1. CHEM1612 - Pharmacy Week 9: Nernst Equation
Dr. Siegbert Schmid
School of Chemistry, Rm 223
Phone:

2. Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille,      Chemistry, John Wiley & Sons Australia, Ltd      ISBN:

3. Electrochemistry Blackman, Bottle, Schmid, Mocerino & Wille:
Chapter 12, Sections 4.8 and 4.9
Key chemical concepts:
Redox and half reactions
Cell potential
Voltaic and electrolytic cells
Concentration cells
Key Calculations:
Calculating cell potential
Calculating amount of product for given current
Using the Nernst equation for concentration cells

4. Recap: Standard cell potential
The measured voltage across the cell under standard conditions is the standard cell potential E0cell (also called emf).
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
This cartoon corresponds exactly to the previous experiment.
Zinc electrode in Zn2+ soln
Copper electrode in Cu2+ soln
Oxidation at anode, Zn is oxidised generating 2 electrons – move thru the wire to the Cu electrode in the Cu2+ soln (cathode). 2e-s reduce Cu2+
Electrons flow left to right.
Salt bridge maintains neutral charge in the electrolyte soln.
Voltmeter registers the electrical output of the cell.

5. Tricks to memorise anode/cathode
1. Anode and Oxidation begin with a vowels,
Cathode and Reduction with consonants.
2. Alphabetically, the A in anode comes before the C in cathode, as the O in oxidation comes before the R in reduction.
3. Think of this picture:
AN OX and a RED CAT
(anode oxidation reduction cathode)

6. Standard cell potential and free energy
For a spontaneous reaction, E0cell and also ΔG0
For a non-spontaneous reaction, E0cell and also ΔG0
So there is a proportionality between E0 and -ΔG0.
You also know that the maximum work done on the surroundings is
-wmax = ΔG
Electrical work done by the cell is w = Ecell × charge
From thermodynamics, know that a spontaneous reaction has a negative free energy change (ΔG<0)
i.e. the signs of ΔG and Ecell are opposite for a spontaneous reaction.

7. Standard cell potential and free energy
The emf E0cell is related to the change in free energy of a reaction:
∆G0 = Standard change in free energy
n = number of electrons exchanged
F = C/mol e- (Faraday constant)
∆G0 = –nFE0cell
∆G = –nFEcell
Also, away from standard conditions:
F = Charge of one mole of electrons
But what is Ecell?

8. Example Calculate ∆G0 for a cell reaction:
Cu2+(aq) + Fe (s)  Cu(s) + Fe2+ (aq)
Is this a spontaneous reaction?
Cu2+ + 2e–  Cu E0 = 0.34 V
Fe2+ + 2e–  Fe E0 = –0.44 V
E0cell =
∆G0 = –nFE0cell
∆G0 =
Electrochemistry and thermodynamics are the same thing
∆G is defined as the maximum useful work obtainable from the system. Substitute Eocell for Eomax (in the real world some free energy is lost as heat so not all the available free energy is converted to work)
This process is spontaneous as indicated by the negative sign of G0 and the positive sign for E0cell.

9. Example: a Dental Galvanic Cell
Al(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg
Al is very easily oxidised,
Al3+ + 3e−  Al E0 = -1.66V.
The filling is an inactive
cathode for the
reduction of oxygen,
O2 + 4H+ + 4e−  2H2O
and saliva is an electrolyte.
Pain experienced if aluminium foil comes into contact with a filling.
Al acts as the active anode (Eo=-1.66 V)
Saliva is the electrolyte
The filling (usually a Ag/Sn/Hg alloy) is an inactive cathode at which the O2 is reduced to water.
The short circuit between the foil in contact with the filling creates a current that is sensed by the nerve of the tooth.
Put the three together (biting on a piece of foil) results in generation of a current and possible pain.

