Redox

1) Element = 0
2) Oxide = -2
3) Hydrogen = +1
4) Ionic element =  charge on ion

 

Redox reactions:

  Oxidation: Reduction
  Loss of e Gain of e
  Ox No increases Ox No decreases
  Are reducing AGENTS Are oxidizing AGENTS

 

Constructing redox equations using relevant half - equations

Example:

Step 1:  Identify the redox half equations:

 

    Fe     Fe3+ + 3e-  
                   
    Cu2+ + 2e- Cu      

 

Step 2:  Balance the electrons:

 

  2Fe     2Fe3+ + 6e- x2
                 
  3Cu2+ + 6e- 3Cu     x3

 

Step 3:  Add the half equations together and cancel out the electrons:

 

  2Fe     2Fe3+ + 6e-  
  3Cu2+ + 6e- 3Cu      
  2Fe + 3Cu2+ + 6e- 3Cu + 2Fe3+ + 6e-  

This gives:

  2Fe + 3Cu2+ 3Cu + 2Fe3+  

Constructing redox equations using oxidation numbers:

Example:

Hydrogen iodide, HI is oxidised to iodine, I2 by concentrated sulphuric acid, H2SO4, which is reduced to hydrogen sulphide, H2S

Step 1:  Identify the redox half equations:

 

    HI     1/2 I2 + e-  
                   
    H2SO4 + 8e- H2S      

 

Step 2:  Balance the electrons:

 

    8HI     4I2 + 8e- x8
                   
    H2SO4 + 8e- H2S      

 

Step 3:  Add the half equations together and cancel out the electrons:

 

  8HI     4I2 + 8e-  
  H2SO4 + 8e- H2S      
  8HI + H2SO4 + 8e- 4I2 + H2S + 8e-  

Step 4: Balance the oxygen's by adding water to the opposite side:

  8HI + H2SO4 4I2 + H2S + 4H2O
O     4       0 so add 4 x waters

Step 5: Balance the hydrogen's by adding hydrogen ions to the opposite side:

  8HI + H2SO4 4I2 + H2S + 4H2O
H   10           10  

Hints for completing redox reactions:

  In acidic conditions Balance using H+
  In alkaline conditions Balance using OH-

Summary:

Step 1:  Identify the redox half equations:

 

Step 2:  Balance the electrons:

Step 3:  Add the half equations together and cancel out the electrons:

Step 4: Balance the oxygen's by adding water to the opposite side:

Step 5: Balance the hydrogen's by adding hydrogen ions to the opposite side:

  In acidic conditions Balance using H+
  In alkaline conditions Balance using OH-

Qu 1-2  P 183

Practical

The feasibility of reactions

Practical

The reactivity series (from GCSE)

Element

Oxidised form

Reduced form

Potassium

K+

K

Sodium

Na+

Na

Lithium

Li+

Li

Calcium

Ca2+

Ca

Magnesium

Mg2+

Mg

Aluminium

Al3+

Al

Zinc

Zn2+

Zn

Iron

Fe2+

Fe

Tin

Sn2+

Sn

Lead

Pb2+

Pb

(Hydrogen)

H+

H

Copper

Cu2+

Cu

Mercury

Hg2+

Hg

Silver

Ag+

Ag

Gold

Au+

Au

 

 

 
  • Those at the top of the reactivity series are so reactive that they prefer to exist in their oxidised form (positive ions).
  • Those at the bottom are so unreactive that they prefer to exist in their reduced form (as metal elements).
  • Those elements in the middle will do the opposite of whatever they are reacting with:

 

Consider the following reactions:

    A)    Zinc powder is added to a solution of copper sulphate:

                          Zn(s)         +          Cu2+(aq)                        Zn2+(aq)       +          Cu(s)

 

The half equations:

 

      Zn(s)                 Zn2+(aq)       +          2e-                 Reaction 1

 

      Cu2+(aq)        +       2e-                           Cu(s)       Reaction 2

    B)    Magnesium powder is added to a solution of zinc sulphate:

      Mg(s)     +             Zn2+(aq)                     Mg2+(s)     +          Zn(s)

 

The half equations:-

 

      Mg(s)                Mg2+(aq)      +          2e-               Reaction 1

 

      Zn2+(aq)         +       2e-                           Zn(s)       Reaction 2

            In these 2 examples (A and B) you have seen that zinc has been oxidised and reduced.

 

Zn(s) Zn2+(aq) + 2e-

 

Applying Le Chateliers Principle:-

 

1.     Add electrons to the system and the equilibrium will shift so as to remove electrons.

2.      Remove electrons from the system and the equilibrium will shift so as to produce electrons.

 

We call this is metal / metal ion system or half cell.

