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ELECTROCHEMISTRY - Chemistry Notes Form 4

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REACTION RATES AND REVERSIBLE REACTIONS - Chemistry Notes Form 4

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METALS: EXTRACTION PROPERTIES AND USES - Chemistry Notes Form 4

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Definition

  • Electrochemistry is the chemistry of electrochemical reactions; which deal with the relationship between electrical energy and chemical reactions.
  • Electrochemical reactions involve transfer of electrons and are essentially REDOX reactions.

Displacement and REDOX Reactions

Experiment 1:- Displacement Reactions among Metals

Procedure

  1. 5 cm3 of 1M CuSO4(aq) is put in a test-tube and its temperature recorded.
    - In the solution, a spatula end-full of iron fillings is added.
    - Any observations and temperature change are determined and recorded.
    - The procedure is repeated with fresh samples of CuSO4 with Zn, Mg, and Cu powders.
  2.  The procedure is repeated with 1M magnesium sulphate solution instead of CuSO4(aq).

Observations:

 Metal solid  Copper (II) sulphate  Magnesium sulphate
 Iron fillings  - A red brown solid (Cu) is formed
- The blue colour of the solution (Cu2+) fades then changes to green (Fe2+)		  - No reaction
 Zinc powder  - A red brown solid, copper metal is deposited.
- The blue colour of the solution (Cu2+ ) fades then turns colourless;		  - No observable reaction (change)
 Copper powder   - No reaction   - No reaction

Explanations

  • Reactions between metals and ions of another metal involve transfer of electrons from the metal to the other metal ion in solution.

Examples:

  • Fe(s) and CuSO4(aq)
    Copper being lower in the electrochemical series accepts electrons easier (than Fe) to form copper atoms (brown solid);
    Half equations
    Fe(s) → Fe2+(aq) + 2e-
    Cu2+(aq) + 2e- → Cu(s)
    Overall reaction
    Cu2+(aq) + 2e- + Fe(s) → Fe2+(aq) + 2e- + Cu(s)
    Then;
    Cu2+(aq)   +   Fe(s)   →    Fe2+(aq)   +    Cu(s)
    (Blue)                                   (Green)             (Brown solid)

Oxidation and Reduction in Terms of Electron Loss and Gain

  • The loss of electrons is oxidation and the species that gains electrons (causes electron loss); Cu2+ in this case is called oxidizing agent ; and is itself reduced.
  • Reduction: refers to gain of electrons, and the species that donates electrons (iron solid in this case) is called a reducing agent and is itself oxidized.
  • Displacement reactions generally involve reduction and oxidation simultaneously and are thus termed Redox Reactions.

    Further examples

    Zinc solid and CuSO4(aq)
    (i) Zn(s) → Zn2+(aq) + 2e- (Oxidation)
    (ii) Cu2+(aq) + 2e- → Cu(s) (Reduction)

    Then:         
    zinc and copper redox                 
    Magnesium solid and copper (II) sulphate
    (i). Mg(s) → Mg2+(aq) + 2e- (Oxidation)
    (ii). Cu2+(aq) + 2e- → Cu(s) (Reduction)

    Then:
    magnesium and copper redox
    Silver nitrate and copper solid.
    (i). Cu(s) → Cu2+(aq) + 2e- (Oxidation)
    (ii). 2Ag+(aq) + 2e- →  2Ag(s) (Reduction)
    Then:
    silver and copper redox
    Note:
  • Amount of heat evolved in these redox reactions depends on the position of the metal in the activity series relative to the metal ion in solution.
  • The closer the metals are in the activity series; the less readily displacement occurs and the lower the heat evolution during the displacement.
    E.g.: Heat evolved Mg//Cu2+ is higher than that evolved between Fe//Cu2+.

Conclusion:

  • Metals displace from solutions, those metals lower than themselves in the activity series.
    Note:
  • The more the reactive a metal is; the stronger a reducing agent it is and the weaker an oxidizing agent it is.

    Example:
  • Potassium is stronger reducing agent; but weaker oxidizing agent than silver, gold etc.

Summary: Strength of Reducing/Oxidizing Agent

strength of oxidising agents

Summary on Displacement Reactions


 Metal ion (in solution)		  Mg  Al  Zn  Fe  Pb  Cu
 K+  X  X  X  X  X  X
 Na+  X  X  X  X  X  X
 Ca2+  X  X  X  X  X  X
 Mg2+  X  X X  X  X  X
 Al3+  √  X  X  X  X  X
 Zn2+  √  √  X  X  X  X
 Fe2+  √  √  √  X  X  X
 Pb2+  √  √  √  √  X  X
 Cu2+ √  √  √  √  √  X
 Ag2+  √  √  √  √  √  √

Key:

  • A cross (x) indicates no reaction hence no redox reaction occurs.
  • A tick (√) indicates redox reaction occurs.

Oxidation Numbers/Oxidation State

  • Is the apparent charge that atoms have in molecules or ions.
  • For monoatomic ions, the oxidation number (state) is the magnitude and sign of charge;
    Example:
    Oxidation no of Aluminium in Al3+ is +3;

Importance of Oxidation Numbers:

  • Helps in keeping track of electron movement in redox reactions; hence determination of the reduced and oxidized species.

Oxidation and Reduction in Terms of Oxidation Numbers

  • Oxidation
    Is an increase in oxidation number.
  • Reduction
    Refers to a decrease in oxidation number

Rules in assigning oxidation numbers

  1. Oxidation number of an uncombined element is zero (0)
  2. The charge on a monoatomic ion is equivalent to the oxidation number of that element;
  3. The oxidation number of hydrogen in all compounds is +1 except in metal hydrides where its 1;
  4. The oxidation number of oxygen in all compounds is 2 except in peroxides where it is 1 and 0F2 where it is +2.
  5. In complex ions the overall charge is equal to the sum of the oxidation states of the constituent elements.
  6. In compounds, the sum of oxidation numbers of all constituent atoms is equal to zero.

Worked examples

  1. Calculate the oxidation number of nitrogen in:
    1. NO3-
      Solution:
      N + (-2 x 3) = -1
      N = -1 + 6
      = +5
      Note: thus nitric acid with a nitrate ion (NO3- ) is called nitric (V) acid since the oxidation number of nitrogen in it is +5;
    2. NO2;
      Solution:
      N + (-2 x 2) = 0
      N = 0 + 4
      = +4
      Note: Thus the gas NO2 is referred to as nitrogen (IV) oxide because the oxidation umber of nitrogen in it is +4
    3. NO2-;
      Solution:
      N + (-2 x 2) = -1
      N = -1 + 4
      = +3
      Note: thus nitrous acid containing nitrite ion is called nitrous (III) acid since the oxidation number of nitrogen in it is +3.
    4. AgNO3;
      Solution:
      1 + N + (-2 x 3) = 0
      1 + N + (-6) = 0
      N = 0 - 1 + 6;
      N = +5
  2. Determine the oxidation number of manganese in each of the following, and hence give the systematic names of the compounds.
    1. MnSO4
      Solution:
      Mn + 6 + (-2 x 4) = 0
      Mn = 0 - 6 + 8;
      Mn = +2
      Systematic name: Manganese (II) sulphate;
    2.  Mn2O3;
      Solution:
      2Mn + (3 x -2) = 0;
      2Mn = 0 + 6
      Mn = ½ x 6;
      Mn = +3;
      Systematic name: Manganese (III) oxide;
    3. KMnO4
      Solution:
      1 + Mn + (-2 x 4) = 0
      Mn = 0 - 1 + 8;
      Mn = +7;
      Systematic name: Potassium manganate (VII) oxide
    4. MnO3-;
      Solution:
      Mn + (-2 x 3) = -1
      Mn = -1 + 6;
      Mn = +5
      Systematic name: Manganese (V) ion;




Experiment: Redox Reactions Involving Halide Ions and Halogens

Procedure:

  • 2 cm3 of chlorine gas are bubbled into each of the following solutions: - KI, KCl, KBr, and KF.
  • The observations are made and recorded.
  • The procedure is repeated using fluorine, bromine and iodine in place of chlorine.