10. Example: a Dental Galvanic Cell
Al(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg
O2 + 4H+ + 4e–  2H2O E0 = 1.23 V
Al3+ + 3e–  Al E0 = –1.66 V
12H+ + 3O2 + 4Al  6H2O + 4Al3+
E0cell =
∆G0 = –nFE0cell=
Pain experienced if aluminium foil comes into contact with a filling.
Al acts as the active anode (Eo=-1.66 V)
Saliva is the electrolyte
The filling (usually a Ag/Sn/Hg alloy) is an inactive cathode at which the O2 is reduced to water.
The short circuit between the foil in contact with the filling creates a current that is sensed by the nerve of the tooth.

11. Nernst Equation Recall ΔG = ΔG0 + RT ln(Q) (1)
Since ΔG0 = -nFE0cell and ΔG = -nFE
Equation (1) becomes
-nFE = -nFE0cell + RTln(Q) Q = [products] / [reactants]
dividing both sides by –nF gives:
E = E0 – RT ln(Q) nF
Image fromnobelprize.org
Walther Nernst
Nobel Prize 1920
From left to right, Nernst, Albert Einstein, Max Planck, Robert Millikan, Max von Laue.
In the photo, L-R
Walther Nernst (Nobel Prize 1920) Walter Nernst derived this equation in 1889 at the age of 25. He also derived the 3rd Law of Thermodynamics and explained the principle of the solubility product. He was the awarded the Nobel Prize in Chemistry in 1920.
Albert Einstein (Nobel Prize 1921)
Max Planck (Nobel Prize 1918)
Robert Millikan (Nobel Prize 1923)
Max von Laue (Nobel Prize 1914)
Because in all practical voltaic cells, such as batteries, reactant concentratioins are far from standard state values, clearly we need to be a able to determine Ecell, the cell potential under non-standard conditions.
We know the relationship between free energy and concentration. Recall…
Can derive an expression for the relationship between cell potential and concentration.
E = actual cell potential
Eo = standard cell potential
R = gas constant (8.314 JK-1mol-1
T= temperature in Kelvins
Natural logarith of reaction quotient (NB, at equilibrium Q=K)
n= no. of electrons transferred
F = Faraday constant=96485Cmol-1 = charge on 1 mole of electrons

12. Nernst Equation E = cell potential E0 = Standard cell potential
R = Real Gas Constant
= JK-1mol-1
T =Temperature (K)
n = no. of e- transferred
F = Faraday constant
= C mol-1
Q = Reaction quotient
(Q = K at equilibrium)
Ecell = E0 – RT ln(Q) nF
Since ln (x) = log (x)
E = E0 – · RT log(Q) nF
At 25°C, (2.303·R·298)/96485 =
At equilibrium, Q = K.
Nernst equation more commonly
written like this (note: only at 25°C)

13. Example calculation 1
Calculate the expected potential for the following cell:
i) [Cu2+] = 1.0 M; [Zn2+] = 10-5M
ii) [Cu2+] = 10-5M; [Zn2+] = 1.0 M
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) E0 = 1.1 V
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Cu2+ + 2e-  Cu
Zn  Zn2+ + 2e-
Firstly, work out the value of n :
You can set up a 1M Cu/Zn cell, then use tap water for the Cu cell. Tap water has typically ~10mM Cu2+ so the observed potential is usually about what you calculate. Of course there are also other ions present, but because of the copper pipes that we use, the Cu seems to be most important.

14. Example calculation Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) (n=2)
[Zn2+] (M)
[Cu2+] (M) Q
log(Q)
Ecell (V)
1.0
10-5
1st line, [Zn2+] = [Cu2+]=1M, i.e. standard state conditions, Q=1 so Ecell=Eocell
2nd line, Q<1, Cell potential is higher than under standard state conditions, i.e. able to do more work.
3rd line, Q>1, cell potential is lower than under standard state conditions, i.e. able to do less work.

