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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 20Electrochemistry Chapter 20
Oxidation-reduction reactions • Oxidation numbers • Oxidation of metals by acids and salts • The activity series • ALL these to be done in class 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
The half-reactions for • Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe3+(aq) • are……… 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
Balancing Equations for Reactions Occurring in Basic Solution • We use OH- and H2O rather than H+ and H2O. • The same method as for acidic solution is used, but OH- is added to “neutralize” the H+ used. Chapter 20
Voltaic Cells • 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
Zn is spontaneously oxidized to Zn2+ by Cu2+. • The Cu2+ is spontaneously reduced to Cu0 by Zn. Chapter 20
“Rules” of voltaic cells: • 1. At the anode electrons are products. (Oxidation) • 2. At the cathode electrons are reagents. (Reduction) • 3. Electrons can’t swim! Chapter 20
Anions and cations move through a porous barrier or salt bridge. • Cations move into the cathodic compartment to neutralize the excess negatively charged ions • Anions move into the anodic compartment to neutralize the excess Zn2+ ions formed by oxidation Chapter 20
A Molecular View of Electrode Processes Chapter 20
Cell EMF • 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
1 V is the potential difference required to impart 1 J of energy to a charge of one coulomb: Chapter 20
Electromotive force (emf) is the force required to push electrons through the external circuit. • Cell potential: Ecell is the emf of a cell. • For 1M solutions at 25 C (standard conditions), the standard emf (standard cell potential) is called Ecell. Chapter 20
Standard Reduction (Half-Cell) Potentials • Standard reduction potentials, Ered are measured relative to the standard hydrogen electrode (SHE). Chapter 20
For the SHE, we assign • 2H+(aq, 1M) + 2e- H2(g, 1 atm) • Ered = 0. Chapter 20
For Zn: • Ecell = Ered(cathode) - Ered(anode) • 0.76 V = 0 V - Ered(anode). • Therefore, Ered(anode) = -0.76 V. • Standard reduction potentials must be written as reduction reactions: • Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V. Chapter 20
Changing the stoichiometric coefficient does not affect Ered. • Therefore, • 2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V. Chapter 20
Reactions with Ered < 0 are spontaneous oxidations relative to the SHE. • The larger the difference between Ered values, the larger Ecell. Chapter 20
Oxidizing and Reducing Agents • The more positive Ered the stronger the oxidizing agent on the left. • The more negative Ered the stronger the reducing agent on the right. Chapter 20
Example: For the following cell, what is the cell 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 CuSO4 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