ELECTROCHEMISTRY

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Electrochemistry can be defined as the study of the effects of electricity on a substance/ compound and how chemical reactions produce electricity. Electrochemistry therefore deals mainly with:

  1. Reduction and oxidation
    1. Electrochemical (voltaic) cell
    1. Electrolysis (electrolytic) cell

(i)REDUCTION AND OXIDATION (REDOX)

1. Interms of oxygen transfer:

    i) Reduction is removal of oxygen.

    ii) Oxidation is addition of oxygen.

    iii) Redox is simultaneous addition and removal of oxygen.

    iv) Reducing agent is the species that undergoes oxidation, therefore gains oxygen.

    v) Oxidizing agent is the species that undergoes reduction, therefore looses/donates oxygen.

 e.g.          When hydrogen is passed through heated copper (II) oxide, it is oxidised to copper metal as in the equation below:

CuO (s)           +           H2 (g)                     ->       Cu (s)        +      H2O (l)

(Oxidising agent)     (Reducing agent)

2. In terms of hydrogen transfer:

   i) Oxidation is the removal of hydrogen.

   ii) Reduction is the addition of hydrogen.

   iii) Redox is simultaneous addition and removal of hydrogen.

   iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates hydrogen.

   v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen.

e.g.    When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid). The chlorine is reduced (gain hydrogen) to hydrogen chlorine gas.

Cl2 (g)           +              H2S (g)     ->               S(S)            +           2HCl (g)

(Oxidizing agent)     (Reducing agent)

3. In terms of electron transfer:

     i) Oxidation is donation/ loss/ removal of electrons.

    ii) Reduction is gain/ accept/ addition of electrons.

    iii) Redox is simultaneous gain/ accept/ addition and donation/ loss/ removal of electrons.

    iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates electrons.

    v) Oxidizing agent is the species that undergoes reduction, therefore gains/ accepts electrons.

Example

  1. Displacement of metals from their solutions:

Place 5cm3 each of Iron (II) sulphate (VI) solution into three different test tubes. Add about 1g of copper tunings / powder into one test tube then zinc and magnesium powders separately into the other test tubes. Shake thoroughly for 2 minutes each. Record any colour changes in the table below.

Metal added to Iron (II) sulphate (VI) solutionColour changes
CopperSolution remains green
ZincGreen colour fades
MagnesiumGreen colour fades

Explanation

-When a more reactive metal is added to a solution of less reactive metal, it displaces it from its solution.

-When a less reactive metal is added to a solution of a more reactive metal, it does not displace it from its solution.

-Copper is less reactive than iron therefore cannot displace iron its solution.

-Zinc is more reactive than iron therefore can displace iron from its solution.

-Magnesium is more reactive than iron therefore can displace iron from its solution.

In terms of electron transfer:

          – the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions

          -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal.

          -displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal.

          Examples

  1. Zn(s) ->     Zn2+(aq) +  2e        (oxidation/donation of electrons)

     Fe2+(aq)  +  2e              ->      Fe(s)     (reduction/gain of electrons)

Fe2+(aq)  +    Zn(s) ->   Zn2+(aq) +  Fe(s)  (redox/both donation and gain of electrons)

  • Mg(s)         ->     Mg2+(aq) + 2e        (oxidation/donation of electrons)

     Fe2+(aq)  +  2e              ->      Fe(s)     (reduction/gain of electrons)

Fe2+(aq)  +    Mg(s) ->   Mg2+(aq) +  Fe(s)  (redox/both donation and gain of electrons)

  • Zn(s) ->     Zn2+(aq) +  2e        (oxidation/donation of electrons)

     Cu2+(aq)  + 2e              ->      Cu(s)     (reduction/gain of electrons)

Cu2+(aq)  +    Zn(s) ->   Zn2+(aq) +  Cu(s)  (redox/both donation and gain of electrons)

  •  Fe(s) ->     Fe2+(aq) +   2e        (oxidation/donation of electrons)