15. Demo: The effect of concentration
0.00 V
Cu|Cu2+||Cu2+|Cu
Both compartments of the voltaic cell are identical.
E0cell = E0copper – E0copper = 0 (in standard conditions, 1M concentrations)
Demo 6.8 concentration cell.
Two identical copper half cells are constructed. On the addition of Sodium sulfide solution to one of them, a precipitate (CuS(s)) forms in the beaker and a voltage is detected.
--on addition of sodium sulfide solution, the voltage should increase from zero to a maximum of 0.6V.
What happens when the concentration of one cell is changed?

16. Demo: Cu Concentration Cell
Low [Cu2+]
Cu  Cu2++ 2e-
Cu2+ + 2e-  Cu
High [Cu2+]
Cu |Cu2+||Cu2+|Cu
E0 same for both half-reactions, so E0cell= 0.
However, we have reduced the concentration of Cu2+ in one cell = non-standard conditions.
Electrical energy is generated until the concentrations in each half-cell become equal (equilibrium is attained).
Electrochemical potential is a function of concentration.
So we have built a voltaic cell using the same half reactions, but with different concentrations in each half cell.
As in any voltaic cell, Ecell decreases until equilibrium is attained, which happens when [Cu2+] is the same in both half cells. The same final concentration would result if we mixed the two solutions, but no electrical work would be done.
Now start to make the connection between observed cell potential and equilibrium.
Can we explain this?…
Add Na2S –precipitate forms.

17. Cu Concentration Cell Cu|Cu2+||Cu2+|Cu Low [Cu2+] High [Cu2+]
Cu  Cu2++ 2e- Cu2+ + 2e-  Cu
E0cell is same on both sides, but the Cu concentrations are different.
More charge carriers in one half-cell.
If we poured the two solutions together, we would expect spontaneous mixing of two solutions of different concentrations to give one of equal concentration.
The electrical connection allows electrons to pour from one half-cell to the other.
Demo 6.8 concentration cell.
Two identical copper half cells are constructed. On the addition of Sodium sulfide solution to one of them, a precipitate (CuS(s)) forms in the beaker and a voltage is detected.
--on addition of sodium sulfide solution, the voltage should increase from zero to a maximum of 0.6V.

18. Concentration cells
“Voltage”
The measured cell potential in our experiment was “Voltage”. Let’s work out what the voltage should be:
Cu  Cu2+ (0.01 M) + 2e-
Cu2+ (0.1M) + 2e-  Cu
Cu2+ (0.1M)  Cu2+ (0.01 M) Ecell =
“Voltage”
0.01
= “Voltage”
Pretty simple calculation for the blackboard:
Concentration of both solutions originally was 0.1M.
Na2S added – concentration of one half-cell is changed.
The half reactions are the same, so Eocell=0. n=2
The cell operates because the half-cell concentrations are different, which makes Ecell>0 in this case.
( The cell operates until the half-cell concentrations are equal.) = my notes
V=1.0 V (Demo book says 0.6 Volts)
1.0 = x log(y/0.1)
-1/ = log (y/0.1)
-25 = log(y/0.1)
y/0.1 = 1E-25
y = 1E-26 M
V=1.0 V => 1E-35 M WHY?????????
these are chemically unrealistic (<1 molecule!)
Challenge students to think what else might be happening…. probably just other dissolved salts… nM concentration is quite clean water/beaker, yet these concentrations are ~20 orders of magnitude more concentrated than we just calculated.
Solve for “Voltage”:
“Voltage” = V

19. Concentration cells
The cell potential depends on the concentration of reactants.
Corollary: It is useful to define a standard concentration, which is 1 M.
Implication: We need to specify concentration when referring to the cell potential.
The overall potential for the Cu/Cu2+ concentration cell is:
E = E0cell – /2 · log [Cu2+]dil / [Cu2+]conc

20. Reference Electrodes Standard Hydrogen Electrode (SHE)
Metal-Insoluble Salt Electrode: Standard Calomel Electrode (SCE) and Silver Electrode
Ion-Specific: pH electrode