The electrochemical series:

 

Element

Oxidised form

Reduced form

Eqcell

 

Potassium

K+

K

-2.92

Sodium

Na+

Na

-2.71

Lithium

Li+

Li

-2.59

Calcium

Ca2+

Ca

-2.44

Magnesium

Mg2+

Mg

-2.37

Aluminium

Al3+

Al

-1.66

Zinc

Zn2+

Zn

-0.76

Iron

Fe2+

Fe

-0.44

Tin

Sn2+

Sn

-0.14

Lead

Pb2+

Pb

-0.13

(Hydrogen)

H+

H

0.00

Copper

Cu2+

Cu

+0.34

Mercury

Hg2+

Hg

+0.79

Silver

Ag+

Ag

+0.80

Gold

Au+

Au

+1.89

 

 

 
  • Eqcell values are arranged with the most negative values at the top.

  • They are arranged with the highest oxidation number on the left.

  • The more negative a value, the greater the tendency for the electrode system to loose electrons.

  • This means that the most negative of the 2 systems will move to the LHS whereas the least negative will move to the RHS.

This means:-

  • The systems at the top of the table have a greater tendency to go from right Left. More negative produces electrons more readily

  • The systems at the bottom of the table have a greater tendency to go from left right.  More positive reacts with the electrons more readily.

 

Will a reaction actually take place:

 

Example 1

 

1)    Write out the half reactions.  The more (-)ve value will produce electrons, the more positive will react with electrons.

 

Mg2+(aq) + 2e- Eq = -2.37

1

+ 2e- Cu(s) Eq = +0.34

2

2)    Draw the direction of each of the half reactions:

  Mg2+(aq) + 2e-   Eq = -2.37

1

+ 2e- Cu(s) Eq = +0.34

2

 

3)    Writing balanced redox equation

Mg(s)              Mg2+(aq)      +       2e-

 

Cu2+(aq)       +       2e-                   Cu(s)

4)    Add the half reactions

                             Mg(s)              Mg2+(aq)      +       2e-

 

Cu2+(aq)       +       2e-                   Cu(s)                                                   

 

Mg(s)     +    Cu2+(aq)    +    2e-                Mg2+(aq)   +    Cu(s)     +    2e-            electrons cancel out

 

               Mg(s)     +    Cu2+(aq)              Mg2+(aq)   +    Cu(s)          

 

This can now be extended to metal ion / metal ion systems and also non metal systems:

 

Example 2:

 

1)    Write out the half reactions.  The more (-)ve value will produce electrons, the more positive will react with electrons.

 

  Sn2+(aq) + 2e-   Eq = -0.14

1

+ e- Fe2+(aq) Eq = +0.77

2

 

2)    Draw the direction of each of the half reactions:

 

  Sn2+(aq) + 2e-   Eq = -0.14

1

+ e- Fe2+(aq) Eq = +0.77

2

 

3)    Writing balanced redox equation

Sn(s)              Sn2+(aq)      +       2e-

 

Fe3+(aq)       +       e-                   Fe2+(aq)                x2

4)    Add the half reactions

                                Sn(s)              Sn2+(aq)      +       2e-

 

2Fe3+(aq)       +       2e-                   2Fe2+(aq)

 

 Sn(s)     +    2Fe3+(aq)    +    2e-                Sn2+(aq)   +    2Fe2+(aq)     +    2e-            electrons cancel out

 

               Sn(s)     +    2Fe3+(aq)              Sn2+(aq)   +    2Fe2+(aq)          

 

Summary steps:

 

1)    Write out the half reactions.  The more (-)ve value will produce electrons, the more positive will react with electrons.

2)    Draw the direction of each of the half reactions:

3)    Writing balanced redox equation

4)    Add the half reactions

 

Limitations of predictions using standard electrode potentials:

 

a)    Effect of changing concentration:

      Cu2+(aq)        +       2e-                           Cu(s)    

b)    Effect of Eqcell value:

Qu 1  P189

 

Cells and half cells

 

Electricity from chemical reactions

Cells and half cells:

                          Zn(s)         +          Cu2+(aq)                        Zn2+(aq)       +          Cu(s)

The half equations:

 

      Zn(s)                 Zn2+(aq)       +          2e-                 half cell:  flows out of the system to the external wire

 

      Cu2+(aq)        +       2e-                           Cu(s)       half cell:  flows into the system from the external wire

 

 

Consider the half cell:

 

      Cu2+(aq)        +       2e-                  Cu(s)                   Ecell = +0.34v  

With a more negative half cell:

  • With a half cell that is more negative, the equilibria will move to the RHS.  Electrons move from the more negative to the more positive half cell.