Precaution:

  • Chlorine and bromine are poisonous.

Observations

 
Halogen		  Potassium fluoride  Potassium chloride  Potassium Bromide  Potassium Iodide
 Fluorine (F2)  No visible, change  Green yellow gas is evolved  Colourless solution changes to red brown  Colourless solution turns black
 Chlorine (Cl2)  No visible colour change  No visible colour change  Colourless solution turns to red brown  Colourless solution turns black
 Bromine (Br2)  No visible colour change  No visible colour change  No visible colour change  Colourless solution turns black
 Iodine (I2)  No visible colour change  No visible colour change  No visible colour change  Colourless solution turns black

Note: Colours of halogens in tetra chloromethane:-

 Halogen  Colour in tetrachloromethane 
 Fluorine  
 Chlorine  Yellow
 Bromine  Red - brown
 Iodine  Purple

Explanations

  • Fluorine displaces all the other halogens; Cl2 , Br2 and I2 because it has a greater tendency to accept electrons than all the rest.
  • Chlorine displaces both Bromine and Iodine from their halide solutions
  • Cl2 takes electrons from the bromide and iodide ions i.e. oxidizes them, to form bromine and iodine respectively.
    Equations:
    1. Chlorine and potassium bromide:
      Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(l)
      Ionically:
      Cl2(g)           +     2Br-(aq)  →   2Cl-(aq)    +    Br2(l)
      Green-yellow                                                Red brown
      Redox equation:
      chlorine and potassium bromine redox
    2. Chlorine and Potasium iodide:
      Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(l)
      Ionically:
      Cl2(g)           +     2I-(aq)  →   2Cl-(aq)     +      I2(l)
      Green-yellow                                                             Black
      Half cell reactions:
      Oxidation: 2I-(aq) →  I2(aq) + 2e-
      Reduction: Cl2(g) + 2e- → 2Cl-(aq)
      Redox equation:
      chlorine and potassium iodide redox
  • Bromine takes electrons form iodide ions but not from fluorine and chlorine.
  • Iodine is formed i.e. due to oxidation of iodide ions by the Bromine.
    Equations
    Bromine and Potasium iodide:
    Br2(g) + 2KI(aq) → 2KBr(aq) + I2(l)
    Ionically:
    Br2(g)        +      2I-(aq) → 2Br-(aq) +  I2(l)
    Red brown                                                Black
    Half cell reactions:
    Oxidation: 2I-(aq) → I2(aq) + 2e-
    Reduction: Br2(g) + 2e- → 2Br-(aq)
    Redox equation: 
    bromine and potassium iodide
  • Note: oxidation number of chlorine decreases from 0 to -1 hence reduction ; while oxidation number of iodine increases from -1 to 0; hence oxidation;

Conclusion:

  • The stronger the tendency of an element to accept electrons, the stronger is its oxidizing power.
  • Fluorine is the strongest oxidizing agent of the 4 halogens considered.

Summary: Order of Oxidizing Power for Halogens

order of oxidizing power of halogens

Further Examples of Redox Reactions

  1. Action of acid on metals
    Mg(s) + 2HCl(aq) → 2MgCl(aq) + H2(g)
    Ionically:
    Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
    Half cell reactions:
    Oxidation: Mg(s) → Mg2+(aq) + 2e-
    Reduction: 2H+(g) + 2e- → H2(g)
    Redox equation: 
    action of acid on metals redox
    Note: oxidation number of hydrogen decreases from 1 to 0 hence reduction ; while oxidation number of magnesium increases from 0 to 2; hence oxidation ;
  2. Reaction of active metals with water
    Example
    2Na(s) + 2H2O(l)  → 2NaOH(aq) + H2(g)
    Ionically:
    2Na(s) + 2H2O(l) → 2Na+(aq) + 2OH-(aq) + H2(g)
    Half cell reactions:
    Oxidation: 2Na(s) → 2Na+(aq) + 2e-
    Reduction: 2H2O(g) + 2e- → 2OH-(aq) + H2(g)
    Redox equation: 
    reaction of active metals with water redox
    Note: oxidation number of water decreases from 0 to -1(total) hence reduction ; while oxidation number of sodium increases from 0 to 1; hence oxidation ;
  3. Reaction of heated Iron with dry chlorine
    2Fe(s) + 3Cl2(g) → 2FeCl3(s)
    Ionically (assumed):
    2Fe(s) + 3Cl2(aq) → 2Fe3+(aq) + 6Cl-(g)
    Half cell reactions:
    Oxidation: 2Fe(s) → 2Fe3+(aq) + 6e-
    Reduction: 3Cl2(g) + 6e- →  H2(g)
    Redox equation:
    reaction of heated iron with chlorine redox
    Note: oxidation number of chlorine decreases from 0 to -1 hence reduction; while oxidation number of iron increases from 0 to 3; hence oxidation;
  4. Reaction between Bromine and Iron (II) ions
    Ionically:
    Br2(l) + 2Fe2+(aq) → 2Fe3+(aq) + 2Br-(aq)
    Half cell reactions:
    Oxidation: Br2(l) → 2Br-(aq) + 2e-
    Reduction: 2Fe2+(g) + 2e- → 2Fe3+(aq)
    Redox equation:
    reaction of bromine and iron II ions redox
    Note: oxidation number of bromine decreases from 0 to -1 hence reduction ; while oxidation number of Fe2+ increases from 2 to 3 (in Fe3+); hence oxidation ;
  5. Oxidation by potassium Manganate (VII) (KMnO4 )
    Procedure:
    • Purple Potassium manganate (VII) is added into a solution containing iron (II) ions in a test tube.
    • A few drops of concentrated sulphuric (VI) acid are added.

    Observations
    • The purple solution (containing Manganate (VII) ions) turns to colourless (manganate (II) ions) i.e. the purple solution is decolourised;

    Explanation
    • The Manganate (VII) ions which give the solution a purple colour are reduced to Manganese (II) ions which appear colourless. This is a redox reaction.

    Equations
    Ionically:
    MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
    Purple                                                       Colourless
    Half cell reactions:
    Oxidation : 5Fe2+(s) → 5Fe3+(aq) + 5e- ; (since number of electrons is 5)
    Reduction: MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
    Redox equation
    oxidation by potassium permangante redox reaction
    Note:
    • oxidation number of Manganate ions in KMnO 4 decreases from 7 to 2 (in Mn2+) hence reduction ; while oxidation number of iron increases from 2 (in Fe2+) to 3 (in Fe3+); hence oxidation
    • The presence of Fe 3+ at the end of the reaction can be detected by adding sodium hydroxide solution to form a red brown precipitate of Fe(OH)3 ;
  6. Action of potassium dichromate (VI) on iron (II) ions (Fe2+):
    Procedure:
    • A solution containing iron (II) ions is added into a solution of oxidized potassium dichromate (VI).

    Observations
    • The orange solution of potassium dichromate turns green.

    Explanations
    • The iron (II) ions are oxidized to iron (III) ions
    • The chromium (VI) ions (orange) are reduced to chromium (III) ions
    • This is thus a REDOX reaction.

