1. REDOX EQUILIBRIA

2. AN EXAMPLE OF A DANIEL CELL (ELECTROCHEMICAL CELL)

3. Redox Equilibria Electrochemical cells
An electrochemical cell has two half–cells.
The two half cells have to be connected with a salt bridge.
Each half cell consists of a metal (acts an electrode) and a 1.00moldm-3 solution of its own ions (eg Cu electrode in 1.00moldm-3 Cu2+(aq)) .
These two half cells will produce a small voltage if connected into a circuit. (i.e. become a Battery or cell).
Standard conditions MUST apply i.e. 298K; 100kPa; 1.00moldm-3 solutions
Electrons flow in the external circuit and NOT through the salt bridge
The voltmeter which measures the potential difference between the two half-cell must be of high resistance

4. In the cell pictured above
Why does a voltage form?
In the cell pictured above
When connected together the zinc half-cell has more of a tendency to oxidise to the Zn2+ ion (as the Zn electrode is more reactive) and release electrons than the copper half-cell.
Zn Zn2+ + 2e-
More electrons will therefore build up on the zinc electrode than the copper electrode.
Zn2+
Zn2+

5. This would ∴ leave a positive charge on the copper electrode
The copper half-cell has more of a tendency to reduce (as copper is less reactive) and ∴ the Cu2+ ions gain electrons from the copper electrode.
Cu2+(aq) + 2e-  Cu(s)
This would ∴ leave a positive charge on the copper electrode
This would leave the Zinc strip with a negative charge (negative electrode)
And the copper strip with a positive charge (positive electrode)
A potential difference is set up between the two half-cells (electrodes)
This potential difference is measured with a high resistant voltmeter, and is given the symbol E.
The E for the electrochemical cell above is E = +1.1V

6. IN GENERAL
When drawing a diagram of a Daniel Cell the MORE reactive half-cell will be placed on the LHS.
OXIDATION always occurs in the LHS half-cell
The LESS reactive half-cell is always placed on the RHS.
REDUCTION always occurs in the RHS half-cell
The LHS electrode is known as the negative electrode
The RHS electrode is known as the positive electrode
Electrons always flow from in the external wires (circuit) from the LHS half-cell to the RHS half-cell

7. Metal atoms lose electrons at the one electrode (oxidation), making it -ve, which travel through the wire to the other electrode, adding to ions to produce metal atoms (reduction).
Remember: oxidation / anode / negative.
Remember also that usually the positive electrode is the right hand electrode.

8. LHS RHS OXIDATION OCCURS REDUCTION OCCURS
This half-cell is MORE reactive
This half-cell is LESS reactive

9. Why use a High resistance voltmeter?
The voltmeter needs to be of very high resistance to stop the current from flowing in the circuit.
In this state it is possible to measure the maximum possible potential difference (E).
The reactions will not be occurring because the very high resistance voltmeter stops the current from flowing.

10. Salt Bridge
The salt bridge is used to connect up the circuit. The free moving ions conduct the charge.
A salt bridge is usually made from a piece of filter paper (or material) soaked in a salt solution, usually Potassium Nitrate.
The salt should be unreactive with the electrodes and electrode solutions.. E.g. potassium chloride would not be suitable for copper systems as Chloride ions can form complexes with copper ions.
Another example is that we would not use potassium chloride with silver systems as chloride ions react with silver ions to form a white precipitate of AgCl.
A wire is not used because the metal wire would set up its own electrode system with the solutions.

11. What happens if current is allowed to flow?
If the voltmeter is removed and replaced with a bulb or if the circuit is short circuited, a current flows.
The reactions will then occur separately at each electrode.
The voltage will fall to zero as the reactants are used up.
The most positive electrode will always undergo reduction.
Cu2+ (aq) + 2e  Cu(s) (positive as electrons are used up)
The most negative electrode will always undergo oxidation.
Zn(s)  Zn2+(aq) + 2e- (negative as electrons are given off)

12. Measuring the standard electrode potential of a cell (EO)
It is not possible to measure the absolute potential of a half electrode on its own.
It is only possible to measure the potential difference between two electrodes.
To measure it, it has to be connected to another half-cell of known potential, and the potential difference between the two half-cells measured.
by convention we can assign a relative potential to each electrode by linking it to a reference electrode (hydrogen electrode), which is given a potential of zero Volts

13. The Standard Hydrogen Electrode (SHE)
The standard electrode potential (E0) of all electrodes are measured by comparing their potential to that of the standard hydrogen electrode.
The standard hydrogen electrode (SHE) is assigned the potential of 0 volts.
NOTE:
Because the equilibrium does not include a conducting metal surface a platinum wire is used which is coated in finely divided platinum. (The platinum black acts as a catalyst, because it is porous and can absorb the hydrogen gas.)

