Revision notes chemical equilibrium: calculations based on Nernst & free energy equations Advanced Level Theoretical-Physical Chemistry

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Doc Brown's Chemistry Advanced A Level Notes - Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 7

Part 7.7 Questions based on Appendix 1. The Nernst Equation

and Appendix 2 Free Energy, Cell Emf and Kc

This is a BIG website, you need to take time to explore it [SEARCH BOX]

Some links to miscellaneous questions on redox titrations, electrolysis products and the redox chemistry of the rusting of iron and its prevention.

The use of the Nernst equation is described i.e. how does the electrode potential of a half–cell reaction vary with the concentration of the electrolyte.

How to use the cell Emf to calculate the free energy change of a cell (redox) reaction and hence the equilibrium constant Kc for a redox equilibrium.

7.7 Exemplar questions

Appendix 1. The Nernst Equation

• The Nernst equation allows the calculation–prediction of what a half–cell potential will be for a different ion concentration (or temperature) from that of the standard conditions i.e. 1.0 mol dm–3 and 298K.

• The Nernst equation is not needed (as far as I know?) for UK GCE–A2 and IB courses, but it may be needed in coursework projects! You do NOT have to present the thermodynamic derivation of the equation!

• For a reduction half–cell equilibrium i.e. oxidised state + electrons reduced state, the Nernst equation for the variation of the half–cell potential is

• E = Eθ + (RT/nF) ln {[ox]/[red]}

• E = half–cell potential in V which varies with the molar concentrations [mol dm–3] of the oxidised and reduced species.

• Eθ is the standard half–cell potential e.g. at 298K when the molarities are 1.0 mol dm–3.

• in the case of two aqueous ions like Fe2+/Fe3+, [red] = [ox] i.e. equal concentrations, so that E varies with the ratio of the two ion concentrations.

• R = 8.314 J mol–1 K–1 (the ideal gas constant).

• T the absolute temperature in K (Kelvin = oC + 273).

• n = moles of electrons transferred per mole of reactants.

• F =96500 C mol–1 (the Faraday constant).

• ln = natural/Napierian logarithm.

• [ox/red] = molar concentration of the oxidised or reduced species. University students might well be using activity values as well as molarity values, in which case the activity of the metal is considered to be unity, i.e. aM(s) = 1.000 (see next paragraph).

• Note the equivalent equations (lg = log to base 10, log or log10)

• E = Eθ – (RT/nF) ln {[red]/[ox]}

• E = Eθ – (2.303RT/nF) lg {[red]/[ox]}

• E = Eθ + (2.303RT/nF) lg {[ox]/[red]}

• For a metal–metal half–cell equilibrium: Mn+(aq) + ne M(s)

• the Nernst equation is  EMn+/M = EθMn+/M + (RT/nF) ln {[Mn+(aq)]/[M(s)]}

• BUT the concentration of the metal cannot change so its molarity or 'thermodynamic activity' is considered to be unity, in which case the Nernst equation for the simple metal–metal ion equilibrium is

• (i) EMn+/M = EθMn+/M + (RT/nF) ln [Mn+(aq)]

• (using natural logarithm ln, watch on calculator!)

• or (ii) EMn+/M = EθMn+/M + (2.303RT/nF) log10 [Mn+(aq)]

• (log to base 10, lg, log or log10)

• at 298K for ln expression: RT/nF = (8.314 x 298)/(n x 96500) = 0.0257/n

• for lg or log10: RT/nF = (2.303 x 8.314 x 298)/(n x 96500) = 0.0591/n

• Note: (i) , (ii)

• Five examples of calculations using the Nernst equation are outlined below.

1. What is the half–cell potential for copper when dipped into a 2.0 mol dm–3 solution of copper(II) sulphate?

• Cu2+(aq) + 2e Cu(s) (Eθ = +0.342V)

• ECu2+/Cu = EθCu2+/Cu + (RT/nF) ln [Cu2+(aq)]

• E = +0.342 + {(8.314 x 298)/(2 x 96500)} ln [2.0]

• E = +0.342 + 0.01284 ln(2)

• ECu2+/Cu = +0.342 + 0.009 = 0.351 V (3sf, 0.35V 2sf)

• Le Chatelier's comment: The increase in Cu2+ concentration increases the oxidising potential of the half–cell i.e. a more positive half–cell potential acting from left to right in the equilibrium.

2. What is the half–cell potential of zinc dipped into a 0.1 molar zinc sulphate solution?

• Zn2+(aq) + 2e Zn(s) (Eθ = –0.763V)

• EZn2+/Zn = EθZn2+/Zn + (2.303RT/nF) lg [Zn2+(aq)]

• E = –0.763 + {(2.303 x 8.314 x 298)/(2 x 96500)} lg [0.1]

• EZn2+/Zn = –0.763 + (–0.029) = –0.792 V (3sf, –0.79 V 2sf)

• Le Chatelier's comment: The decrease in Zn2+ concentration increases the reducing potential of the half–cell i.e. a more negative potential half–cell potential acting from right to left in the equilibrium.