    Equations
    Ionically:
    Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
    Orange                                                          Green
    Half cell reactions:
    Oxidation:     6Fe2+(s) → 6Fe3+(aq) + 6e-; (since number of electrons is 6)
    Reduction: Cr2O72-(aq) + 14H+(aq) + 5e- → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
    Redox equation: 
    action of potassium dichromate and Iron II redox
    Note:
    • The oxidation number of dichromate ions in K2Cr2O7 decreases from 6 to 3 (in Cr3+) hence reduction; while oxidation number of iron increases from 2 (in Fe2+) to 3 (in Fe3+); hence oxidation ;
    • The presence of Fe3+ at the end of the reaction can be detected by adding sodium hydroxide solution to form a red brown precipitate of Fe(OH) 3 ;
  7. Action of acidified potassium permanganate on Hydrogen Peroxide.
    • The overall reaction is a Redox reaction
      Redox equation
      2MnO4-(aq) + 5H2O(l) + 6H+(aq) → 2Mn2+(aq) + 5H2O(l) + 5O2(g)
      Purple                                                     Colourless
      Half cell reactions:
      Oxidation: 5H2O2(aq) → 10H+(aq) + 5O2(g) + 10e-;
      Reduction: 2MnO4-(aq) + 16H+(aq) + 10e- → 2Mn2+(aq) + 8H2O(l)
      Redox equation: 
      action of acidified potassium permanaganate on hydrogen peroxide
      Note :
      • The oxidation number of manganate ions in KMnO4 decreases from 7 to 2 (in Mn2+) hence reduction ; while oxidation number of hydrogen increases from -1 (in H2O2) to 1 (in H+); hence oxidation ;
      • Thus the acidified potassium manganate (VII) oxidizes hydrogen peroxide to water and hydrogen;
  8. H2O2 oxidizes Iron (II) salts to Iron (III) salts in acidic medium
    Oxidation:
    2Fe2+(aq) →2Fe3+(aq) + 2e-
    Reduction:
    2H+(aq) + H2O2(aq) + 2e- → 2H2O(l)
    Overall redox:
    2Fe2+(aq) + H2O2(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l)

Terms Used in Describing Oxidation-reduction

 Term  Electron change  Oxidation number change
 Oxidation
 Reduction
 Oxidising agent
 Reducing agent
 Substances oxidized
 Substances reduced		  Loss of electrons
 Gain of electrons
 Receives electons
 Loses electrons
 Loses electrons
 Gains electrons		  Increases
 Decreases
 Decreases
 Increases
 Increases
 Decreases

The Tendency of Metals to Form Ions

  • When a metal is placed in an aqueous solution of its ions, some of the metal dissolves;
    Equation:
    M(s) → Mn+(aq) + ne-
  • Dissolution of the metal causes electron build up on its surface; making it negatively charged, while the surrounding solution becomes positively charged.
    dissolution of metal and electron build up
    Diagram: Dissolution of metal and electron build up.

  • The positive charge of the solution increases and some of the cations start recombining with the electrons on the metal surface to form atoms.
    Equation:
    Mn+(aq) + ne- → M(s)
  • Consequently, an electric potential difference is created between the metal rod and the positively charged ions in solution.
  • This arrangement of a metal rod (electrodes) dipped in a solution of its ions constitutes a half cell.

    Note:
  • The tendency of a metal to ionize when in contact with the ions differs form one metal to another.
  • This difference can be measured by connecting two different Half cells to make a full cell .
  • The electrodes of the 2 half cells are connected by a metallic conductor; while the electrolytes (solutions) of the half cells are connected through a salt bridge .
    full cell
    Diagram: Connection of two half cells to a full cell.

Experiment: - To Measure the Relative Tendency of Metals to Ionize

Procedure:

  • A Zinc rod is placed into 50 cm3 of 1M zinc sulphate in a beaker
  • Into another beaker containing 50 cm3 of 1M CuSO4(aq), a copper rod is dipped
  • The two solutions are connected using a salt bridge.
  • The two metal rods are connected through a connecting wire connected to a voltmeter
  • The experiment is repeated using the following half-cells instead of the Zinc-half cells:-
    • Mg rod dipped in 1M MgSO4(aq)
    • Lead dipped in 1M lead Nitrate
    • Copper dipped in 1M CuSO4(aq)

Apparatus

apparatus to measure relative tendency of metals to ionize

Observation

  • The zinc rod in the zinc-zinc ions half cell dissolves;
  • The blue colour of the copper (II) Sulphate solution fades/ decrease;
  • Red-brown deposits of copper appear on the copper rod in the copper-copper ions half-cell.
  • A voltage of 1.10 V is registered in the voltmeter.

Equations/reactions, at each half cell

  • In zinc-zinc ions half cell
    Zn(s) →  Zn2+(aq) + 2e-
  • In the copper-copper ion half cell
    Cu2+(aq) + 2e- → Cu(s);

Explanations

  • Zinc rod has a higher tendency to ionize than the copper rod, when the metal rods are placed in solutions of their ions.
  • Thus the zinc rod has a higher accumulation of electrons than the copper rod.
  • This makes it more negative compared to the relatively more positive copper rod, which has a lower accumulation of electrons.
  • On connecting the 2 half cells; electrons will flow form the zinc rod to the copper rod through the external wire.
  • The copper rod gains the electrons lost by the Zinc rod.

    Roles of the slat bridge
    • Note: It forms a link between the 2 half cells, thereby completing the circuit
    • It completes the circuit by:
      1. Allowing its ions to carry charge from one half - cell to the other.
      2. Maintaining the balance of charge in the two half-cells; by providing the ions which replace those used up at the electrodes.
  • The overall reaction in the cells is a Redox reaction
    Half cell reactions:
    Zn(s) → Zn2+(aq) + 2e- (oxidation)
    Cu2+(aq) + 2e- → Cu(s); (reduction)
    Overall redox equation
    zinc and copper redox
  • The voltage of 1.10 V registered in the voltmeter is a measure of the difference between the electrode potential (Eθ ) of Zinc and Copper electrodes, i.e. the potential difference/the Electromotive force.
  • Thus: An electrochemical/ voltaic cell;
    •  Is the combination of two half – cells to give a full cell capable of generating an electric current from a redox reaction.

Cell Diagram for a Voltaic Cell

Rules/Conventions for Cell Representation

  1. A vertical continuous line (/); represents the metal-metal ion or metal ion-metal interphase.
  2. Vertical broken line (|); between the 2 half-cells; represents the salt bridge.
    Note : The salt bridge may also be represented by two unbroken parallel lines (//).

    Example: - cell diagram for the above cell;
    cell diagram with broken line as salt bridge
    Alternatively:
    Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s);

Electrode potential Eθ , values of other metal – metal ions relative to the Cu/Cu2+ half cell

 Metal/metal ion half cell  Electrode potential Eθ relative to Cu2+/Cu half cell
 Mg(s)/Mg2+(aq)  +2.04
 Zn(s)/Zn2(aq)  +1.10
 Pb(s)/Pb2+(aq)  +0.78
 Cu(s)/Cu2+(aq)  +0.00
 Ag(s)/Ag+(aq)  -0.46

Positive and negative E values

  1. Positive E values -
    • If the Eθ value for a metal/metal ion is positive then the metal undergoes oxidation (loses electrons) while the reference electrode undergoes reduction (accepts electrons)
      Example :
    • The Eθ value for Zn(s)/Zn2+(aq) relative Cu2+(aq)/Cu(s) is positive because the zinc metal is oxidized to zinc ions while the copper ions are reduced to copper metal.
  2. Negative E values :
    • Implies that the reference half cell undergoes oxidation (donates electrons) while the other metal ions in the other half cell undergoes reduction (accepts electrons)
      Example :
    • The E value for Ag(s)/Ag+(aq) is negative because Cu is more reactive than silver and gives out electrons (oxidation); while the less reactive Ag has its ions accepting electrons (reduction) to form Ag solid.
  3. The 0 (Zero) E value:
    • Always indicate the reference electrode / half-cell; in which case there would be no potential difference (with itself)
      Example:
    • The 2 half cells of Cu(s)/Cu2+(aq) or Cu2+(aq)/Cu(s) have no potential difference in between them hence a zero (0) E value.
      Note:
      1. Any other element could be chosen as the reference electrode in place of copper Cu and difference electrode potentials values would be obtained for the same elements.
      2. The electrode potential of a single element is usually determined by measuring the difference between the electrode potential of the element and a chosen standard electrode.
      3. This gives the standard electrode potential (E0 ) of the element.

The standard Electrode Potential (E0 )

Definition

  • Is the potential difference for a cell comprising of a particular element in contact with 1 molar solution of its ions and the standard hydrogen electrode.
  • It is denoted with the symbol E0 .

Importance

  • It is useful in comparing the oxidizing and reducing powers of various substances.

The standard Hydrogen Electrode

  • Is the hydrogen half-cell, which has been conventionally chosen as the standard reference electrode.
  • It has an electrode potential of zero at:-
    • A temperature of 25oC
    • A hydrogen pressure of 1 atmosphere
    • A concentration of 1M hydrogen ions
  • Note: The ions in the other half-cell must also be at a concentration of 1 molar.