14. The hydrogen electrode equilibrium is:
H2 (g) ⇌ 2H+ (aq) + 2e-
In a cell diagram the hydrogen electrode is represented by:
Pt|H2(g)|H+(aq)
Components of a standard hydrogen electrode.
To make the electrode a standard reference electrode some conditions apply:
1. Hydrogen gas at pressure of 1atm
2. Solution containing the hydrogen ion at 1 moldm-3(solution is usually 1moldm-3 HCl)
3. Temperature at 298K

15. These other standards are themselves calibrated against the SHE.
Secondary standards
The Standard Hydrogen Electrode is difficult to use, so often a different standard is used which is easier to use.
These other standards are themselves calibrated against the SHE.
This is known as using a secondary standard - i.e. a standard electrode that has been calibrated against the primary standard.
The common ones are: silver / silver chloride E = +0.22V
calomel electrode E = +0.27V

16. Standard Electrode Potentials (EO)
When an electrode system is connected to the hydrogen electrode system, and standard conditions apply the potential difference measured is called the standard electrode potential (EO)
REMEMBER - The standard conditions are :
All ion solutions at 1moldm-3
temperature 298K
gases at 100kPa pressure
No current flowing

17. Measuring EO for the Cu2+/Cu half-cell against the SHE.
When finding the potential of a half-cell under test, the standard electrode is always the left hand electrode.
Measuring EO for the Cu2+/Cu half-cell against the SHE.
Note: in the electrode system containing two solutions it is necessary to use a platinum electrode and both solutions must be of a 1M concentration.
This diagram shows how we can measure EO for the Cu2+/Cu half-cell
EO cell = E (Cu2+/Cu) = V

19. A diagram showing how the EO value of Zn2+/Zn half-cell is worked out
EO cell = EO(Zn2+/Zn) = – 0.76V
Pt(s)|H2(g)|H+(aq)||Zn2+(aq)|Zn(s)

21. Standard electrode potentials are found in data books and are quoted as:
Li+(aq)|Li(s) EO= -3.03V
more oxidised form on left
They may also be quoted as half equations
Li+(aq) + e-  Li(s) EO= -3.03V
but again the more oxidised form is on the left

22. BACK TO ELECTROCHEMICAL CELLS
RULES:
The half-cell placed on the LHS
- is the more reactive system
- is the system which is always OXIDISED
- is always the negative electrode
2. The half-cell placed on the RHS
- is the less reactive system
- is the system which is always REDUCED
- is always the positive electrode

23. HOW DO WE KNOW WHICH SYSTEM IS MORE REACTIVE?
ANSWER
From the EO data of each half-cell
The more negative the EO value the more reactive the system ∴ this system will be placed on the LHS and is OXIDISED

24. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
From the data book:
EO Zn2+/Zn = -0.76V
EO Cu2+/Cu = +0.34V
∴ The Zn2+/Zn half-cell is more reactive
Zinc Half Cell (Negative electrode)
Oxidation occurs
Zn(s)  Zn2+(aq) + 2e-
Copper Half Cell (Positive electrode)
Reduction occurs
Cu2+(aq) + 2e-  Cu(s)
OVERALL EQUATION
Zn  Zn e-
Cu e-  Cu
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

25. Cell notation (cell diagrams)
Rather than drawing complicated diagrams of electrochemical cells each time we can write a shorthand form instead in the form of a cell diagram
In a cell diagram the symbols
where | represents a phase boundary (i.e. between species in different states)
|| represents a salt bridge
Also we can use the word
ROOR (Red/Ox/Ox/Red)
To know the order of the symbols

26. REMEMBER – there are 3 types of electrodes
Metal electrodes - These are the type met above, which consist of a metal surrounded by a solution of its ions,
e.g. Zn(s) | Zn2+(aq).
Gas electrodes This is for a gas and a solution of its ions. Here an inert metal (usually platinum) is the actual electrode,
e.g. Pt(s) | H2(g) | H+ (aq)
Redox electrodes This is for two different ions of the same element (e.g. Fe2+ and Fe3+), where the two types of ions are present in solution with an inert electrode (usually Pt).
e.g. Pt(s) | Fe2+(aq), Fe3+(aq)

27. So for the Zn/Zn2+ and Cu/Cu2+ electrochemical cell:
The cell digram is
Zn(s)/Zn2+(aq)//Cu2+ (aq)/Cu(s)
R O R

28. Calculating the EMF of a cell
In order to calculate the Ecell, we must use ‘standard electrode potentials’ for the half cells.
So for this electrochemical cell:
EMF = ERHS – ELHS
= 0.34 – (-0.76)
= 1.10V
A positive value indicates that a reaction is feasible.