With a more positive half cell:

  • With a half cell that is more positive, the equilibria will move to the LHS.  Electrons move from the more negative to the more positive half cell.

 

 

So what determines where the copper half cell fits in the electrochemical series?

M(s) Mz+(aq) + ze-
         
  • As the equilibrium is set up electrons are deposited on the metal foil as metal ions are released into solution.

  • The solution will become positively charged due to the excess metal ions.

  • The metal will become negatively charged as the electrons are deposited upon it.

  • An electrode potential difference is set up between the metal and the ions in solution.

  • Note the overall difference in potential between the ions and the metal is zero.

 

 

  • The position of the equilibrium between different metals and their ions will be different.

  • This means that the electrode potential differences set up will also be different.

  • These potentials are called absolute potentials and cannot be measured.

  • A reference half cell is used for which all standard electrode potentials are measured against (later).

  • By joining 2 of these metal/ion systems together we have a difference in potential between the 2 systems which can be measured

  • The potential difference between the 2 metal electrodes is proportional to the difference in the 2 absolute potentials.

  • This is measured by the voltage between the 2 metal electrodes.

 

 

Zn(s)     +      Cu2+(aq)                  Zn2+(aq)     +        Cu(s)

 

The half equations:

 

Zn2+(aq)       +          2e-                        Zn(s)        

 

Most (-)ve half cell - e's flow out of the half cell to the external wire - Equilibrium moves to LHS

 

Cu2+(aq)        +       2e-                        Cu(s)      

 

Most (+)ve half cell - e's  flows into the half cell from the external wire - Equilibrium moves to RHS

 

Current flows in the opposite direction to the flow of electrons

Non - metal / non - metal ion systems:

      2H+(aq)          +       2e-                     H2(g)      

 

The hydrogen electrode:
  • There is an obvious problem here, there is no metal for the electrons to be transferred from or to the half cell.

  • This is overcome by using a platinum wire coated with finely divided platinum (platinum black), which dips into the acid (H+ ions).

  • A slow stream of hydrogen gas is passed over the platinum black surface, the above equilibrium is gradually set up.

  • Platinum black is porous and retains a lot of hydrogen gas.

  • Platinum metal is also convenient for electron transfer.

This is the reference electrode against which all other half cell electrode potential are measured (later)

 

For this reason the hydrogen electrode has a potential = 0

 

Conditions:

  • 1M HCl (as the source of H+ions)

  • 1 Atmosphere H2 gas (100KPa)

  • 298 K

  • Inert platinum electrode

 

 

Other non - metal / non - metal ion half cells:

 

Fe2+(aq)        +       e-                        Fe3+(s)   
  • This system contains iron in its 2 oxidation states.
  • Again there is no metal to allow electron transfer.
  • Platinum is used as an inert electron carrier.
  • Equimolar solutions are used (usually 1M)

 

Qu 1  P185

 

Cell potentials

 

Standard electrode potentials

Measuring standard electrode potentials:

 

Standard electrode potentials and cell reactions

Example 1:  A magnesium - copper cell is made by connecting 2 half cells together:  Mg2+ / Mg and Cu2+ / Cu

 

1)    Write out the half reactions.  The more (-)ve value will produce electrons, the more positive will react with electrons.

 

Mg2+(aq)

+

2e-

Mg(s)

Eq = -2.37

1

Cu2+(aq)

+

2e-

Cu(s)

Eq = +0.34

2

 

2)    Draw the direction of each of the half reactions:

  Mg2+(aq) + 2e-   Eq = -2.37

1

+ 2e- Cu(s) Eq = +0.34

2

3.  Balanced chemical equation - Write out the 2 half reactions and the overall equation balancing with electrons

Mg(s)              Mg2+(aq)      +       2e-

 

Cu2+(aq)       +       2e-                   Cu(s)

4)    Add the half reactions

                             Mg(s)              Mg2+(aq)      +       2e-

 

Cu2+(aq)       +       2e-                   Cu(s)                                                    add the half reactions

 

Mg(s)     +    Cu2+(aq)    +    2e-                Mg2+(aq)   +    Cu(s)     +    2e-            electrons cancel out

 

               Mg(s)     +    Cu2+(aq)              Mg2+(aq)   +    Cu(s)          

 

      5)  Calculating emf of electrochemical cells:

 

Emf   =    Eqpos        -       Eqneg

Emf   =    + 0.34       -        - 2.37

Emf   =    + 2.71v

 

Rules:

1)    Write out the half reactions.  The more (-)ve value will produce electrons, the more positive will react with electrons.