Components of the Hydrogen Half Cell

  • Consist of an inert platinum electrode immersed in a 1M solution of Hydrogen ions
  • Hydrogen gas at 1 atmosphere is bubbled into the platinum electrode.
  • The hydrogen is adsorbed into the platinum surface and an equilibrium (state of balance) is established between the adsorbed layer of molecular hydrogen and hydrogen ions in the solution.
    Equation:
    ½H2(g) → H+(aq) + e-
    Platinised platinum
    • Is platinum loosely coated with finely-divided platinum.
    • This enables it to retain comparatively large quantity of hydrogen due to its porous state.
    • Platinised platinum also serves as a route by which electrons leave or enter the electrode.
  • The hydrogen electrode is represented as: H2(g)/H+(aq); 1M
    standard hydrogen electrode
    Diagram: The standard hydrogen electrode:
  • The electrode potential of any metal is taken as the difference in potential between the metal electrode and the standard hydrogen electrode.

Negative and positive electrode potentials

  1. Negative electrode potential
    • If the metal electrode has a higher/greater tendency to loose electrons than the hydrogen electrode; then the electrode is negative with respect to hydrogen electrode; and its electrode potential is negative.
      Examples: Zinc, Magnesium etc.
  2. Positive electrode potential
    • If the tendency of an electrode to loose electrode is lower than the hydrogen electrode, then the electrode is positive with respect to the hydrogen electrode; and its potential is positive.
      Examples: copper, silver etc

Reduction Potentials

  • Is a standard electrode potential measured when the electrode in question is gaining electrons.
  • The lower the tendency of an electrode to accept/gain electrons; the lower (more negative) the reduction potential and vise versa.
    Examples:
    K+(aq) + e-  → K(s);  E0 = -2.92V
    F2(g) + e- → 2F-(aq) ; E0 = +12.87V

    Mg(s) + 2e- → Mg2+(aq); Eθ = -2.71 V
  • Thus potassium ions with E0 = -2.92V have a lesser tendency to gain electrons than magnesium ions.
  • Thus Potassium is the weakest oxidizing agent; but the strongest reducing agent, since it has the greatest tendency to donate electrons.
    Note:
  • Oxidation potentials will be the potentials of electrodes measured when they are losing electrons hence undergoing oxidation.

Standard electrode potentials (reduction potentials) of some elements

   Reduction equation  Eθ volts

 

 

↑Increasing strength of oxidising agent 

↓Decreasing strength of oxidising agent

  F2(g) + 2e- → 2F-(aq)  +2.87
 Cl2(g) + 2e- → 2Cl-(aq)  +2.87
 Br2(g) + 2e- → 2Br-(aq)  +1.36
 Ag+(aq) + 2e- → Ag(s)  +0.80
 I2(g) + 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+(g) + 2e- → Zn(s)  -0.76
 Al3+(aq) + 3e- → Al(s)  -1.66
 Mg2+(aq) + 2e- → Mg(s)  -2.71

Note:

  • The standard electrode potentials in the above table are reduction potentials.
  • The greater the tendency to undergo reduction, the higher (more positive) the Eθ value.
  • The reverse reaction (oxidation ) would have a potential value equal in magnitude but opposite in sign to the reduction potential.
    Example :
    Zinc
    Reduction potential
    Zn2+(aq) + 2e- → Zn(s); E0 = -0.76V;
    Oxidation potential
    Zn(s) → Zn2+(aq) + 2e-; Eθ = + 0.76V;

Uses of E0 values

  1. Comparing the reducing powers and oxidizing powers of various substances;
  2. Predicting whether or NOT a stated REDOX reaction will take place.

The E0 value for a REDOX reaction

  • Is usually calculated as the sum of the E 0 value for the half cells involved.
    Note:
  • If the sum is positive then the reaction can occur simultaneously;
  • If the value of the sum is negative the reaction cannot occur;

Sample calculations

  1.  Cu2+ can oxidize Zinc but Zn2+ cannot oxidize Cu;
    1. Cu2+(aq)/Cu(s)//Zn(s)/Zn2+(aq)
      Cu2+(aq) + 2e- → Cu(s); E0 = + 0.34 V
      Zn(s) → 2e- + Zn2+(aq); E0 = + 0.76 V
      Cu 2+(aq) + Zn(s) → Cu(s) + Zn(aq) ; E0 = - 1.10 V
      The overall reaction is positive, hence zinc can be oxidized by copper (II) ions, and hence reaction occurs;
    2. Zn2+(aq)/Zn(s)//Cu(s)/Cu2+(aq)
      Zn2+(aq) + 2e-→ Zn(s); E0 = - 0.76 V
      Cu(s) → 2e- + Cu2+(aq); E0 = - 0.34 V
      Zn2+(aq) + Cu(s) → Zn(s) + Cu2+(aq) ; E0 = - 1.10 V
      The overall E0 is negative; thus Zn2+ cannot oxidize Cu to Cu2+ (Cu cannot reduce Zn2+ to Zn); and hence the reaction cannot occur.
  2. Can chlorine displace bromine form bromide solution?
    Cell diagram : Cl2(g)/Cl-(aq)//2Br-(aq)/Br2(g)
    Cl2(g) + 2e- → 2Cl-(aq); E0 = + 1.36 V
    2Br-(aq) → 2e- + Br2(g) ; E0 = - 1.09 V
    Cl2(g) + 2Br-(aq) → 2Cl-(aq) + Br2(g) ; E0 = +0.27 V
    The overall E0 for the reaction is positive, so chlorine can displace bromine from a bromide solution.

Worked examples

  1. The diagram below represents part of the apparatus to be used for the determination of the standard electrode potential of Aluminum, Eθ Al3+(aq)/Al(s)
    1. Name the solutions which could be placed in beakers A and B; specifying their concentrations.
      Answer:
      • In beaker A; 1M HCl(aq) i.e. any solution with 1M hydrogen ions
      • In beaker B; 1M Al (NO3)(aq); i.e. any aqueous solution with 1M Al3+.
    2. One essential part of the cell has been omitted. Name the missing part and give its functions.
      Answer:-
      Missing part : salt bridge
      Function: completes the circuit by;
      1. Allowing its ions to carry charge form one half cell to another.
      2.  Providing ions which repose those used up of the electrodes, hence maintaining a balance of charge in the 2 half cells.
    3. The voltmeter reading was found to be 1.66 V.
      1. Give the standard electrode potential for the aluminum electrode.
        Solution: it is -1.66 since the Hydrogen-hydrogen ions half-cell = 0.00V.
      2. Show the direction of flow of electrons in the circuit;
        Solution: From Al3+/Al half cell to the H2/2H+(aq) half cell
    4. Give the half-cell equations and the overall cell equation
      Solution:
      3H+(aq) + 3e- → 1½H2(g); E0 = + 0.00V;
      Al(s) → 3e- + Al3+(aq); E0 = + 1.66V;
      Overall:
      Al(s) + 3H+(aq) →Al3+(aq) + 1½H2(g); E0 = +1.66 V
  2. You are given the following half-equations;
    Mg2+(aq) + 2e- →Mg(s); Eθ = - 2.71 V
    Zn2+(g) + 2e-→ Zn(s) ; Eθ = - 0.76 V
    1.  Obtain an equation for the cell reaction.
      Mg(s) → 2e- + Mg2+(aq); E0 = +2.71V
      Zn2+(aq) + 2e- → Zn(s) ; E0 = - 0.76V
      Mg(s) + Zn2+(aq) →Mg2+(aq) + Zn(s); E0 = +1.95V
      Thus equation:
      Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s); E0 = +1.95V
    2. Calculate the E0 value for the cell.
      Mg(s) → 2e- + Mg2+(aq); E0 = +2.71V
      Zn2+(aq) + 2e- → Zn(s); E0 = - 0.76V
      Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s); E0 = +1.95V
    3. Give the oxidizing species and the reducing species
      - Reducing species
      Magnesium i.e. The species that undergoes oxidation since its oxidation number increases (from 0 to 2); as it reduces the other;
      - Oxidizing species
      Zinc/Zinc ions; - the species that undergoes reduction; since its oxidation number decreases (from 2 to 0) as it oxidizes the other species (Mg).
  3. Given the following half-equations
    I2(g) + 2e- → 2I-(aq); Eθ = + 0.54V
    Br2(g) + 2e- → 2Br-(s); Eθ = +1.09 V
    1. Obtain an equation for the all reaction
      2I-(aq) → 2e- + I2(g); E0 = -0.54V
      Br2(g) + 2e- → 2Br-(s); E0 = +1.09V
      Br2(g) + 2I-(aq) → 2Br-(aq) + I2(g); E0 = +0.55V
    2. Calculate the E0 value for the cell
      2I-(aq) → 2e- + I2(g); E0 = - 0.54V
      Br2(g) + 2e- → 2Br-(s); E0 = +1.09V
      Br2(g) + 2I-(aq) → 2Br-(aq) + I2(g); E0 = +0.55V
    3. Give the oxidizing species;
      Oxidizing species : Bromine; Br2(aq)
  4. Consider the following list of electrodes and electrode potential values.
     Electron reaction  Eθ volts
     A2+/A   +0.34
     B2+/B  -0.71
     C2+/C  -0.76
     D+/D  +0.80
     E2+/E  -2.87
     F2+/F  -2.92
    1. Which of the ions is the strongest oxidizer?
      D+; because it is most readily reduced/have the highest tendency to accept electrons as evidenced by its highest positive Eθ value when the ions change to element (D+/D-)
    2. Which of the ions is the strongest reducer?
      - Is least readily reduced hence lowest E0 value (-2.92 V); accepts electrons least readily i.e. shows the lowest E0 when its ions gain electrons/ are reduced (F+/F, = -2.92 V)
  5.  The following is a list of reduction standard electrode potentials.
     Metal  Eθ volts (standard electrode potential)
     Magnesium  -2.36
     Zinc  -0.76
     Iron  -0.44
     Hydrogen  0.00
     Copper  +0.34
     Silver  +0.79