29. Write appropriate half equations for each half cell
EXAMPLE 1 – Given the following E0 values:
K+ (aq) + e-  K(s) EO = V
Ag+ (aq) + e-  Ag(s) EO =+ 0.80V
Draw an electrochemical cell to show how these half-cells are connected
Write appropriate half equations for each half cell
Give the overall equation
Which way does the current flow in this electrochemical cell
How is the salt bridge made
What is the purpose of having a salt bridge
Write a cell diagram for this electrochemical cell
Give the value of the EMF
In reality why would we NOT carry out this electrochemical process

30. High resistant voltmeter
298K
100kPa
Salt bridge
Silver
1moldm-3 K+ ions
Potassium
1 moldm-3 Ag+ ions 2.
K(s)  K+(aq) + 1e-
Ag+(aq) + 1e-  Ag(s) 3.
Overall equation K(s) + Ag+(aq)  K(aq) + Ag(s) 4.
From left to right in the external wires 5.
Normally made by soaking filter paper in aqueous KCl BUT in this case we cant use KCl as it precipitates with Ag+ ions ∴ KNO3 is used instead

31. 6. The salt bride allows IONS to be conducted between the two half cells ∴ allowing electrical connection to be made between the two half-cells
K(s)/K+(aq)//Ag+(aq)/Ag(s)
EMF = ERHS –ELHS
= 0.8 – (-2.93) = +3.73V
9. Since K is highly explosive in water and would explode in a 1.00moldm-3 solution of its own ions

32. Example 2 – Given
MnO4 - (aq) + 8H+ (aq) + 5 e-  Mn2+(aq) + 4H2O(l) EO = +1.51
Fe3+(aq) + e-  Fe2+(aq) EO = +0.77
Draw an electrochemical cell to show how these half-cells are connected
Write appropriate half equations for each half cell
Give the overall equation
Write a cell diagram for this electrochemical cell
Give the value of the EMF

33. MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
High resistance voltmeter
Platinum electrode
Platinum electrode
Salt bridge
1 moldm-3 MnO4-(aq)
1moldm-3 Mn2+(aq)
1moldm-3 H+(aq)
1 moldm-3 Fe2+(aq)
1 moldm-3 Fe3+(aq)
Fe2+(aq)  Fe3+(aq) + 1e-
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
OVERALL : 5Fe2+(aq) + MnO4-(aq) + 8H+(aq)  5Fe3+(aq) + Mn2+(aq) 4H2O(l)
Pt(s)/Fe2+(aq);Fe3+(aq)//MnO4-(aq);Mn2+(aq)/Pt(s)
R O O R

34. EMF = ERHS – ELHS
= 1.51 – 0.77
= 0.74V

35. OXIDATION AND REDUCING AGENTS
The most powerful reducing agents will be found at the most negative end of the series on the right, since it will itself be OXIDISED in the process (ie the one with the lower oxidation number)
The most powerful oxidising agents will be found at the most positive end of the series on the left since it will itself be REDUCED in the process (ie the one with the higher oxidation number)

36. So,
In general
the more positive the EO value the more powerful the oxidising agent
The more negative the EO value the more powerful the reducing agent

37. Example 1
Use electrode data to explain why fluorine reacts with water. Write an equation for the reaction that occurs.
ANSWER
First apply idea that more positive Eo will reduce (go forward) and more negative Eo will oxidise (go backwards
Explanation to write
As EO F2/F- > EO O2/H2O, and Ecell is a positive value of +1.64V,
F2 will oxidise H2O to O2
work out Ecell and quote it as part of your answer
Ecell = Ered - Eox = = V
Remember to cancel out electrons in full equation
Equation
2F2(g) + 2H2O(I) → 4F–(aq) + O2(g) + 4H+(aq)