     2Ag+(aq)  + 2e              ->      2Ag(s)     (reduction/gain of electrons)

2Ag+(aq)  +    Fe(s) ->   Fe2+(aq) +  2Ag(s)  (redox/both donation and  gain of electrons)

  •     Zn(s)      ->     Zn2+(aq) +  2e        (oxidation/donation of electrons)

Cl2(g)  +     2e              ->      2Cl(aq)     (reduction/gain of electrons)

Cl2(g)   +    Zn(s) ->   Zn2+(aq) +  2Cl(aq) (redox/both donation and  gain of electrons)

  •     2Mg(s)   ->     2Mg2+(aq) +         4e        (oxidation/donation of electrons)

    O2(g)  +  4e              ->      2O2-(aq)     (reduction/gain of electrons)

O2(g)   +   2Mg(s) ->   2Mg2+(aq)2O2-(aq)   (redox/both donation and  gain of electrons)

Note

(i)The number of electrons donated/lost MUST be equal to the number of electrons gained/acquired.

(i)During displacement reaction, the colour of ions /salts fades but does not if displacement does not take place. e.g

a)Green colour of Fe2+(aq) fades if Fe2+(aq) ions are displaced from their solution. Green colour of Fe2+(aq) appear if Fe/iron displaces another salt/ions  from their solution.

b)Blue colour of Cu2+(aq) fades if  Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions  from their solution.

c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution. Brown colour of Fe3+(aq) appear if Fe/iron displaces another salt/ions  from their solution to form Fe3+(aq).

(iii)Displacement reactions also produce energy/heat. The closer/nearer the metals in the reactivity/electrochemical series the less energy/heat of displacement.

(iv)The higher the metal in the reactivity series therefore the easier to loose/donate electrons and thus the stronger the reducing agent. 

4. (a)In terms of oxidation number:

          i) Oxidation is increase in oxidation numbers.

          ii) Reduction is decrease in oxidation numbers.  

iii) Redox is simultaneous increase in oxidation numbers of one species/substance and a decrease in oxidation numbers of another species/substance. 

iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number.

v) Oxidizing agent is the species that undergoes reduction, therefore increases  its oxidation number.

(b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules:

Guidelines /rules applied in assigning oxidation number

1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1

2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1

3.All atoms and molecules of elements have oxidation number 0 (zero)

AtomOxidation numberMoleculeOxidation number
Na0Cl20
O0O20
H0H20
Al0N20
Ne0O30
K0P30
Cu0S80

4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g.

Metal/non-metal ionValencyOxidation stateOxidation number
 Fe2+2-2-2
 Fe3+3-3-3
 Cu2+2-2-2
 Cu+1+1+1
 Cl1-1-1
 O2-2-2-2
Na+1+1+1
Al3+3+3+3
 P3-3-3-3
 Pb2+2+2+2

5.Sum of oxidation numbers of atoms of elements making a compound is equal  zero(0) e.g.

Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below:

a) CuSO4 has-

-one atom of Cu with oxidation number  +2( refer to Rule 4)

-one atom of  S with oxidation number  +6 ( refer to Rule 4)

-six atoms of O each with oxidation number  -2( refer to Rule 4)

Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0

b) H2SO4 has-

-two atom of H each with oxidation number  +1( refer to Rule 2)

-one atom of  S with oxidation number  +6 ( refer to Rule 4)

-four atoms of O each with oxidation number  -2( refer to Rule 4)

Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0

c) KMnO4 has-

-one atom of K with oxidation number  +1( refer to Rule 4)

-one atom of  Mn with oxidation number  +7 ( refer to Rule 4)

-four atoms of O each with oxidation number  -2( refer to Rule 4)

Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0

Determine the oxidation number of:

I.Nitrogen in;

