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.
E0cell = E0cathode – E0anode = ( )= = 1.10 V

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 < 0
For a non-spontaneous reaction, E0cell <0 and also ΔG0 > 0
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 = (-0.44) = 0.78 V
∆G0 = –nFE0cell
∆G0 = – 2 · · 0.78 = – 1.5 · 105 J
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 Eo = -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 = 2.89 V
∆G0 = –nFE0cell=
–12 · · 2.89 = –3346 kJmol–1
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 But what is Ecell? 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. 2 mol e- transferred per mole of reaction: n = 2
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-
2 mol e- transferred per mole of reaction: n = 2
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) 1.0 10-5
[Zn2+] (M)
[Cu2+] (M) Q
log(Q)
Ecell (V)
1.0
10-5
1.0
10-5
105
0.0
-5.0
5.0
1.10
1.25
0.95
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
anode
cathode

22. Metal-insoluble salt electrodes
The Standard Hydrogen Electrode (SHE) isn't 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)

27. Silver electrode Ag+ + e–  Ag AgCl  Ag+ + Cl–
Overall: AgCl + e–  Ag + Cl–
E0 = 0.22 V
AgCl (s) + e–  Ag (s) + Cl– (sat)
A thin coating of AgCl is deposited
on the pure metal surface.
Ag | AgCl | Cl–

28. Cell Potentials 2
A Fe3+/Fe2+ cell with [Fe3+]=[Fe2+] =1 M has a potential of 0.55V respect to the Ag/AgCl electrode (E0= 0.22 V). What is the potential of this electrode with respect to the SHE?
Answer.
The reactions that occur in the two half-cells are:
Fe3+ + e- → Fe2+ at the cathode E = 0.55 V; E0 = ?
Ag + Cl- +e-→ AgCl at the anode E0= 0.22 V
The potential of this electrode with respect to the SHE is the difference of the two electrode potentials:
E = E E0Fe3+/Fe2+ = = 0.77 V
0.22 Ag+/Ag
0.0V H2/H+ ?

29. Measurement of pH We could construct a concentration cell,
using the standard hydrogen electrode (SHE)
and a hydrogen electrode:
H2(g, 1 atm)  2H+(aq, unknown) + 2e- anode
2H+ (aq, 1M) + 2e-  H2(g, 1 atm) cathode
2H+ (1M)  2H+ (unknown) Ecell = ?
Using Nernst equation:
i.e. the measurement of the cell potential provides pH directly!
unknown
Once conditioned, [H+]glass in the hydration layer is essentially unaffected by aqueous solutions of pH ≤ 14.
The mobility of Na+ within the inner part of the glass permits electrical conduction.
The 'reference solution' (which is often enclosed within the electrode) is a buffered, known aH+ solution.
The electrode then depends on the diffusion of H+ into or out of the hydration layer as a result of contact with a solution of unknown aH+.
at 25°C,

30. pH electrode Eglass electrode= E’ + RT/2.303F log [H+]
The pH electrode potential is typically measured versus a fixed reference calomel electrode.
E’ is the sum of the constant offset potentials of the inner glass surface/solution, the Ag/AgCl electrode, and the calomel electrode.
Eglass electrode= E’ + RT/2.303F log [H+]
Based on a thin glass membrane: a modified glass enriched in H+ and resulting in a hydration layer a few micrometers thick.
Inside the membrane is a 'reference solution' of known [H+] (1M HCl).
The potential difference relevant to pH measurement builds up across the outside glass/solution interface marked ||.
The glass electrode contains the silver/silver chloride metal/insoluble salt electrode immersed in HCl soln of known concentration.
The saturated calomel electrode acts as the reference.
B – most modern labs use the combination electrode – both are housed in the one tube.