    1. Which two metals, if used together in a cell would produce the largest e.m.f?
      Magnesium-silver cell;
      Mg2+(aq) + 2e- → Mg(s); Eθ = - 2.36 V
      2Ag+(g) + 2e- → 2Ag(s); Eθ = - 0.79 V
      Thus;
      Mg(s) → 2e- + Mg2+(aq); E0 = +2.36V
      2Ag+(aq) + 2e- → 2Ag(s); E0 = - 0.79V
      Mg(s) + 2Ag+(aq) → Mg2+(aq) + 2Ag(s); E0 = +3.15V
    2. What would be the voltage produced by:-
      1. Zinc-copper cell
        Cu2+(aq) + 2e- → Cu(s); Eθ = + 0.34 V
        Zn2+(g) + 2e- → Zn(s) ; E θ = - 0.76 V
        Thus; Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s);
        Zn(s) → 2e- + Zn2+(aq); E0 = + 0.76V
        Cu2+(aq) + 2e- → Cu(s); E0 = + 0.34V
        Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s); E0 = +1.10V
      2. Copper-silver cell
        Cu2+(aq) + 2e- → Cu(s); Eθ = + 0.34 V
        Ag+(g) + 2e- → Ag(s); Eθ = + 0.79 V
        Thus; Cu(s)/Cu2+(aq)//2Ag+(aq)/2Ag(s);
        Cu(s) → 2e- + Cu2+(aq); E0 = - 0.34V;
        2Ag+(aq) + 2e- → 2Ag(s); E0 = + 0.79V;
        Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s); E0 = + 0.45V
    3. Explain the meaning of the positive and negative signs;
      Positive signs
      - The metal in question has a lower tendency to loose electrons than hydrogen hence more relatively positive to hydrogen;
      - They are stronger oxidizing agents but weaker reducing agents but weaker reducing agents than hydrogen;
      Negative signs
      - The particular metal has a higher tendency to loose electrons than hydrogen; hence relatively more negative than hydrogen.
      - They are weaker oxidizing agents but stronger reducing agents than Hydrogen.
  6. The following are some half-cell electrode potentials of some elements.
    Reaction
    1. Select two half cells which when oxidized give the burst E value; and fill the cell representation.
      Solution: The silver copper cell; i.e. Cell representation
    2. Calculate the E0 value
    3. Give the strongest reducing agent and strongest oxidizing agent.
      Strongest reducing agent
      Strongest oxidizer - silver - has highest reduction potential
  7. Study the table below and answer the questions that follow.
    1. Which two metals would form a metallic couple with the highest EMS.
    2. Calculate the e.m.f. of the cell that would be produced by (i) above.
    3. Write down the cell representation for the cell above.
    4. Which metal is the strongest reducing agent in the above list
      Metal A - have the lowest reduction potential.
  8.  
    1. The table below gives reduction potentials obtained when the half-cells for each of the metals represented by letters J, K, L, M and N where connected to a copper half- cell as the reference electrode.
      1. What is metal L likely to be? Give a reason
        Copper It has an E0 value/ reduction potential of 0.00, with copper as the reference electrode
      2. Which of the metals cannot be displaced from the solution of its salt by any other metal in the table? Give a reason.
        Metal J:Has the lowest reduction potential; meaning it least readily accepts electrons (most readily donates electrons) than any other metal.
      3. Metal K and M were connected to form a cell as shown in the diagram below;
        1.  Indicate on the diagram, the direction of flow of electrons. Explain. 
          from K to M.
          K is a stronger reducing agent than M, as evidenced by its lower reduction potential.
          It thus loses electrons faster becoming more Negative than M; hence electrons move from K through external wires to M.
        2. Write the equations for the half-cell reactions that occur at:-
          Metal K electrode:-
          Metal M electrode
        3. If the slat bridge is filled with saturated sodium Nitrate solution, explain how it helps to complete the circuit.
          Answer
          It allows its ions (Na+(aq) , and NO3-(aq)) to carry charge from one half cell to another. Providing ions which replace those used up at the electrodes.


Voltaic Cells

  • Are also called electrochemical cells.
  • Are cells in which electrical energy is generated from chemical reactions.

Types of Electrochemical Cells

  1. Primary cells -
    • Electrochemical cells which are not rechargeable
  2. secondary cells-
    • Voltaic/electrochemical cells which are rechargeable.

Primary cells/dry cells

  • Are of various types and an example is the Le clanche dry cell.
    leclanche dry cell
    Diagram: The Le clanche dry cell

    Structure
  • Consist of a Zinc can with carbon rod at the centre.
  • The central graphite/carbon rod is surrounded by powdered Manganese (IV) oxide and carbon; which are inturn surrounded by a paste of NH4Cl (s) and Zinc Chloride.
  • The protruding portion of the carbon rod is covered with a brass cap; and the zinc can covered with a sealing material.
    Chemical reaction
  • The Zinc can is the negative terminal; while the carbon/graphite rod is the positive terminal;
    1. a the Zinc can/negative terminal
    2. positive terminal /brass cap
      (NH3(g) + 2e- → 2NH3(g) + H2(g)
  • Note:These gases (NH3(aq) and H2(g)) are NOT used immediately but are used in more complex reactions.
  • The ammonia gas forms a complex with the zinc chloride in the paste.
  • The hydrogen gas is oxidised to water by the Manganese (IV) oxide.

    Functions of the various components.
    1. Brass cap -
      • Functions as the positive terminal where the reduction reaction occurs.
    2. Zinc can
      • Is the Negative terminal; where the oxidation reaction occurs.
    3. Carbon rod
      • It serves as the positive electrode.
      • Acts as the connecting wire between the positive and negative terminal through it electron flow form the Zinc can to the brass cap.
    4. Manganese (IV) Oxide
      • To oxidise the Hydrogen gas produced at the anode/positive terminal/brass cap to water.
      • A single dry cell can produce a potential of 1.5 V
  • Note:Dry ammonium chloride does not conduct electric current. This explains why a paste, which is a conductor, is used.
  • The dry cells cannot provide a continuous supply of electricity for an undulate period of time.
    Reason -The reactants (electrolytes) are used up and cannot be replaced.