3. What is the theoretical half–cell potential of aluminium dipped into a 0.001 molar aluminium sulphate solution?

• Al3+(aq) + 3e Al(s) (Eθ = –1.662V)

• EAl3+/Al = EθAl3+/Al + (RT/nF) ln [Al3+(aq)]

• E = –1.662 + {(8.314 x 298)/(3 x 96500)} ln [0.001]

• EAl3+/Al = –1.662 + (–0.059) = –1.721 V (4sf, –1.72V 3sf)

• Le Chatelier's comment: The decrease in Al3+ concentration increases the reducing potential of the half–cell i.e. a more negative potential half–cell potential acting from right to left in the equilibrium.

4. What is the half–cell potential for a solution of iron(II) and iron(III) ions, containing 0.20 mol dm–3 of  Fe2+ and 0.10 mol dm–3 of Fe3+?

• Fe3+(aq) + e Fe2+(aq) (Eθ = +0.771V)

• EFe3+/Fe2+ = EθFe3+/Fe2+ + (RT/nF) ln [ox]/[red]

• EFe3+/Fe2+ = EθFe3+/Fe2+ + (2.303RT/nF) lg [Fe3+(aq)]/[Fe2+(aq)]

• E = +0.771 + (0.0591/1) log10 (0.1/0.2)

• EFe3+/Fe2+ = +0.771 + (–0.018) = +0.753 V (3sf, 0.75V 2sf)

• Le Chatelier's Principle comment: The lower Fe3+ concentration compared to Fe2+, decreases the oxidising potential of the half–cell i.e. a less positive potential half–cell potential acting from left to right in the equilibrium.

5. Very accurate measurement half–cell electrode potentials can be used to determine the concentration of an ion (e.g. pH with glass electrode).

When a copper strip (+ve pole) is dipped into a copper(II) ion solution, and combined with a saturated KCl–calomel electrode (–ve pole, E = 0.244V at 298K), the cell voltage measured 0.088 V at 298K.

From the information deduce the concentration of copper(II) ions in the solution.

• (i) You first need to calculate the copper  half–cell potential.

• Ecell = E+pole – E–pole

• Ecell = ECu/Cu2+ – Ecalomel

• Ecell = 0.088 = ECu/Cu2+ – (+0.244) = ECu/Cu2+ – 0.244

• therefore ECu/Cu2+ = Ecell + 0.244 = 0.088 + 0.244 = +0.332 V

• (ii) You then need to rearrange the Nernst equation to calculate the concentration.

• EθCu2+/Cu = +0.342V, and from example 1.

• ECu2+/Cu = EθCu2+/Cu + (RT/2F) ln [Cu2+(aq)]

• rearranging (with care!) gives

• (RT/2F) ln [Cu2+(aq)] = ECu2+/Cu – EθCu2+/Cu

• ln [Cu2+(aq)] = (ECu2+/Cu – EθCu2+/Cu)/(RT/2F)

• [Cu2+(aq)] = e{(ECu2+/Cu – EθCu2+/Cu)/(RT/2F)}

• [Cu2+(aq)] = e{(0.332 – 0.342)/0.01284}

• [Cu2+(aq)] = e{(–0.01)/0.01284} = e–0.7788 = 0.459 mol dm–3

• NOTE that even a small and expected error in the cell Emf can lead to a large concentration calculation error e.g. if the ECu2+/Cu of 0.332 is assumed to be absolutely correct, but was actually measured as 0.328 (≈ 1.2% measurement error) the calculation would give e{(–0.014)/0.01284} = e–1.0903 = 0.336 mol dm–3 which is only 73% of the 'real' concentration, i.e. a 27% calculation error! Check it out for yourself and note I've used 3sf Emf's in the calculation, so you need a very accurate voltmeter or potentiometer system to have any hope of accurate analytical data.

Appendix 2. Free Energy, cell Emf and equilibrium constant Kequilibrium

• The concentration equilibrium constant Kc

• for a reaction aA + bB ... dD + eE ...

• Kc = [A]a [B]b .../[D]d [E]e ...

• The standard Gibbs free energy change ∆Gθ = ∆Hθ – T∆Sθsys

• and ∆Gθ = –RT ln(Kc) = –RT ln([A]a [B]b .../[D]d [E]e ...)

• The free energy change for a reversible electrode reaction is ∆Gθ = –nEθF

• n is the number of electrons transferred in the theoretical cell reaction and F is the Faraday constant.

• therefore for a redox equilibria ∆Gθ = –nEθF = –RT ln(Kc)

• so nEθF = RT ln(Kc) from which the equilibrium constant Kc can be calculated.

• REMEMBER ...

• ∆Gθ must be negative, and Eθ positive, for the theoretical overall redox reaction, for it to be feasible.

• For more details see Thermodynamics Part 3 section 3.5b

WHAT NEXT?

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4 Partition between two phases, solubility product Ksp, common ion effect, ion–exchange systems * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria–vapour pressure, boiling point and intermolecular forces watch out for sub-indexes to multiple sections or pages

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