-NO   => x + -2 = 0 thus  x = 0 – (-2) = + 2

The chemical name of this compound is thus Nitrogen(II)oxide

-NO2   => x + (-2 x2)= 0 thus  x = 0 – (-4) = + 4

The chemical name of this compound is thus Nitrogen(IV)oxide

-N2O  => 2x + -2 = 0 thus  2x = 0 – (-2) = +2/2= +1

The chemical name of this compound is thus Nitrogen(I)oxide

II. Sulphur in;

-SO2   => x + (-2 x2)= 0 thus  x = 0 – (-4) = + 4

The chemical name of this compound is thus Sulphur(IV)oxide

-SO3   => x + (-2 x3)= 0 thus  x = 0 – (-6) = + 6

The chemical name of this compound is thus Sulphur(VI)oxide

-H2SO4  = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6

The chemical name of this compound is thus Sulphuric(VI)acid

-H2SO3  = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4

The chemical name of this compound is thus Sulphuric(IV)acid

III. Carbon in;

-CO2   => x + (-2 x2)= 0 thus  x = 0 – (-4) = + 4

The chemical name of this compound is thus carbon(IV)oxide

-CO   => x + -2 = 0 thus  x = 0 – -2 = + 2

The chemical name of this compound is thus carbon(II)oxide

-H2CO3  = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4

The chemical name of this compound is thus Carbonic(IV)acid

IV.Manganese in;

-MnO2   => x + (-2 x2)= 0 thus  x = 0 – (-4) = + 4

The chemical name of this compound is thus Manganese(IV)oxide

-KMnO4  = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7

The chemical name of this compound is thus Potassium manganate(VII)

 V.Chromium in;

– Cr2O3   => 2x + (-2 x 3)= 0 thus  2x = 0 – (-6) = +6 / 2= +3

The chemical name of this compound is thus Chromium(III)oxide

                   -K2Cr2O7   => (+1 x 2) + 2x + (-2 x7)= 0

thus 2x = 0 – +2 +-14 = +12 / 2= +6

The chemical name of this compound is thus Potassium dichromate(VI)

                   -K2CrO4   => (+1 x 2) + x + (-2 x4)= 0

thus 2x = 0 – +2 +-8 = +12 / 2= +6

The chemical name of this compound is thus Potassium chromate(VI)

6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge.

Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below;

a) SO42- has-

-one atom of S with oxidation number  +6( refer to Rule 4)

-four atoms of O each with oxidation number  -2( refer to Rule 1)

Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2

The chemical name of this radical is thus sulphate(VI) ion

 b) NO3 has-

-one atom of N with oxidation number  +4( refer to Rule 4)

-three atoms of O each with oxidation number  -2( refer to Rule 1)

Sum of oxidation numbers of atoms in NO3 = ( +4 + (-2 x 3)) = -1

The chemical name of this radical is thus nitrate(IV) ion.

Determine the oxidation number of:

I.Nitrogen in;

-NO2   => x + (-2 x2)= -1 thus  x = -1 – (-4) = + 3

The chemical name of this compound/ion/radical is thus Nitrate(III)ion

II. Sulphur in;

-SO32-  => x + (-2 x3)= -2 thus  x = -2 – (-6) = + 4

The chemical name of this compound/ion/radical is thus Sulphate(IV)ion

III. Carbon in;

-CO32-  = x + (-2 x 3) = -2 thus  x = -2 – (-6) = + 4

The chemical name of this compound/ion/radical is thus Carbonate(IV)ion

IV.Manganese in;

-MnO4 =  x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7

The chemical name of this compound/ion/radical is thus manganate(VII) ion

 V.Chromium in

-Cr2O72-   => 2x + (-2 x7)= -2

thus 2x = -2 – +2 +-14 = +12 / 2= +6

The chemical name of this compound/ion//radical is thus dichromate(VI) ion

                   -CrO42-   => x + (-2 x4)= -2

thus x =  -2 + (-2  x 4) = +6

The chemical name of this compound/ion//radical is thus chromate(VI) ion

(c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined  as in the following examples;

