activities. Concentration is an approximation with a stated standard concentration and activity coefficient.
Use Eright−Eleft and deltaG=−nFE with the balanced electron count. Equal activities must give zero potential for the same reduction couple.
One demonstration · 10 minutes
Open the concentration-cell preset on the projector. On paper, make three rows for the original, exchanged and equal activities. If your device is offline, use the projected inputs and complete the same hand check. Predict the sign, exchange left and right activities, then set them equal. Keep T and reference conventions fixed.
Keep one saved baseline for the pair investigation.
Pair investigation · 70 minutes
- 10 min: predict and record the baseline.
- 25 min: change one input and calculate.
- 20 min: exchange roles and check the result by hand.
- 15 min: prepare one plot or table and a short explanation.
In Lab 13, compare a concentration cell with the activities exchanged. Predict and explain the sign of E, then independently check deltaG=−nFE.
Discussion · 15 minutes
What determines the sign of a concentration-cell potential?
Compare results and discuss unexpected behavior. Use one numerical check to support your explanation.
After class · 30–40 minutes
In Lab 13, compare a concentration cell with the activities exchanged. Predict and explain the sign of E, then independently check deltaG=−nFE.
Assessment 6 and independent checkpoint
Retain one study JSON, one comparison and a short explanation.
Optional: Session appendix.
Exit ticket + feedback · 8 minutes
5 min · Explain and check
Answer the session question in the form. Give one result or independent check from today’s lab, with units.
3 min · Help improve the next class
Rate the pace, report lab access, and suggest one improvement.
Thai or English. No name, ID or email requested.
Wait for “Response received.” We will use the final 2 minutes to close the class.