38. Example 2
Use data from the table to explain why chlorine should undergo a redox reaction with water. Write an equation for this reaction
First select relevant half equations by considering the EO values and applying the idea that more positive EO will reduce (go forward) and more negative EO will oxidise (go backwards)
Explanation to write As EO Cl2/Cl- > EO O2/H2O, and Ecell is a positive value of +0.13V, Cl2 will oxidise H2O to O2
Equation
2Cl2(g) + 2H2O(I) → 4Cl–(aq) + O2(g) + 4H+(aq)

39. Example 3
Suggest what reactions occur, if any, when hydrogen gas is bubbled into a solution containing a mixture of iron(II) and iron(III) ions. Explain your answer.
Fe3+ (aq) + e– → Fe2+ (aq) EO = V
2H+(aq) + 2e– → H2(g) EO = 0.00V
Fe2+ (aq) + 2e– → Fe(s) EO = –0.44V
First select relevant half equations by considering the EO values and applying the idea that more positive EO will reduce (go forward) and more negative EO will oxidise (go backwards)
Explanation to write
Fe3+ will be reduced to Fe2+ by H2 oxidising to H+ because EO Fe3+ /Fe2+ > EO H+/H2 and Ecell is a positive value of +0.77V
Equation
2Fe3+ (aq) + H2(g) → 2Fe2+ (aq) + 2H+(aq)

40. Example 4
Use the half-equations to explain in terms of oxidation states what happens to hydrogen peroxide when it is reduced.
Explanation to write
As EO H2O2/H2O > EO O2/H2O2 and Ecell is a positive value of +1.09V , H2O2 disproportionates from -1 oxidation state to 0 in O2 and -2 in H2O
Equation
2H2O2(aq) → O2 + 2H2O(I)

41. Example 5
Will a reaction take place when zinc is added to silver nitrate solution?
Relevant half equations are
Zn2+ + 2e-  Zn Eθ = -0.76V
Ag+ + e-  Ag Eθ = +0.80V
If a reaction takes place the zinc will become zinc ions.
This involves losing electrons so it corresponds to the left hand cell. (This is the reverse of the usually written half equation, so this will be the left hand cell.)
So the Ag is the RHS and Zn is the LHS
Eθ CELL = EθRHC - EθLHC Eθ CELL = = +1.56
The positive value indicates that the reaction is feasible:
2Ag+ (aq) + Zn(s) → Zn2+(aq) + 2Ag(s)

42. Half equation 1: Br2 + 2e-  2Br- Eθ = +1.07
Example 6
Will a reaction take place when acidified hydrogen peroxide is added to bromide ions? Relevant half equations are:
Half equation 1: Br2 + 2e-  2Br Eθ = +1.07
Half equation 2: O2 + 2H+ + 2e-  H2O2 Eθ = +0.68V
Half equation 3 H2O2 + 2H+ + 2e-  2H2O Eθ = +1.77V
The oxidation of bromide ions is 2Br- → Br2 + 2e-
Since this process involves losing electrons, it is the left hand cell
There are two half equation for hydrogen peroxide.
Looking at half equation 2 first;
EθCELL = EθRHS - Eθ LHS Eθ CELL = = -0.39
The negative value tells us that this reaction is not feasible.
Inspection of the half equation would also tell us this of course since in a redox equation one half equation always has to be the reverse of the usual form.

43. H2O2(aq) + 2H+ (aq) + 2Br- (aq) → 2H2O(l) + Br2(aq)
Looking now at the other hydrogen peroxide half equation
EθCELL = EθRHS - EθLHS EθCELL = = +0.70
The positive value indicates that the reaction is feasible:
H2O2(aq) + 2H+ (aq) + 2Br- (aq) → 2H2O(l) + Br2(aq)

44. Effect of conditions on Cell voltage Ecell - Limitations of Eθ values
The electrode potential values have limitations because;
• they refer to standard conditions
• they indicate the energetic feasibility of a reaction, not the kinetics.
If the conditions are different from the standard, the emf can change.
For example, the measurement of Eθ is made at a concentration of 1moldm-3.
If the concentration of one of the solutions is changed, this will change the emf of the reaction.
For example
2Ag+ (aq) + Cu(s) → Cu2+(aq) + 2Ag(s)
For the Cu and Ag cell, the standard value for the reaction is:
Eθ CELL = Eθ RHS - Eθ LHS
Eθ CELL = = +0.46V

45. Values of emf for different silver ion concentrations are shown in the table below
At low silver ion concentration, the reaction will tend to go in the opposite direction
2Ag+ (aq) + Cu(s) → Cu2+(aq) + 2Ag(s)