(i)                                  Cu2+   (aq) +      Zn(s)  ->      Zn2+   (aq) +     Cu(s)

Oxidation numbers ->   +2                         0                 +2                         0

Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)

Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s)

(ii)                                 2Br  (aq) +      Cl2(g) ->      2Cl  (aq)   +     Br2 (l)

Oxidation numbers ->   -1                         0                 -1                         0

Oxidizing agent =>Cl2(g);its oxidation number decrease from 0 to-1 in 2Cl  (aq)

Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s)

(iii)                                Br2 (l)         +      Zn(s)  ->      Zn2+   (aq) +     2Br(aq)

Oxidation numbers ->   0                          0                 +2                         -1

Oxidizing agent =>Br2 (l);its oxidation number decrease from 0 to-1 in 2Br(aq)

Reducing agents => Zn(s);its oxidation number increase from 0 to +2 in Zn2+

(iv)                                 2HCl (aq)   +      Mg(s)   ->   MgCl2  (aq)      +   H2(g)

Oxidation numbers ->    2 (+1  -1)             0                 +2  2(-1)                 0

Oxidizing agent => H+  in HCl;its oxidation number decrease from +1to 0 in H2(g)

Reducing agents => Mg(s);its oxidation number increase from 0 to +2 in Mg2+

(v)                             2H2O (l)   +      2Na(s)   ->         2NaOH  (aq)      +   H2(g)

Oxidation numbers -> +1  -2                 0           +1 -2  +1                  0

Oxidizing agent => H+  in H2O;its oxidation number decrease from +1to 0 in H2(g)

Reducing agents => Na(s);its oxidation number increase from 0 to +1 in Na+

(vi)  5Fe2+ (aq)   +   8H+ (aq)  +  MnO4   ->  5Fe3+ (aq)   +   Mn2+ (aq)  +  4H2O (l)                                           +2               +1                 +7   -2            +3                  +2               +1   -2

Oxidizing agent => Mn in MnO4;its oxidation number decrease from +7to+2 in Mn2+  

Reducing agents => Fe2+;its oxidation number increase from +2  to +3 in Fe3+

(vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq)  +  Cr3+ (aq)  +  7H2O (l)                                           +2               +1             +6  -2                 +3                        +3                +1   -2

Oxidizing agent:

 Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+  

Reducing agents => Fe2+;its oxidation number increase from +2  to +3 in Fe3+

(viii) 2Fe2+ (aq)  +   2H+ (aq)   +   H2O2(aq)   ->   2Fe3+ (aq)  +    2H2O (l)                                           +2                 +1                +1  -1                +3                   +1   -2

Oxidizing agent:

 O in H2O2;its oxidation number decrease from -1 to -2 in H2O

Reducing agents => Fe2+;its oxidation number increase from +2  to +3 in Fe3+

(ix) Cr2O72-(aq)  +   6H+ (aq)   +  5H2O2(aq)   ->  2Cr3+ (aq)  +   2H2O (l) + 5O2(g)                                                                                                       

      +6   -2              +1                +1  -1                +3                   +1   -2       0

Oxidizing agents:

 O in H2O2;its oxidation number decrease from -1 to -2 in H2O

 Cr in Cr2O72- itsoxidation number decrease from +6 to +3 inCr3+

Reducing agents

 O in H2O2;its oxidation number increase from -1 to O in O2(g)

 O in Cr2O72- itsoxidation number increase from -2 to O inO2(g)

(x) 2MnO4(aq)  +   6H+ (aq)   +  5H2O2(aq)   ->  2Mn2+ (aq)  +   8H2O (l) + 5O2(g)                                                                                                       

      +7   -2              +1                +1  -1                +2                   +1   -2       0

Oxidizing agents:

 O in H2O2;its oxidation number decrease from -1 to -2 in H2O

 Mn in MnO4 itsoxidation number decrease from +7 to +2 inMn2+

Reducing agents

 O in H2O2;its oxidation number increase from -1 to O in O2(g)

 O in MnO4 itsoxidation number increase from -2 to O inO2(g)

(ii)ELECTROCHEMICAL (VOLTAIC) CELL

1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e.