21. Standard Hydrogen Electrode (SHE)
Finely divided surface Pt electrode.
HCl solution with [H+] =1,
H2 p= 1 atm bubbling over the electrode.
H2 absorbs on the Pt, forming the equivalent of a 'solid hydrogen‘ electrode in equilibrium with H+.
Platinum – gas electrode
H2 electrode
H2  2H+ + 2 e- Eo = 0.00 V
Metal – Metal ion Electrode
Cu2+ + 2e-  Cu Eo = 0.34 V
RHS standard hydrogen electrode. Hydrogen gas at 1 atm pressure bubbles over an inert platinum electrode that is immerse in a soln containing 1M H+ ion at 25C. The potential for this electrode is defined as exactly 0 V.
The half cell potentials of many different half reactions can be measured by comparing them to the hydrogen electrode.
Copper is reduced ( the oxidising agent).
The standard hydrogen electrode is the anode. The reaction at the standard hydrogen electrode is the opposite to that in the reaction with zinc.
Pt|H2(g)|H+(aq)||Cu2+(aq)|Cu(s) Eocell = 0.34 – 0 = 0.34 V

22. Metal-insoluble salt electrodes
The Standard Hydrogen Electrode (SHE) is not convenient to use in practice (can be contaminated easily by O2 or organic substances).
There are more practical choices, like metal - insoluble salt electrodes.
The potential of these electrodes depends on the concentration of the anion X- in solution.
In practice 2 interfaces:
1. M / MX insoluble salt: MX (s) + e-  M (s) + X-(s)
2. coating/solution: X- (s)  X- (aq)
Overall: MX (s) + e-(metal)  M (s) + X- (aq) M X- MX C+
Calomel electrode – consists of Pt wire immersed in a paste of Hg2Cl2 (calomel), liquid Hg and 1M or saturated KCl soln
Silver/silver chloride electrode – Ag/AgCl half reaction immersed in 1M HCl soln.

23. Metal-insoluble salt electrodesX- MX C+
The concentration of anions in solution is controlled
by the salt's solubility:
[Ag+] [X-] = Ksp
Normal calomel electrode, Pt | Hg | Hg2Cl2 | KCl (1M) E 0 = 0.28 V
Saturated calomel electrode Pt | Hg | Hg2Cl2 | KCl(sat.) E0 = 0.24 V
Silver/Silver chloride, Ag | AgCl | Cl- (1M) E 0 = 0.22 V
(used in pH meters)
Calomel electrode – consists of Pt wire immersed in a paste of Hg2Cl2 (calomel), liquid Hg and 1M or saturated KCl soln
Silver/silver chloride electrode – Ag/AgCl half reaction immersed in 1M HCl soln.

24. Saturated Calomel Electrode
The ‘saturated calomel electrode’ (SCE) features the reduction half-reaction:
Hg+ + e–  Hg
Hg2Cl2  2Hg+ + 2Cl–
Overall:
Hg2Cl2 (s) + 2e–  2 Hg (s) + 2Cl– (sat)
Pt | Hg | Hg2Cl2 | KCl ||
Standard cell potential of
E0 = 0.24 V. 2
5 M

25. Cell Potentials 1
Q: The standard reduction potential of Zn2+/Zn is V. What would be the observed cell potential for the Zn/Zn2+ couple when measured using the SCE as a reference?
0.24 Hg+/Hg
0.0V H2/H+
-0.76 Zn2+/Zn
Calomel: Hg2Cl2(s) + 2e-  2Hg(l) + 2Cl-(aq) E0 = 0.24V
Zn  Zn2+ + 2e- E0 = V
(reversed because it is written as an oxidation)
Ans: The Zn will be oxidised (lower reduction potential), so
E (cell) = = 1.00 V respect to the SCE
However with these standard electrodes it isn’t as easy to calculate the standard half cell potential.
So to get the oxidation half-reaction E0 using the SCE as cathode, subtract 0.24 V from the volt meter reading.

26. Summary CONCEPTS Concentration cells Nernst equation CALCULATIONS
Work out cell potential from reduction potentials;
Work out cell potential for any concentration (Nernst equation)