2)    Draw the direction of each of the half reactions:

3)    Writing balanced redox equation

4)    Add the half reactions

5)    Calculating emf:  Eqcell   =    Eqpos        -       Eqneg

 

 

Example 2:  A tin - iron cell is made by connecting 2 half cells together:  Sn2+ / Sn and Fe3+ / Fe2+

 

1)    Write out the half reactions.  The more (-)ve value will produce electrons, the more positive will react with electrons.

 

  Sn2+(aq) + 2e- Sn2+(s)   Eq = -0.14

1

Fe3+(aq) + e- Fe2+(aq) Eq = +0.77

2

 

2)    Draw the direction of each of the half reactions:

 

  Sn2+(aq) + 2e-   Eq = -0.14

1

+ e- Fe2+(aq) Eq = +0.77

2

 

3.  Balanced chemical equation - Write out the 2 half reactions and the overall equation balancing with electrons

Sn(s)              Sn2+(aq)      +       2e-

 

Fe3+(aq)       +       e-                   Fe2+(aq)                x2

4)    Add the half reactions

                                Sn(s)              Sn2+(aq)      +       2e-

 

2Fe3+(aq)       +       2e-                   2Fe2+(aq)

 

 Sn(s)     +    2Fe3+(aq)    +    2e-                Sn2+(aq)   +    2Fe2+(aq)     +    2e-            electrons cancel out

 

               Sn(s)     +    2Fe3+(aq)              Sn2+(aq)   +    2Fe2+(aq)          

 

5)    Calculating emf:  Eqcell   =    Eqpos        -       Eqneg

 

Emf   =    Eqpos        -       Eqneg

Emf   =    + 0.77       -        - 0.14

Emf   =    + 0.91v

 

Practical

 

Qu 1  P187

 

Storage and fuel cells

Electrochemical cells

 

The half equations:

 

Zn2+(aq)   +     2e-                 Zn(s)           -0.76v   

 

Cu2+(aq)  +     2e-                   Cu(s)          +0.34v

 

Overall equation: 

 

Zn(s)     +      Cu2+(aq)                  Zn2+(aq)     +        Cu(s)

 

Emf   =    Eqpos        -       Eqneg

Emf   =    + 0.34       -        - 0.76

Emf   =    + 1.10v

 

Modern cells and batteries:

1) Non - rechargable cells: Provides electricity until the chemicals have reacted away.
2) Rechargable cells: The chemicals react providing electricity until they have reacted away.  The difference is that the chemicals can be regenerated by reversing the flow of electrons during charging.
3) Fuel cells: The chemicals react providing electricity but the chemicals needed are constantly supplied.

Fuel cells:

Practical

The hydrogen oxygen fuel cell:

  • The modern fuel cell uses hydrogen and oxygen to create a voltage.
  • The difference is stationary alkaline electrolyte giving a large voltage.
  • The fuel (hydrogen) and oxygen flow into the cell.
  • This produces electricity:

The half equations:

 

2H2O(l)   +     2e-                 H2(g)    +   2OH-(aq)           -0.83v   

 

1/2O2(g)  +    H2O(l)    +   2e-                   2OH-(aq)         +0.40v

 

Overall equation: 

 

                   H2(g)     +      1/2O2(g)                  H2O(l)

 

Emf   =    Eqpos        -       Eqneg

Emf   =    + 0.40       -        - 0.83

Emf   =    + 1.23v

  • The products flow out of the cell.

Qu 1-2  P191

Hydrogen cells for the future

Development of fuel cell vehicles (FCV's):

                            CH3OH    +    H2O        3H2    +    CO2

Methanol as an alternative to hydrogen:

Advantages:

        1)  Liquids are easier to store than gases

        2)  Methanol can be produced from biomass

Problems:

        1)  Only generate a small amount of power

        2)  Produce CO2

Advantages of fuel cell vehicles:

        1)  Less CO2 produced

        2)  Normal hydrocarbons produce CO which needs to be removed by catalytic converters

        3)  Fuel cells are about 40 - 60% efficient / hydrocarbons are about 20%

Storage of hydrogen

  Under pressure Adsorbed onto a solid surface Absorbed within a solid
Hydrogen can be stored as a liquid under pressure.  Very low temperatures are required to keep them as liquids.  A large 'thermos flask' would be needed to prevent it from boiling. Like a catalyst holds molecules in place.  This means that the hydrogen molecules occupy a smaller volume then as a gas. Similar to before, the hydrogen molecules absorb into the material meaning that the hydrogen occupies a smaller volume.

Limitations of hydrogen fuel cells:

The hydrogen economy:

Qu 1 - 2  P193  /  Qu 1 - 7  P195  /  Qu 1 - 6  P196 - 197