                   M(s)     ->   M+(aq)    +   e   ( monovalent metal)

                   M(s)     ->   M2+(aq)   +  2e  ( divalent metal)

                   M(s)     ->   M3+(aq)   +  3e  ( Trivalent metal)

The ions move into the solution leaving electrons on the surface of the metal rod/plate.

2.The metal rod becomes therefore negatively charged while its own solution positively charged. As the positive charges of the solution increase, some of them recombine with the electrons to form back the metal atoms

M+(aq)        +   e     ->   M(s)  ( monovalent metal)

M2+(aq)        +   2e   ->   M(s)          (divalent metal)

M3+(aq)        +   3e   ->   M(s)          (Trivalent metal)

3. When a metal rod/plate is put in a solution of its own salt, it constitutes/forms a half-cell. The tendency of metals to ionize differ from one metal to the other. The difference can be measured by connecting two half cells to form an electrochemical/voltaic cell as in the below procedure:

To set up an electrochemical /voltaic cell

To compare the relative tendency of metals to ionize

 Place 50cm3 of 1M Zinc(II) sulphate(VI) in 100cm3 beaker. Put a clean zinc rod/plate into the solution. Place 50cm3 of 1M Copper(II) sulphate(VI) in another 100cm3 beaker. Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers. The whole set up should be as below

Repeat the above procedure by replacing:

(i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution

(ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution

(iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution

Record the observations in the table below

Changes on the  1st  metal rod (A)Changes on the  2nd   metal rod (B)Changes on the  1st  solution  (A(aq))Changes on the  2nd solution  (B(aq))Voltage/voltmeter reading(Volts)
Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes    -copper rod /plate increase in size /mass/ deposited    Zinc(II)sulphate (VI)colour remain colourless    Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit      0.8 (Theoretical value=1.10V)
Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes    -copper rod /plate increase in size /mass/ deposited    Magnesium(II) sulphate(VI) colour remain colourless    Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit      1.5 (Theoretical value=2.04V)
Using Ag/Cu half cell -The rod increase in size /mass /deposited    -silver coin/ rod /plate increase in size /mass/ deposited    Blue Copper (II)sulphate (VI)colour remains      Silver(I)nitrate (V)colour remain colourless      0.20 (Theoretical value=0.46V)
Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes    -copper rod /plate increase in size /mass/ deposited    Iron(II)sulphate (VI)colour becomes more green    Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit      0.60 (Theoretical value=0.78V)

From the above observations ,it can be deduced that:

(i)in the Zn/Cu half-cell the;

          -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+

          Ionic equation      Zn(s)          ->       Zn2+(aq)    +       2e

          -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms

          Ionic equation      Cu2+(aq)    +       2e     ->      Cu(s)

This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod.

When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell.

The purpose of the salt bridge therefore is:

          (i)complete the circuit

          (ii)maintain balance of charges /ions on both half cells.

 For the negative half cell the NO3 /Cl  from salt bridge decrease/neutralise  the increased positive(Zn2+) ion.

For the positive half cell the Na+ / K+  from salt bridge increase  the decreased positive(Cu2+) ion.

The voltmeter should theoretically register/read a 1.10Volts as a measure of the electromotive force (e.m.f) of the cell .Practically the voltage reading is lowered because the connecting wires have some resistance to be overcomed.

A combination of two half cells that can generate an electric current from a redox reaction is called a voltaic/electrochemical cell.

By convention a voltaic/electrochemical cell is represented;

              M(s)       /  M2+(aq)   //   N2+ (aq)   /   N(s)

   (metal rod of M)(solution ofM)(solution ofN)(metal rod ofN)

Note;

a)(i)Metal M must be the one higher in the reactivity series.

