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Chapter 20 Electrochemistry. Oxidation-reduction reactions Oxidation numbers Oxidation of metals by acids and salts The activity series ALL these to be done in class. Oxidation-Reduction Reactions. In the reaction Zn( s ) + 2H + ( aq )  Zn 2+ ( aq ) + H 2 ( g ).

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Chapter 20 electrochemistry
Chapter 20Electrochemistry

Chapter 20


Chapter 20


Oxidation-Reduction Reactions

  • In the reaction

  • Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g).

  • Which element is oxidised and which one is reduced?

  • Oxidation – loss of e-

  • Reduction – gain of e-

Chapter 20


Balancing Oxidation-Reduction Reactions

  • Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end.

  • Conservation of charge: electrons are not lost in a chemical reaction.

Chapter 20


Half Reactions

Half-reactions are a convenient way of separating oxidation and reduction reactions.

Chapter 20


Chapter 20


Balancing Equations by the Method of Half Reactions

The two incomplete half reactions are

MnO4-(aq)  Mn2+(aq)

C2O42-(aq)  2CO2(g)

Balance the overall reaction equation in an acidic solution

Chapter 20


Chapter 20


Voltaic Cells Solution

  • If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+.

Chapter 20


Chapter 20


  • “Rules” of voltaic cells: Solution

    • 1. At the anode electrons are products. (Oxidation)

    • 2. At the cathode electrons are reagents. (Reduction)

    • 3. Electrons can’t swim!

Chapter 20


Chapter 20



Cell EMF bridge.

  • e- flow from anode to cathode because the cathode has a lower electrical potential energy than the anode.

  • 1 V is the potential difference required to impart 1 J of energy to a charge of one coulomb:

Chapter 20


Chapter 20


Chapter 20


Chapter 20


Chapter 20 electrons through the external circuit.


  • For the SHE, we assign electrons through the external circuit.

  • 2H+(aq, 1M) + 2e- H2(g, 1 atm)

  • Ered = 0.

Chapter 20


  • For Zn: electrons through the external circuit.

  • Ecell = Ered(cathode) - Ered(anode)

  • 0.76 V = 0 V - Ered(anode).

  • Therefore, Ered(anode) = -0.76 V.

  • Standard reduction potentials must be written as reduction reactions:

  • Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V.

Chapter 20


Chapter 20


  • Reactions with electrons through the external circuit.Ered < 0 are spontaneous oxidations relative to the SHE.

  • The larger the difference between Ered values, the larger Ecell.

Chapter 20


  • Oxidizing and Reducing Agents electrons through the external circuit.

  • The more positive Ered the stronger the oxidizing agent on the left.

  • The more negative Ered the stronger the reducing agent on the right.

Chapter 20


Chapter 20 electrons through the external circuit.


Chapter 20


Example: For the following cell, what is the cell electrons through the external circuit.

reaction and Eocell?

Al(s)|Al3+(aq)||Fe2+(aq)|Fe(s)

Al3+(aq) + 3e-→ Al(s); EoAl = -1.66 V

Fe2+(aq) + 2e- → Fe(s); EoFe = -0.41 V

Chapter 20


Example: When an aqueous solution of CuSO electrons through the external circuit.4 is electrolysed, Cu metal is deposited:

Cu2+(aq) + 2e-→ Cu(s)

If a constant current was passed for 5.00 h and 404 mg of Cu metal was deposited, what was the current?

Ans: 6.81 x 10-2 A

Chapter 20


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