31. Nernst Equation
The Nernst equation describes the effect of concentration on cell potential.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Q is plotted on a logarithmic scale so get a straight line for the zinc-copper cell.
Shows a linear decrease.
When Q<1, [reactant, Cu2+] is relatively high and the cell can do relatively more work, i.e. the cell potential is higher than Eocell
When Q=1, Ecell = Eocell
When Q>1, [reactant,Cu2+] is relatively low, and the cell can do relatively less work i.e. the cell potential is lowerthan Eocell.
As the cell operates, the [Zn2+] will increase, while [Cu2+] will decrease, i.e. as the cell operates the potential decreases.
When Q < 1, [reactants] >[products] and the cell can do more work.
When Q = 1, Ecell = E0cell (standard conditions [x] = 1 M).
When Q > 1 , [products] > [reactants] and Ecell is lower.

32. Difference between Q and K
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
Breakdown of N2O4 to NO2:
N2O4 (g, colourless) → 2 NO2 (g, brown)
Q =K

33. Potential of an electrochemical cell
0 V
~1037
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) -
Ecell decreases as the reaction proceeds, until at equilibrium Ecell =0 and . K
Ecell (Volts)

34. Recap: Examining Q To K Ratios
If Q/K < 1
Ecell is positive for the reaction as written. The smaller the Q/K ratio, the greater the value of Ecell and the more electrical work the cell can do.
If Q/K = 1,
Ecell = 0. The cell is at equilibrium and can no longer do work.
If Q/K > 1,
Ecell is negative for the reaction as written. The cell will operate in reverse – the reverse reaction will take place and do work until Q/K = 1 at equilibrium.

35. Link between E 0 and K
Q: What happens if the reaction proceeds until equilibrium is reached?
A: The reaction stops, therefore the voltage, or electrical potential, is zero (the battery is flat). In mathematical terms:
At equilibrium Q=K
At equilibrium Ecell = 0. When Q=K.
No more free energy is released, so the cell can do no more work. At this point we say that a battery is ‘dead’
Derived the expression relating standard cell potential Eo to K.
So the equilibrium constant determines the cell potential.
Large K  products favoured  large standard cell potential, E0

36. Redox reactions are special
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
For redox reactions there is a direct experimental method to measure K and ΔG°.

37. Relation between E 0 and K
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
K is plotted on a logarithmic scale to give a straight line.
K is plotted on a logarithmic scale to give a straight line.
K=1, then Eocell = 0
K>1, Eocell is positive for the reaction as written
K<1, Eocell is negative for the reaction as written.

38. Example question 2
Q: A voltaic cell consisting of a Ni/Ni2+ half-cell and Co/Co2+ half-cell is constructed with the following initial concentrations: [Ni2+] = 0.80 M; [Co2+]=0.2 M.
a) What is the initial Ecell?
b) What is the [Ni2+] when the voltage reaches V?
c) What are the equilibrium concentrations of the ions?
Given: E 0 Ni2+/Ni = V; E 0 Co2+/Co = V
Ni2+ + 2e- → Ni E 0 = -0.25V
Co → Co2+ + 2e- E 0 = +0.28V
Write the standard half-cell equations, reversing the one with more negative potential (Co).
Can see that number of electrons transferred = 2 = n.
Write the equation for the overall cell reaction and the associated standard potential.
Co(s) + Ni2+(aq)  Co2+(aq) + Ni(s) E 0 = 0.03 V

39. Example question 2a a) What is the initial Ecell?
Co(s) + Ni2+(aq)  Co2+(aq) + Ni(s) E 0 = 0.03 V
Determine Q and substitute values.
Substitute values into the Nernst equation gives the initial Ecell.

40. Example question 2b
b) What is the [Ni2+] when the voltage reaches 0.025V?
Q = 1.47
Co(s) + Ni2+(aq)  Co2+(aq) + Ni(s)
x x
So when Ecell = V
[Co2+] = M
[Ni2+] = 0.40 M
When the voltage reaches V, Ecell = V.
N=2, Eo=0.03. Substitute values into the nernst equation and determine Q.
Set up a reaction table. Equilibrium concentrations are 0.8-x, 0.2+x
Substitute into Q, solve for x.
Determine concentrations..
x = 0.40

41. Example question 2c
c) What are the equilibrium concentrations of the ions?
K = 10.24
Co(s) + Ni2+(aq)  Co2+(aq) + Ni(s)
0.80-x x
x = 0.71
So at equilibrium,
[Co2+] = M
[Ni2+] = 0.09 M
Use the equation which relates Eo and K. (because at eqm, Eobs (Ecell) = 0, and Q=K)
Know Eo and n., solve for K.
Set up reaction table.
Initial concentrations are 0.80 and 0.20
Change in concentration = x
Final concentrations shown.
Substitute into equation for K, and solve for x.
Determine equilibrium concentrations.