   (ii)It forms the negative terminal of the cell.

   (iii)It must diagrammatically be drawn first  on the left hand side when illustrating the voltaic/electrochemical cell.

b)(i)Metal N must be the one lower in the reactivity series.

   (ii)It forms the positive terminal of the cell.

   (iii)It must diagrammatically be drawn second/after/ right hand side when illustrating the voltaic/electrochemical cell.

Illustration of  the voltaic/electrochemical cell.

(i)Zn/Cu cell

1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal

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

2. Copper ions in solution gain the donated electrons to form copper atoms/metal

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

3.Overall redox equation

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

4.cell representation.

 Zn(s) / 1M, Zn2+(aq) // 1M,Cu2+(aq) / Cu(s)  E0  = +1.10 V

5.cell  diagram

ii)Mg/Cu cell

1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal

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

2. Copper ions in solution gain the donated electrons to form copper atoms/metal

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

3.Overall redox equation

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

4.cell representation.

 Mg(s) / 1M, Mg2+(aq) // 1M,Cu2+(aq) / Cu(s)  E0  = +2.04 V 5.cell  diagram.

(iii)Fe/Cu cell

1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal

          Fe(s)           ->       Fe2+(aq)    +       2e

2. Copper ions in solution gain the donated electrons to form copper atoms/metal

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

3.Overall redox equation

Cu2+(aq)    +    Fe(s)      ->     Fe2+(aq)    +       Cu(s)

4.cell representation.

 Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s)  E0  = +0.78 V 5.cell  diagram.

(iv)Ag/Cu cell

1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal

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

2. Silver ions in solution gain the donated electrons to form silver atoms/metal

2Ag+(aq)    +       2e     ->      2Ag(s)

3.Overall redox equation

2Ag+(aq)    +    Cu(s)      ->     Cu2+(aq)   +       2Ag(s)

4.cell representation.

 Cu(s) / 1M, Cu2+(aq) // 1M,2Ag+(aq) / 2Ag(s)  E0  = +0.46 V

5.cell  diagram.

Standard electrode potential  (Eᶿ)

The standard electrode potential  (Eᶿ) is  obtained if the hydrogen half cell is used as reference. The standard electrode potential  (Eᶿ) consist of inert platinum electrode immersed/dipped in 1M solution of (sulphuric(VI) acid) H+ ions. Hydrogen gas is bubbled on the platinum electrodes at:

(i)a temperature of 25oC

(ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg

          (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell.    

          Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell.

                   ½ H2 (g)       ====        H+ (aq)     +  e

The half cell representation is:

                   Pt,½ H2 (g)           / H+ (aq), 1M

The standard electrode potential (Eᶿ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode.

If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (Eᶿ) values.

If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (Eᶿ) values.

 Table showing the standard electrode potential (Eᶿ) of some reactions

Reaction(Eᶿ) values in volts
F2 (g)+ 2e -> 2F(aq)+2.87
H2 O2 (aq)+ H+ (aq)   +2e -> H2 O (l)+1.77
Mn O4 (aq)+ 4H+ (aq)   +3e -> MnO2 (s) +H2 O (l)+1.70
2HClO (aq)+ 2H+ (aq)   +2e -> Cl2(aq) +2H2 O (l)+1.59
Mn O4 (aq)+ 4H+ (aq)   +5e -> Mn2+ (aq) +H2 O (l)+1.51
Cl2 (g)+ 2e -> 2Cl(aq)+1.36
Mn O2 (s)+ 4H+ (aq)   +2e -> Mn2+ (aq) +2H2 O (l)+1.23
Br2 (aq)+ 2e -> 2Br(aq)+1.09
NO3 (aq)+ 2H+ (aq)   +  e -> NO2(g) + H2 O (l)+0.80
Ag+ (aq)  +  e    ->  Ag(s)+0.80
Fe3+ (aq)  +  e    ->  Fe2+ (aq)+0.77
2H+ (aq)+ O2 (g)  -> H2 O2 (aq)+0.68
I2 (aq)+ 2e -> 2I(aq)+0.54
Cu2+ (aq)  +  2e    ->  Cu(s)+0.34
2H+ (aq)  +  2e    ->  H2(g)+0.00
Pb2+ (aq)  +  2e    ->  Pb(s)-0.13
Fe2+ (aq)  +  2e    ->  Fe(s)-0.44
Zn2+ (aq)  +  2e    ->  Zn(s)-0.77
Al3+ (aq)  +  3e    ->  Al(s)-1.66
Mg2+ (aq)  +  2e    ->  Mg(s)-2.37
Na+ (aq)  +  e    ->  Na(s)-2.71
K+ (aq)  +  e    ->  K(s)-2.92