Secondary cells

  • Are voltaic/electrochemical cells that are rechargeable.
  • A common example is the lead acid accumulator.
    lead acid accumulator
    Diagram: The lead acid accumulator.

    Structure:
  • The positive plate is a lead grill filled with lead (IV) oxide; while the negative plate consists of a similar lead grill filled with spongy lead.
  • The grills are immersed in sulphuric acid; which serves as the electrolyte.
    Reaction
    1. During discharge/when in use
      • At the negative terminal/lead - The lead dissolves forming lead (II) ions 
      • At the positive terminal (lead (IV) oxide; lead (IV) oxide reacts with the Hydrogen ions (H+) in sulphuric acid; also forming lead (II) ions.
      • Then; the lead (II) ions formed at both electrodes react instantly with the Sulphate ions to form lead (II) Sulphate.
      • The insoluble lead Sulphate adheres to the electrodes.
        Overall reaction
        discharging reaction in a lead acid accumulator
      • Note: The Lead Sulphate should NOT be left for too long to accumulate on the electrodes
        Reason: The fine PbSO4(s) will charge to coarse non reversible and inactive form and the accumulator will become less efficient.
      • During use/discharge; the lead and the lead (IV) oxide are depleted, and the concentration of sulphric acid declines.
    2. During recharging
      • Is usually done by applying a suitable voltage to the terminals of the cell.
      • At the negative terminal - the lead ions accept electrons to form lead solid.
        Overall reaction
        charging reaction in a lead acid accumulator
      • This process restores the original reactants.

Electrolysis

Definition of Electrolysis

  • Is the decomposition of molten or aqueous solutions by passage of electric current through it.

Terminologies Used in Electrolysis

  1. Electrolyte
    • Is a solution which allows electric current to pass through while it gets decomposed.
    • Electric current transfer in electrolyte occur through ions.
    • The electrolyte can be aqueous solutions or molten solutions
    • Electrolyte with may ions are called strong electrolyte; while those with few ions are called weak electrolytes.
      Examples
  2.  Electrodes
    • Are the solid conductors, usually roots, which usually complete the circuit between electrolytes and cell/battery
    • Are of two types
      1. Anode - The electrode connected to the positive terminal f a battery/cell
      2. Cathode -Electrode connected to the negative terminal of the battery/cell.
    • Note: Graphite rods are commonly preferred as electrodes in most access.
      Reasons:
      • They are inert/unreactive
      • Are cheap
    • Platinum is also relatively inert; but not less preferred to Graphite because its expensive.

Preferential discharge of ions

  • The products of electrolysis of any given electrolyte depend on the ions present in an electrolyte.
  • Commonly most molten electrolytes have only two ions; a cation and an anion and are termed Binary electrolytes.
  • As the electrolyte decomposes, ions collect/move to the opposite poles.
  • Negatively charged ions move to the Anode; the positive electrode, while the positively charged ions move to the cathode, the negative electrode;
  • Regardless of how many ions move to an electrode; only one can be discharged ot give a product.
  • Both cations and anions have a preferential discharge series.
    1. Discharge for cations
      • Cations are discharged by reduction (accepting electrons) to form their respective products.
      • The ease of reduction of cations depends on their position of electrochemical series.
      • Thus Ag+ is most readily discharged as it s the weakest reducing agent.
    2. Discharge for anions
      • Anions are discharged by oxidation (electron loss) to form their respective products.
      • Discharge of anions is viz.

Electrolysis of Various Substances

  1. Electrolysis of Dilute Sulphuric Acid
    Apparatus.
    electrolysis of dilute sulphuric acid
    Procedure
    Anelectric current is passed through the dilute sulphuric acid.
    Observation
    At the Anode
    • A colourless gas; collects
    • The gas collected relights a glowing splint; and its volume is half the volume of the gas at cathode. The gas is oxygen.

    At the cathode
    • A colourless gas collects
    • The collected gas burns with a pop-sound; and its volume is double the volume of gas at the anode.
    • The gas is Hydrogen gas.

    Explanations
    • Ions present in the electrolyte
    • Hydrogen ions and Sulphate ions form sulphuric acid.
    • Hydrogen ions and hydroxide ions from water.


      At the Anode (Positive electrode)

    • The negatively charged Sulphate ions and hydroxide ions migrate to the anode.
      Reason: OH-(aq) ions have a greater tendency to loose electrons than the SO42-(aq) ions
      Anode equation
      4OH-(aq) → O2(g) + 2H2O(l) + 4e-

      At the cathode (negative electrodes)

    • The positively charged hydrogen ions migrate to the cathode 
      Equation
      2H+(aq) + 2e- → H2(g)

      Note:

      1. The volume of oxygen produced at the anode is half the volume of hydrogen produced at the cathode.
        Reason:The 4 electrons lost by the hydroxide ions to form 1 mole (1 volume) of oxygen molecules are gained by the four hydrogen ions which form 2 molecules (2 volumes) of hydrogen molecules.
      2.  During the electrolysis, the concentration of the electrolyte (H2SO4(aq), increases
        Reasons:The volumes of hydrogen and oxygen gas liberated are in the same ratio as they are combined in water.
        Thus the amount of water in the electrolyte progressively decrease; hence the increased electrolyte concentration.

    Conclusion
    • Electrolysis of dilute sulphuric acid is thus the electrolysis of water.
      Note: The Hoffmans voltmeter can be used instead of the circuit above. Viz.
  2. Electrolysis of dilute sodium chloride

    Apparatus
    electrolysis of dilute sodium chloride
    Procedure
    • An electric current is passed through dilute sodium chloride solution; with carbon rods as the electrodes.
    • Gases evolved of each electrode are collected and tested.

    Observations
    At the anode:
    • A colourless gas is collected
    • The gas relights a glowing splint, and its volume is half the volume of the gas collected at the cathode.
    • The gas is Oxygen, O2

    At the cathode
    • A colourless gas collects
    • The gas burns with a pop sound; and its volume is twice the volume of the gas collected at the anode.
    • The gas is hydrogen gas;

    Explanations
    • The ions present in the electrolyte are:-
      - Na+(aq) and Cl- from sodium chloride
      - H+(aq) and OH-(aq) from water.
      At the Anode
      • Cl- and OH- migrate to the anode
      • OH- are preferentially discharged coz they have greater tendency to lose electrons than the chloride ions; - the OH-(aq) lose electrons to form water and O2(g) at anode.
        Anode equation
        4OH-(aq) → 2H2O(l) + O2(g) + 4e-

      At the Cathode
      • The positively charged Na+(aq), and H+(g) migrate to the cathode
      • the H+(aq) are preferentially discharged.
        Reason:They H+(aq) have a greater tendency to gain electrons than Na+(aq) ions. The H+(aq) gain electrons to form Hydrogen atoms (H) which then form molecules of hydrogen which bubble off at the electrode.
        Cathode equation
        2H+(aq) + 2e- → H2(g)

      Conclusion
    • Ratio of the volumes of H2(g) and O2(g) evolved at cathode and anode is 2:1 respectively.
    • Electrolysis f dilute NaCl is thus the electrolyze of water since only water is decomposed.
  3. Electrolysis of Brine/concentrate sodium chloride.

    Apparatus
    electrolysis of concentrated sodium chloride

    Procedure
    • An electric current is passed though concentrated sodium chloride/brinc

    Observation
    At the Anode
    • A greenish yellow gas is evolved.
    • The gas has a pungent irritating smell; and its volume is equal to the volume of the gas evolved at he cathode.
    • The gas is chlorine Cl2(g)

    At the Cathode
    • A colourless gas is liberated
    • The gas burns with a pop sound; and its volume is equal to volume of gas evolved at the anode.
    • The gas is Hydrogen gas; H2(g)

    Explanations
    • The ions present in the electrolyte are:-
      • Na+(aq) and Cl- from sodium chloride
      • H+(aq) and OH-(aq) from water

      At the anode
      • Cl-(aq), and OH-(aq), migrate to the anode
      • The chloride ions are preferentially discharged.
        Reason; -OH-(aq) have higher tendency to lose electrons than Cl- ions.
      • However coz of the higher concentration Cl-(aq), relative to OH-(aq), the Cl-(aq), are preferentially discharged hence the evolution of Chlorine gas.