42. Concentration Cells in Nature
Concentration cells are present all around us, e.g.
nerve signalling: concentration gradients produce electrical current
ion pumps across cell membranes: Na+ / K+ pump, Ca2+ pump
energy production and storage in cells: ATP
Ion selective electrodes- sensitive toonly one kind of ion – e.g., H+,Na+, I-, BF4-

43. Nerve cells
Energy from ATP hydrolysis is used by ion pumps, so that across the nerve cell membrane concentration gradients are maintained:
Ion Concentration Gradient:
Inside Outside
K+ High Low
Na+ Low High
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
The membrane potential is more positive outside than inside the cell.
On nerve stimulation, Na+ enters cell, the inside cell membrane becomes > +ive, then K+ ions leave cell to re-equilibrate the outside.
These rapid (ms) changes in charge across the membrane stimulate the neighbouring region and the electrical impulse moves down the length of the cell.
Specialised cells in our nervous system – neurons can be understood using electrochemical principles.
Communication between neurons occurs through electrical signals generated by millisecond alterations in voltage due to changes in ion concentration.
Cell membrane – separates different concentrations of Na+, K+, Cl- and Ca2+ inside and outside the cell.
The resting membrane potential of a neuron is about -70 mV (mV=millivolt) - this means that the inside of the neuron is 70 mV less than the outside Since , by convention, the potential outside the cell is arbitrarily defined as zero
Nerve is stimulated, ion channels open, large changes in charge stimulates neighbouring region electrical impulse moves down the length of the cell.
Action potentials are caused by an exchange of ions across the neuron membrane. A stimulus first causes sodium channels to open. Because there are many more sodium ions on the outside, and the inside of the neuron is negative relative to the outside, sodium ions rush into the neuron. Remember, sodium has a positive charge, so the neuron becomes more positive and becomes depolarized. It takes longer for potassium channels to open. When they do open, potassium rushes out of the cell, reversing the depolarization. Also at about this time, sodium channels start to close. This causes the action potential to go back toward -70 mV (a repolarization).

44. Nerve cells
Nernst Equation gives the membrane potential generated by the differing extracellular versus intracellular concentrations of each ion:
(Eo = 0)
Consider K+: [K+]outside = 3 mM,
[K+]inside = 135 mM mV
Substituting n = 1, T = 37oC:
( in mV)
1 mV = 10-3 V
Membrane potential caused by differing concentrations on either side of membrane – selectively permeable.
Membrane potential due to K+ concentration difference:
The cell membrane has a potential 102 mV more negative on the inside than the outside due to the much higher K+ concentration inside the cell
since , by convention, the potential outside the cell is arbitrarily defined as zero

45. Cellular Electrochemistry
Biological cells apply the principles of electrochemical cells to generate energy in a complex multistep process.
Bond energy in food is used to generate an electrochemical potential.
The potential is used to create the bond energy of the high-energy molecule adenosine triphosphate (ATP) (energy currency for the cell).
Bond energy in food is used to generate an electrochemical potential.
The potential is used to create the bond energy of the high energy molecule ATP.
The hydrolysis of the high energy molecule ATP (adenosine triphosphate) to ADP (a spontaneous reaction) provides the free energy to drive a wide variety of non-spontaneous reactions, i.e. enables biological processes, eg synthesis of complex molecules.
Explain the prime on the Gibbs free energy symbol.
In biochemical systems, the standard-state concentration of H+ is 10-7 M (pH = 7), not the usual 1 M (would destroy a typical cell). the standard free energy change has the symbol ΔGo’ (solution at pH 7 and at human body temperature 37oC.)
ATP4- + H2O → ADP3- + HPO42- + H+
ΔG °’ = kJ mol-1
ΔGo’ (solution at pH 7 and at human body temperature 37oC.)