Note:

(i)Eᶿ values generally show the possibility/feasibility of a reduction process/oxidizing strength.

(ii)The element/species in the half cell with the highest negative Eᶿ value easily gain / acquire electrons.

It is thus the strongest oxidizing agent and its reduction process is highly possible/feasible. The element/species in the half cell with the lowest positive Eᶿ value easily donate / lose electrons.

 It is thus the strongest reducing agent and its reduction process is the least possible/feasible.

(iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ.

If the overall redox reaction is not possible/ not feasible/ forced, it has a  negative (-) Eᶿ

Sample standard electrochemical cell

Calculation examples on Eᶿ 

 Calculate the Eᶿ value of a cell made of:

a)Zn and Cu

From the table above:

Cu2+ (aq) +  2e ->  Cu(s)  Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram)

Zn2+ (aq) +  2e ->Zn(s)    Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram)

Zn(s) ->Zn2+ (aq) +  2e    Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation)

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced

Substituting:

 Overall Eᶿ = +0.34 – (- 0.77) = +1.10V

Overall redox equation:

Cu2+ (aq) + Zn(s)   ->   Zn2+ (aq)   +  Cu(s)    Eᶿ  = +1.10V

Overall conventional cell representation:

 Zn(s) / Zn2+ (aq) 1M,  // 1M,Cu2+ (aq)  / Cu(s)    Eᶿ  = +1.10V

Overall conventional cell diagram:

Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution.

b)Mg and Cu

From the table above:

Cu2+ (aq) +  2e ->  Cu(s)  Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram)

Mg2+ (aq) +  2e ->Mg(s)    Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram)

Mg(s) ->Mg2+ (aq) +  2e    Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation)

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced

Substituting:

 Overall Eᶿ = +0.34 – (- 2.37) = +2.71V

Overall redox equation:

Cu2+ (aq) + Mg(s)   ->   Mg2+ (aq)   +  Cu(s)    Eᶿ  = +2.71V

Overall conventional cell representation:

 Mg(s) / Mg2+ (aq) 1M,  // 1M,Cu2+ (aq)  / Cu(s)    Eᶿ  = +2.71V

c)Ag and Pb

From the table above:

2Ag+ (aq) +  2e ->  2Ag(s)  Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram)

Pb2+ (aq) +  2e ->Pb(s)    Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram)

Pb(s) ->Pb2+ (aq) +  2e    Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation)

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced

Substituting:

 Overall Eᶿ = +0.80 – (- 0.13) = +0.93V

Overall redox equation:

2Ag+ (aq) + Pb(s)   ->   Pb2+ (aq)   +  2Ag(s)    Eᶿ  = +0.93V

Overall conventional cell representation:

 Pb(s) / Pb2+ (aq) 1M,  // 1M,2Ag+ (aq)  / Ag(s)    Eᶿ  =  +0.93V

d)Chlorine and Bromine

From the table above:

2e  +  Cl2(g) ->2Cl(aq)   Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram)

2e  +  Br2(aq) ->2Br(aq)   Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram)

2Br(aq) -> Br2(aq) +  2e    Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation)

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced

Substituting:

 Overall Eᶿ =  – 0.13 – (- 1.36) = +1.23V

Overall redox equation:

2Br(aq) + Cl2(g)   ->   2Cl(aq)   +  Br2(aq)    Eᶿ  = +1.23V

Overall conventional cell representation:

 Cl2(g) / 2Cl(aq) 1M,  // 1M, 2Br(aq)  / Br2(aq)    Eᶿ  =  +1.23V

Chlorine displaces bromine  from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value. 

e)Strongest oxidizing agent and the strongest reducing agent.