      At the Cathode
      • Na+(aq) and H+(aq) migrate to the cathode.
      • H+ with a higher tendency to gain electrons are preferentially discharged; hence the evolution of hydrogen gas at the cathode.
        NB: 
        1. The pH of the electrolyte becomes alkaline/increases with time.
          Reason: The removal of H+(aq) which come form water leaves excess hydroxide ions (OH-(aq), hence the alkalinity.
        2. Evolution of chlorine gas at anode soon stops after sometime and is replaced by O2(g)
          Reason: Evolution of Cl2(g) decrease/lowers the concentration of Cl-(aq) in the electrolyte. 
        3. As soon as the Cl-(aq) concentration becomes equal to that of OH-(aq)

      The Mercury Cathode Cell
    • Is an electrolytic arrangement commonly used for the large scale manufacture of chlorine and sodium hydroxide.
      Apparatus
      mercury cell
    • Electrolyte in the mercury cell is Brine (concentrated NaCl)
    • Anode is carbon or titanium
    • Cathode is a moving mercury film.

      Reactions

      At the Anode
    • Both Chloride and Hydroxide ions are attracted
    • Due to their high concentrations the chloride ions are preferentially discharged.
    • The Cl-(aq) lose electrons to form Chlorine gas. (greenish yellow)
      Equation
      2Cl-(aq) → Cl2(g) + 2e-

      Cathode (moving mercury
    • The Na+(aq) and H+(aq) are attracted
    • The discharge of H+(aq) is more difficult than expected
    • Hydrogen has a high over voltage at the moving mercury electrode and so sodium is discharged.
      Equation
      Na+(aq) + e- → Na(s)
    • The discharged sodium atoms combine with mercury to form sodium amalgam
      Equation
      Na(s) + Hg(l) → NaHg(l)
    • The sodium amalgam reacts with water to form sodium hydroxide, hydrogen and mercury.
      Equation
      2NaHg(l) + 2H2O(l) → 2NaOH(aq) + H2(g) + 2Hg(l)
    • Hydrogen is pumped out while the Mercury is recycled.
    • The resultant NaOH is of very high party.

      Limitations/disadvantages of the Mercury Cathode cell.
    • Its expensive due to the high cost of mercury.
    • Pollution form Mercury; i.e. Mercury is poisonous and must be removed from the effluent.
  4. Electrolysis of Copper (II) Sulphate solution.
    NOTE: The products of electrolysis of copper (II) Sulphate solution depends on the nature of the elctrodes used.

    Apparatus
    electrolysis of copper II sulphate using carbon electrodes
    Electrolysis of copper sulphate solution using copper electrodes
    Ions present in the elctrolyte:
    • From copper (II) Sulphate
    • From water
    • During electrolysis
  1. Using carbon/platinum electrodes

    Observations

    At the Anode:
    • A colourless gas is liberated
    • The gas relights a glowing splint; hence its oxygen.

    At the cathode
    • A reddish brown coating (of Cu solid) is deposited.

    In the Electrolyte
    • The blue colour of the solution (CuSO4)(aq)/ becomes pale and finally colourless after a long time.
      Reason : The blue colour is due to Cu2+. As the Cu2+(aq) are continuously being discharged at the cathode; the concentration of CU2+ decreases i.e. decrease in the concentration of Cu2+(aq) in the solution
    • The electrolyte become acidic/pH decreases (declines)
      Reason:Accumulation of H+(aq) in the solution since only OH- (from water) are being discharged (at the anode).

    Explanation

    At the anode
    • The SO42-(aq) and OH–(aq) migrate to the anode.
    • The hydroxide ions have a higher tendency to lose electrons than the SO42-(aq)
    • They (OH-) easily loose electrons to form the neutral and unstable hydroxide radical (OH-)
    • The hydroxide radical (OH) decomposes to form water and Oxygen.
      Anode equation
      4OH-(aq) → O2(g) + H2O(l) + 4e-

    At the cathode
    • Copper ions and H+(aq) migrate to the cathode
    • Cu2+(aq) have a greater tendency to accept electrons than H+(aq)
    • The Cu2+(aq) are thus reduced to form copper metal which is deposited as a red-brown coating on the cathode.
      Cathode equation.
      Cu2+(aq) + 2e- → Cu(s)
  2. Using copper rods electrodes

    Observations

    At the anode
    • Mass of the anode (Copper anode) decreases

    At the Cathode
    • reddish brown deposit
    • cathode increases in mass

    Electrolyte
    • no apparent change

    Note    : The gain in mass of the cathode is equal to the loss in mass of the anode.

    Explanations

    At the anode
    • The SO42-(aq) and OH-(aq) are attracted to the anode.
    • However, none of them is discharged;
    • Instead; the copper anode itself gradually dissolves; hence the loss in mass of the anode;
      Reason: it s easier to remove electrons form the copper anode itself than format the hydroxide ions
      Anode equation
      Cu(s) → Cu2+(aq) + 2e-

    At the cathode
    • H+(aq) migrate to the cathode
    • The Cu2+(aq) are preferentially discharged; because they have a greater tendency to accept electrons
    • The copper cathode is thus coated with a reddish brown deposit of copper metal hence increase in mass.

      Cathode equation
      Cu2+(aq) + 2e- → Cu(s)

Factors Affecting Electrolysis and Electrolytic Products

  1. Electrochemical series
    • Electrolytic products at the anode and cathode during electrolysis depends on its position in the Electrochemical series.
    • Cations:The higher the cation in the electrochemical series; the lower the tendency of discharge at the cathode.
      Reason:Most electropositive cations require more energy in order to be reduced and therefore are more difficult to reduce.
      Anions:       Discharge is through oxidation and is as follows.
      electrochemical series
  2. Concentration of electrolytes
    • A cation or anion whose concentration is higher is preferentially discharge if the ions are close in the electrochemical series.
      Example : dilute and concentrated NaCl
      Product at the anode .
  3. The electrodes used:Products obtained at electrodes depend on the types of electrodes used
    Examples:in the electrolysis of CuSO4(aq) using carbon and copper rods separately.

Applications of Electrolysis

  1. Extraction of reactive metals
    • Reactive metals/elements like sodium, magnesium, aluminum are extracted form their compounds by electrolysis.
      Example :Sodium is extracted from molten sodium chloride using carbon electrodes.
  2. Purification of metals
    • It can be used in refining impure metals
      Examples: Refining copper
      • The impure copper is made of the anode.
      • Their strips of copper are used as the cathode
      • Copper (II) Sulphate are used as the electrolyte.
      • During the electrolysis the anode dissolves and pure copper is deposited on the cathode.
      • The impurities (including valuable amounts of silver and gold) from the crude copper collect as a sludge become the anode.
  3. Electroplating
    • Is the process of coating one metal with another, using electrolysis so as to reduce corrosion or to improve its appearance.
    • During electrolysis:
      • the item to be electroplated is made the cathode
      • the metal to be used in electroplating is used as the anode
      • the electrolyte is made from a solution containing the ions of the metal to be sued in electroplating.
    • Examples
      • Gold-plated watches; silver plated utensils
      • Steel utensils marked EPNS. I.e. Electroplated Nickel Silver.
  4. Anodizing Aluminum
    • Is the reinforcement of the oxide coating on Aluminum utensils/articles
    • Is done by electrolysis of dilute sulphuric acid using Aluminum articles as anode.
    • Importance: Prevention arrosion of Aluminum articles
  5. Manufacture of sodium hydroxide, chlorine and hydrogen
    • Sodium hydroxide is prepared by the electrolysis of brine, for which 3 methods are available
    • The method depends on the type of electrolytic cell.
    • These cells are
      • The mercury cell
      • The diaphragm cell
      • The membrane cell

    1. The mercury cell
      manufacture of sodium hydroxide

      Components
      • The electrolyte is concentrated sodium chloride
      • The anodes are made of graphite or titanium, which are placed above the cathode.
      • The cathode consists of mercury, which flows along the bottom of the cell.

