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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 K_{c}
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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 = E^{θ}
+ (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.

E^{θ}
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 Fe^{2+}/Fe^{3+}, [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 = ^{o}C + 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. a_{M(s)} = 1.000 (see next paragraph).

Note the equivalent
equations (lg = log to base 10, log or log_{10})

E = E^{θ} – (RT/nF) ln {[red]/[ox]}

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

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

For a
metal–metal half–cell equilibrium:
M^{n+}_{(aq)}
+ ne^{–}
M_{(s)}

the Nernst
equation is E_{Mn+/M} = E^{θ}_{Mn+/M} + (RT/nF) ln
{[M^{n+}_{(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)
E_{Mn+/M} = E^{θ}_{Mn+/M}
+ (RT/nF) ln [M^{n+}_{(aq)}]

(using natural
logarithm ln, watch on calculator!)

or (ii) E_{Mn+/M} = E^{θ}_{Mn+/M}
+ (2.303RT/nF) log_{10} [M^{n+}_{(aq)}]

(log to base 10, lg, log or log_{10})

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

for lg or log_{10}:
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.

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

Cu^{2}^{+}_{(aq)}
+ 2e^{–}
Cu_{(s)} (E^{θ} =
+0.342V)

E_{Cu2+/Cu} = E^{θ}_{Cu2+/Cu}
+ (RT/nF) ln [Cu^{2+}_{(aq)}]

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

E =
+0.342 + 0.01284 ln(2)

E_{Cu2+/Cu}
= +0.342 + 0.009 = 0.351 V (3sf, 0.35V 2sf)

Le
Chatelier's comment: The increase in Cu^{2+}
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.

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

Zn^{2}^{+}_{(aq)}
+ 2e^{–}
Zn_{(s)} (E^{θ} =
–0.763V)

E_{Zn2+/Zn} = E^{θ}_{Zn2+/Zn}
+ (2.303RT/nF) lg [Zn^{2+}_{(aq)}]

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

E_{Zn2+/Zn}
= –0.763 + (–0.029) = –0.792 V (3sf, –0.79 V
2sf)

Le
Chatelier's comment: The decrease in Zn^{2+}
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.

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

Al^{3}^{+}_{(aq)}
+ 3e^{–}
Al_{(s)} (E^{θ} =
–1.662V)

E_{Al3+/Al} = E^{θ}_{Al3+/Al}
+ (RT/nF) ln [Al^{3+}_{(aq)}]

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

E_{Al3+/Al}
= –1.662 + (–0.059) = –1.721 V (4sf, –1.72V
3sf)

Le
Chatelier's comment: The decrease in Al^{3+}
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.

What is the
half–cell potential for a solution of iron(II) and iron(III)
ions, containing 0.20 mol dm^{–3} of Fe^{2+}
and 0.10 mol dm^{–3} of Fe^{3+}?

Fe^{3+}_{(aq)} + e^{–}
Fe^{2+}_{(aq)} (E^{θ} = +0.771V)

E_{Fe3+}_{/Fe2+} = E^{θ}_{Fe3+}_{/Fe2+}
+ (RT/nF) ln [ox]/[red]

E_{Fe3+}_{/Fe2+} = E^{θ}_{Fe3+}_{/Fe2+}
+ (2.303RT/nF) lg [Fe^{3+}_{(aq)}]/[Fe^{2+}_{(aq)}]

E =
+0.771 + (0.0591/1) log_{10} (0.1/0.2)

E_{Fe3+}_{/Fe2+}
= +0.771 + (–0.018) = +0.753 V (3sf, 0.75V
2sf)

Le
Chatelier's Principle comment: The lower Fe^{3+}
concentration compared to Fe^{2+}, 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.

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.

E_{cell}
= E_{+pole} – E_{–pole}

E_{cell}
= E_{Cu/Cu2+} – E_{calomel}

E_{cell}
= 0.088 = E_{Cu/Cu2+} – (+0.244) = E_{Cu/Cu2+}
– 0.244

therefore E_{Cu/Cu2+} = E_{cell} + 0.244 =
0.088 + 0.244 = +0.332 V

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

E^{θ}_{Cu2+/Cu}
= +0.342V, and from example 1.

E_{Cu2+/Cu} = E^{θ}_{Cu2+/Cu}
+ (RT/2F) ln [Cu^{2+}_{(aq)}]

rearranging (with care!) gives

(RT/2F)
ln [Cu^{2+}_{(aq)}] = E_{Cu2+/Cu}
– E^{θ}_{Cu2+/Cu}

ln [Cu^{2+}_{(aq)}]
= (E_{Cu2+/Cu} – E^{θ}_{Cu2+/Cu})/(RT/2F)

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

[Cu^{2+}_{(aq)}]
= e^{{(0.332 – 0.342)/0.01284}}

[Cu^{2+}_{(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
E_{Cu2+/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 K_{equilibrium}

The
concentration equilibrium constant K_{c}

The standard
Gibbs free energy change ∆G^{θ} =
∆H^{θ} – T∆S^{θ}_{sys}

and ∆G^{θ}
= –RT ln(K_{c}) = –RT ln([A]^{a} [B]^{b} .../[D]^{d}
[E]^{e} ...)

The free energy
change for a reversible electrode reaction is ∆G^{θ} =
–nE^{θ}F

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

therefore for a
redox equilibria ∆G^{θ}
= –nE^{θ}F = –RT ln(K_{c})

so nE^{θ}F
= RT ln(K_{c}) from which the equilibrium constant K_{c}
can be calculated.

REMEMBER ...

∆G^{θ}
must be negative, and E^{θ} 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. K_{c} and K_{p} equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * Part 4
Partition between two
phases, solubility product K_{sp}, 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 subindexes
to multiple sections or pages
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