From the table above:

2e  +  F2(g) ->2F(aq)   Eᶿ = +2.87V(highest Eᶿ /strongest oxidizing agent)

2e  +  2K+ (aq) ->2K(aq)   Eᶿ = -2.92V(lowest Eᶿ/ strongest reducing agent)

2K(aq) -> 2K+ (aq) +  2e    Eᶿ = +2.92V (reverse lower Eᶿ to derive cell reaction / representation)

Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS – Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced

Substituting:

 Overall Eᶿ =  +2.87 – (-2.92) = +5.79V

Overall redox equation:

F2(g) + 2K(s)   ->   2F(aq)   +  2K+ (aq)    Eᶿ  = +5.79V

Overall conventional cell representation:

 2K(s) / 2K+ (aq),1M,  // 1M, 2F(aq)  / F2(g)    Eᶿ  = +5.79V

The redox reactions in an electrochemical/voltaic is commercially applied to make the:

          (a)Dry /primary/Laclanche cell.

          (b)Wet /secondary /accumulators.

(a)Dry/primary/Laclanche cell

Examine a used dry cell.

 Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other.

The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of;

                   -Ammonium chloride

-Zinc chloride

                   -powdered manganese (IV) oxide mixed with Carbon.

Zinc acts/serve as the negative terminal where it ionizes/dissociates:

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

Ammonium ions in ammonium chloride serve as the positive terminal where it is converted to ammonia gas and hydrogen gas.

          2NH4+(aq)  +   2e  -> 2NH3(g) + H2(g)

Ammonia forms a complex salt / compound /(Zn(NH3) 4)2+ (aq) / tetramminezinc(II) complex with the Zinc chloride in the paste.

Manganese (IV) oxide oxidizes the hydrogen produced at the electrodes to water preventing any bubbles from coating the carbon terminal which would reduce the efficiency of the cell.

Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile.

Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging.

b)Wet/Secondary/Accumulators

1. Wet/Secondary/Accumulators are rechargeable unlike dry /primary /Laclanche cells.Wet/Secondary/Accumulators are made up of:

          (i)Lead plate that forms the negative terminal

          (ii)Lead(IV) oxide that forms the positive terminal

2.The two electrodes are dipped in concentrated sulphuric(VI) acid of a relative density 1.2/1.3

3.At the negative terminal,lead ionizes /dissolves;

              Pb(s)     ->  Pb2+   +  2e 

4.At the positive terminal,

(i)Lead(IV) oxide reacts with the hydrogen ions in sulphuric(VI)acid to form Pb2+ (aq) ions;

          PbO2(s)    + 4H+(aq) + 2e ->  Pb2+ (aq)  + H2O(l)

          (ii) Pb2+ (aq) ions formed instantly react with sulphate (VI) ions/SO42- (aq) from sulphuric (VI)acid to form insoluble Lead(II) sulphate (VI).

Pb2+ (aq)   +  SO42- (aq) ->  PbSO4(s)

5.The overall cell reaction is called discharging

PbO2(s)  +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) Eᶿ = +2.0V

6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient.

As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases.

  1. During recharging, the electrode reaction is reversed as below:

2PbSO4(s) + 2H2O(l) ->PbO2(s)  +Pb(s) + 4H+(aq) + 2SO42- (aq)

          8. A car battery has six Lead-acid cells making a total of 12 volts.