46. Bond Energy to Electrochemical Potential
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
Inside mitochondria, redox reactions are performed by a series of proteins that form the electron-transport chain (ETC) which contain redox couples, such as Fe2+/Fe3+.
Large potential differences provide enough energy to convert ADP into ATP.

47. Bond Energy to Electrochemical Potential
In nature, the most important reducing agent is a complex molecule named nicotinamide adenine dinucleotide, abbreviated NADH, which functions as a hydride donor (H-).
= NADH = biological reducing
agent
NAD+ = biological oxidising agent

48. Electron Transport Chain (ETC)
ETC consists of three main steps, each of which has a high enough potential difference to produce one ATP molecule.
The reaction that ultimately powers ETC is the reduction of oxygen in the presence
of NADH:
2H+ + 2e- + ½ O2 → H2O Eo’ = +0.82
NADH + H+ → NAD+ + 2H+ + 2e- Eo’ = -0.32
NADH(aq) + H+(aq) + ½O2(aq) → NAD+(aq) + H2O(l)
Eo’overall = 1.14 V
Driving force for ATP synthesis is the reduction potential
of NADH relative to that of O2:
This is the amount of energy released by the reduction of O2 with NADH..
THIS IS A SUBSTANTIAL RELEASE OF FREE ENERGY (only 31.4 kJmol-1 were released for the hydrolysis of ATP)
THIS ENERGY IS USED TO CREATE A PROTON GRADIENT!!
Substantial energy release!
ΔGo’ = -nFEo’ = -2 · C mol-1· 1.14 V = kJ mol-1

49. ATP Synthesis
In short: e- are transported along the chain, while protons are forced into the intermembrane space.
This creates a H+ concentration cell across the membrane.
In this step, the cell acts as an electrolytic cell, i.e. uses a spontaneous process to drive a non-spontaneous process.
When [H+]intermembrane/[H+]matrix ~ 2.5 a trigger allows protons to flow back across membrane, and ATP is formed.
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
electron transport generates a proton gradient across the inner mitochondrial membrane
Proton gradient is built up as a result of NADH (produced from oxidation reactions) feeding electrons into electron transport system.
proton gradients - a type of potential energy available to do work in a cell
The chemical energy that started with glucose is converted to the energy of a concentration gradient.
Explain prime on standard potential indicates pH 7!!! Explain the prime on the Gibbs free energy symbol.
In biochemical systems, the standard-state concentration of H+ is 10-7 M (pH = 7), not the usual 1 M (would destroy a typical cell). the standard potentials are designated Eo’ (solution at pH 7.) Biochemical reactions take place at 37oC and pH 7. Under these conditions the Gibbs free energy change is specified as ΔGo’
STANDARD POTENTIAL REQUIRED FOR THIS SYNTHESIS – supplied by the proton gradient!!
When the protons flow (via ATP Synthase) back down their concentration gradient, the energy of the gradient drives ATP synthesis.
Energy of the proton concentration gradient is recaptured by ATP synthase in the inner mitochondrial membane.
The energy released pumps H+ out of the mitochondrial matrix into the intermembrane space: a H+ concentration cell is generated across the inner mitochondrial membrane.
At a critical [H+] difference, H+ ions flow back into the matrix, and free energy released drives ATP synthesis:
Eo’ = -ΔGo’/nF = /(3 x 96485) = V
It’s not simple: Noble prize in 1997 to Boyer and Walker for elucidating this.

50. Summary CONCEPTS Concentration cells Nernst equation E 0 and K
Link between E , Q and K
Applications of concentration cells
CALCULATIONS
Work out cell potential from reduction potentials;
Work out cell potential for any concentration (Nernst equation)
Work out K from E 0
Work out pH from concentration cell

51. Pop Quiz 1 Balance the following reaction in basic solution:
MnO CN-  MnO2 + CNO-
Answer: H2O + 2 MnO CN- --> 2 MnO2 + 3 CNO- + 2 OH-

52. Pop Quiz 2 Balance the following reaction in basic solutions:
NO3- + Zn   Zn2+ + NH3
Answer: NO Zn + 6 H2O →  4 Zn2+ + NH3 + 9 OH-