AP Chemistry · Free-response skill

From entropy to free energy to cell potential

The move. Predict the sign of an entropy change from the moles of gas, calculate it, decide favorability from free energy, and convert free energy into a cell potential.

How to do it

  1. For the sign of an entropy change, count the moles of gas on each side first, then say what that change means for how freely the particles can move.
  2. Name the particles you mean (gas molecules, ions in solution, atoms in a solid) rather than saying "the system gets more ordered".
  3. For favorability, write ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ, convert ΔS∘\Delta S^\circ to kJ/(mol·K), put in T in kelvins, and state the sign and what it means.
  4. For a cell, find n from the balanced half-reactions, then use Ecell∘=−ΔG∘/(nF)E^\circ_{\text{cell}} = -\Delta G^\circ/(nF) with ΔG∘\Delta G^\circ in joules. Report volts to the significant figures of the data.

Practice prompt Try it

A student studies the reaction 2Ag(s)+Cl2(g)→2AgCl(s)2\text{Ag}(s) + \text{Cl}_{2}(g) \rightarrow 2\text{AgCl}(s) at 298 K. She uses these standard values: ΔH∘=−254.0\Delta H^\circ = -254.0 kJ per mole of reaction as written; standard molar entropies S∘S^\circ: Ag(s), 42.6 J/(mol·K); Cl2(g)\text{Cl}_2(g), 223.1 J/(mol·K); AgCl(s), 96.3 J/(mol·K). F = 96,485 C/mol.

Work through parts a to e with the student’s data, showing each step of your arithmetic.

  1. Before any arithmetic: will ΔS∘\Delta S^\circ for the reaction come out positive or negative?
  2. Support your part a answer by counting the moles of gas on each side of the equation.
  3. Work out the number for ΔS∘\Delta S^\circ from the table values (per mole of reaction as written).
  4. Combine the enthalpy change with your part c result to get ΔG∘\Delta G^\circ at 298 K. Is the reaction thermodynamically favorable at 298 K, and which feature of your value tells you?
  5. Silver and chlorine can be made to react in a galvanic cell that moves 2 mol of electrons per mole of reaction as written. Turn your part d value into the voltage of this cell under standard conditions.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 6 points

Model answer, annotated

(a) ΔS∘\Delta S^\circ is negative: the entropy of the system decreases.

  • Point 1 States the sign plainly.

(b) There is 1 mol of gas, Cl2\text{Cl}_2, on the left and no gas on the right, so the moles of gas fall from 1 to 0. Freely moving gas molecules become part of a solid whose ions only vibrate in place, so the particles have far fewer ways to arrange themselves.

  • Point 2 Counts the gas on each side: 1 mol before, 0 after.

(c) ΔS∘=2(96.3)−[2(42.6)+223.1]=192.6−308.3=−115.7\Delta S^\circ = 2(96.3) - [2(42.6) + 223.1] = 192.6 - 308.3 = -115.7 J/K per mole of reaction.

  • Point 3 Products minus reactants with every coefficient.

(d) ΔG∘=ΔH∘−TΔS∘=−254.0−(298)(−0.1157)=−254.0+34.5=−219.5\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -254.0 - (298)(-0.1157) = -254.0 + 34.5 = -219.5 kJ/mol. Because ΔG∘\Delta G^\circ is negative, the reaction is thermodynamically favorable at 298 K.

  • Point 4 Converts the entropy change to kJ before multiplying by T.
  • Point 5 States the sign and what it means for favorability.

(e) Ecell∘=−ΔG∘nF=−−219,500 J(2)(96,485 C)=+1.14E^\circ_{\text{cell}} = -\frac{\Delta G^\circ}{nF} = -\frac{-219{,}500 \text{ J}}{(2)(96{,}485 \text{ C})} = +1.14 V.

  • Point 6 Uses n = 2 and the free energy in joules, giving a positive potential that matches a favorable reaction.