46. Effect of concentration on Ecell
Ecell is a measure of how far from equilibrium the cell reaction lies. The more positive the Ecell the more likely the reaction is to occur.
The effects of changing conditions on E cell can be made by applying le Chatelier’s principle
If current is allowed to flow, the cell reaction will occur and the Ecell will fall to zero as the reaction proceeds and the reactant concentrations drop.
Zn2+(aq) + 2e-  Zn(s) EO= V
Fe2+(aq) + 2e-  Fe(s) EO= -0.44V
Overall: Zn + Fe2+  Fe + Zn2+ EO= +0.32
Increasing the concentration of Fe2+ and decreasing the concentration of Zn2+ would cause Ecell to increase.
Effect of concentration on Ecell
Looking at cell reactions is a straight forward application of le Chatelier. So increasing concentration of ‘reactants’ would increase Ecell and decreasing them would cause Ecell to decrease.

47. Effect of temperature on Ecell
Most cells are exothermic in the spontaneous direction so applying Le Chatelier to a temperature rise to these would result in a decrease in Ecell because the equilibrium reactions would shift backwards.
If the Ecell positive it indicates a reaction might occur. There is still a possibility, however, that the reaction will not occur or will occur so slowly that effectively it does not happen. If the reaction has a high activation energy the reaction will not occur.

48. Ecell is directly proportional to the total entropy change and to lnK (where K is equilibrium constant)for a reaction
A positive Ecell will lead to a positive total entropy change

49. Breathalysers
The earliest breathalysers used the colour change that occurs when dichromate(VI) ions react with ethanol to measure the amount of alcohol. They could measure the extent to which dichromate turns green
Fuel cells measure the current from an ethanol fuel cell. More alcohol means larger current measured. Fuel cell breathalysers are portable
Infrared breath analysers can determine the amounts of alcohol from an infrared spectrum. More alcohol means greater absorbance. The infrared breathalysers do not use the OH absorption to detect the amount of alcohol on the breath because Water in the breath also has an OH bond.
Many countries will use evidence from both fuel cells and infrared breath analysers for evidence for prosecution, because additional evidence is more reliable

50. Hydrogen Fuel cell (potassium hydroxide electrolyte)
A fuel cell uses the energy from the reaction of a fuel with oxygen to create a voltage
Hydrogen Fuel cell (potassium hydroxide electrolyte)
4e- + 4H2O  2H2 +4OH- E=-0.83V
4e- + 2H2O +O2  4OH- E=+0.4V
Overall reaction: 2H2 + O2  2H2O E=1.23V
Alkaline conditions
In acidic conditions these are the electrode potentials. The Ecell is the same as alkaline conditions as the overall equation is the same
2e- + 2H+  H2 E=0V
4e- + 4H+ +O2 2H2O E=1.23V
Overall 2H2 + O2  2H2O E=1.23V
Using standard conditions: The rate is too slow to produce an appreciable current.
Higher temperatures are therefore used to increase rate but the reaction is exothermic so by applying le chatelier would mean the emf falls. A higher pressure can help counteract this

51. Advantages of Fuel cells over conventional petrol or diesel-powered vehicles
less pollution and less CO2. (Pure hydrogen emits only water whilst hydrogen-rich fuels produce only small amounts of air pollutants and CO2).
(ii) greater efficiency;
Limitations of hydrogen fuel cells
expensive
(ii) storing and transporting hydrogen, in terms of safety, feasibility of a pressurised liquid and a limited life cycle of a solid ‘adsorber’ or ‘absorber’
limited lifetime (requiring regular replacement and disposal) and high production costs,
use of toxic chemicals in their production
Hydrogen is readily available by the electrolysis of water, but this is expensive. To be a green fuel the electricity needed would need to be produced from renewable resources
Hydrogen can be stored in fuel cells (i) as a liquid under pressure, (ii) adsorbed on the surface of a solid material, (iii) absorbed within a solid material;

52. Ethanol fuel cells
Equation that occurs at oxygen electrode
4e- + 4H+ +O2  2H2O E=1.23V
Equation that occurs at ethanol electrode
C2H5OH + 3H2O → 2CO2 + 12H+ + 12e
Overall Equation
C2H5OH + 3O2 → 2CO2 + 3H2O
Ethanol fuel cells have also been developed. Compared to hydrogen fuel cells they have certain advantages including. Ethanol can be made from renewable sources in a carbon neutral way Raw materials to produce ethanol by fermentation are abundant Ethanol is less explosive and easier to store than hydrogen. New petrol stations would not be required as ethanol is a liquid fuel