      Chemical reactions

      Anode
      • Both chloride and hydroxide ions are attracted.
      • Chloride ions are preferentially discharged due to their high concentration
      • The chloride ions undergo oxidation to form green yellow chlorine gas.
        Equation

      At the cathode (flowing mercury)
      • Na+(aq) migrate to the cathode
      • Sodium ions are preferentially discharged.
      • They undergo reduction to form sodium solid.
        Equation
        Na+(aq) + e- → Na(s)
      • the discharged sodium atoms combine with mercury to form sodium amalgam
        Equation
        Na(s) + Hg(l) → NaHg(l)
      • The sodium amalgam is then passed into another reactor containing water.
      • The amalgam reacts with water forming hydrogen and sodium hydroxide.
        Equation
        2NaHg(l) + 2H2O(l) → 2NaOH(aq) + H2(g) + 2Hg(l)
      The mercury is regenerated and it is recycled into the main cell.
      Main product:
      • Sodium hydroxide

      By products
      • Sodium and chlorine.

      Advantages of mercury cathode cell
      • The resultant sodium hydroxide is very pure; as it has no contamination from sodium chloride.
      • It is highly concentrated; i.e. about 50%.

      Disadvantages
      • Some of the mercury said its way into the environment leading to mercury pollution; a common case of brain damage in humans.
      • At the operating temperatures (70o C - 80o C), mercury vapours escape into the atmosphere and cause irritation and destruction of lungs tissues.
      • Its operation requires highly skilled man power.
    2. Diaphragm cell
      Components
      • An asbestos diaphragm; to separate the electrolytic cell into two compartments; thus preventing mixing of H2 and Cl2 molecules
      • The anode compartment contains a graphite rod/Titanium.
      • The cathode compartment contains a stainless steel cathode.
      diaphragm cell1
      Diagram: The diaphragm cell
      Chemical reactions
      • The asbestos diaphragm is permeable only to ions, but not to the hydrogen or chlorine molecules.
      • It thus prevents H2(g) and Cl(g) form mixing and reacting to yield HCl(g)
      • It also separates NaOH and Cl2 which would otherwise react.

      At the anode
      • Chloride ions undergo oxidation to form chlorine gas.
        Equation
        2Cl-(aq) → Cl2(g) + 2e-

      At the cathode
      • H+ and Na+(aq) migrate to the cathode compartment.
      • The H+ are preferentially discharged.
      • They (H+(aq)) undergo reduction to form hydrogen gas.
        Equation
        2H2O(l) + 2e- → H2(g) + 2OH-(aq)

      The discharge of H+ causes more water molecules to dissociate, thus increasing the concentration of OH- in the solution. Therefore, the Na+ and OH- ions also react in the cathode compartment to form sodium hydroxide.

      Equation
      2H2O(l) + 2Cl-(aq) + 2Na+(aq) → 2Na+(aq) + 2OH-(aq) + H2(g) + Cl2(g)

      Advantage
      • Does not result into pollution

      Disadvantages
      • The resultant NaOH is dilute (12% NaOH)
      • It is also not pure due to contamination with NaCl (12% NaOH + 15% NaCl by mass.

      - Note: -The concentration of the NaOH can be increased by evaporating excess water, during which NaCl with a lower solubility crystallizes out first, leaving NaOH at a
      higher concentration.
      - The solid NaCl (Crystals) are then filtered off.
      - This is a case of fractional crystallization .
    3. The membrane cell
      • Is divided into 2 compartments by a membrane
      • Most commonly used type of Membrane is the cation exchange membrane.
      • This membrane type allows only cations to pass through it.

        Components
      • A cation exchange that divides the cell into 2 compartments; an anode and a cathode compartments.
      • Both electrodes are made of graphite
      • The electrolyte in the anode compartment is purified brine
      • The electrolyte in the cathode compartment is pure water.
        Diagram
        membrane cell
        Chemical reactions
        The anode
      • Chloride ions undergo oxidation to form green yellow chlorine gas.

        Equation
        2Cl-(aq) → Cl2(g) + 2e-

        The cathode
      • As current passes through the cell H+ and Na+ pass across the membrane to the cathode
        • H+ are preferentially discharged.
        • They undergo reduction to liberate hydrogen gas.
        • Continuos discharge of H+ leaves the OH- at a higher concentration
        • The OH- react with Na+ to form sodium hydroxide

          Equation
          2H2O(l) + 2e- → H2(g) + 2OH-(aq

        Advantages
        1.  Resultant sodium hydroxide is very pure, since it has no contamination from NaCl.
        2. The sodium hydroxide has a relatively high concentration; at about 30 35% NaOH by mass.

Uses of sodium hydroxide, chlorine, and hydrogen

  1. Sodium Hydroxide
    • React with chlorine to form sodium chlorate sodium hypochlorite or I, NaOCl. This is a powerful oxidising agent which is used for sterilization and bleaching in textiles, paper and textile industries.
      i.e. 2NaOH(aq) + Cl2(g) → NaOCl(aq) + NaCl(g) + H2O(l)
    • Manufacture of sodas, detergents and cosmetics.
    • Neutralization of acidic solutions in the laboratories.
  2. Hydrogen
    • For hydrogenation in the manufacture of margarine
    • Manufacture of ammonia
    • Production of hydrochloric acid
  3. Chlorine
    • Formation of sodium chlorate I; for bleaching in pulp, textile and paper industries.
    • Sewage and water treatment
    • Manufacture of polymers such as polyvinyl chloride.
    • Manufacture of pesticides.

Quantitative Aspects of Electrolysis

Basic Terminologies and Concepts

  1. Ampere
    • Is the standard unit used to measure an electric current; the flow of electrons
    • Is usually abbreviated as amps.
  2. Coulomb
    • Is the quantity of electricity, when a current of 1 ampere flows for one second. I.e. 1 Coulomb = 1 Ampere x 1 second
    • Generally:
      Quantity of Electricity = current x Time in seconds
      A = It; Where
      Q = Quantity of electricity in coulombs
      I = Current in Amperes
      T = time in seconds
  3.  Faraday
    • Is the quantity of electricity produced by one mole of electrons; and is usually a constant equivalent to 96487 (approx. 96500) coulombs

Faradays laws of Electrolysis.

First law;

  • The mass of substance liberated during electrolysis is directly proportional to the quantity of electricity passed.

Worked examples

  1. A current of 2.0 Amperes was passed through dilute potassium sulphate solution for two minutes using inert electrodes.
    1. Write the equation for the reaction at anode.
    2. Work out the mass of the product formed at the cathode. (H = 1.0, Faraday = 96,000 C)
      Solution; quantity of electricity = current x time
      = 2 x 2x 60
      = 240 coulombs
      cathode reaction
      1 mole of electrons = 96000 C
      4 moles of e- = 4 x 96000 C
      = 384,000 C
  2. What mass of copper would be coated on the cathode from a solution of copper (II) Sulphate by a current of 1 amp flowing for 30 minutes. (Cu = 63.5; Faraday constant = 96487 Cuo/-)
    solution
    Cathode reaction
    Cu2+(aq) + 2e- → Cu(s)
    • 1 mole of Cu requires 2 moles of electrons
    • Quantity of electricity passed; = 1 x 30 x 60 coulombs; = 1800C
    • 1 mole of electrons carriers a charge of 96487 coulombs = 192974 coulombs
    • 192.974 coulombs deposit 63.5 g of Cu.
    • Thus 1800 C will deposit 63.5 x 1800/192974  = 0.592 grams
  3. An element x has relative atomic mass of 88g. when a current of 0.5 amperes was passed through a solution of x chloride for 32 minutes, 10 seconds; 0.44 g of x was deposited at the cathode. (1 faraday = 96500 C) Calculate the charge on the ion of x.
  4. In the electrolysis of dil CuSO4 solution, a steady current of 0.20 Amperes was passed for 20 minutes. (1 Faraday = 96, 500 C Mol-, Cu = 64)
    Calculate
    1. The number of Coulombs of electricity used
    2. The mass of the substance formed at the cathode.
      2 moles of electrons liberate 1 mole of Cu. i.e. Cu 2+ + 2e- → Cu(s)
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