ELECTROCHEMICAL CELLS

ELECTROCHEMISTRY The reason Redox reactions are so important is because they involve an exchange of electrons If we can find a way to make those electrons flow, we will create electric current Electric Current – The flow of charged particles One of the major ways that we can create electric current with a redox reaction is using an Electrochemical Cell

ELECTROCHEMICAL CELLS Electrochemical cells use a spontaneous chemical reaction to convert chemical energy into electrical energy An electrochemical cell is the basic unit of a battery One battery may have several electrochemical cells to create electric current

ELECTROCHEMICAL CELL EXAMPLE Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) This reactions involves two half-reactions: Oxidation: Zn → Zn e - Reduction: Cu e - → Cu If these were placed in the same container, the electrons would immediately exchange between the copper and zinc atoms Zinc metal strip placed in a Cu(NO 3 ) 2 solution The electrons cannot be used for anything If we want to use the electrons to do something, we need to separate the reactants and force the electrons to transfer another way

SETTING UP AN ELECTROCHEMICAL CELL To set up an electrochemical cell, the 2 half-reactions are separated in separate containers The cells are then connected with 2 separate parts The electrodes are connected with a wire to transfer electrons. This is called the External Circuit. A salt bridge connects the electrolyte solutions so ions can be transferred This is called the Internal Circuit Remember charges must be balanced. In the zinc half- cell, Zn 2+ ions are entering the solution. This buildup of positive charge attracts the negative ions in the salt bridge and balances the charges

ELECTRODES The electrode where oxidation happens is called the anode The anode is the source of electrons, making it the negative post of the electrochemical cell. Anode = oxidation “An. Ox.” The electrode where reduction occurs is called the cathode The cathode is the positive post of the electrochemical cell as it consumes electrons. Cathode = reduction “Red. Cat.” What is the anode is the following reaction? What is the cathode? Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s)

WHY ARE SOME MATERIALS OXIDIZED AND SOME REDUCED? Metals tend to lose their electrons easy because they have few valence electrons Easily oxidize The activity series determines which lose their electrons easiest The fact that different substances are oxidized more readily than others is the driving force behind electrochemical cells, and it is this force that forces electrons through the external circuit from the anode (site of oxidation) to the cathode (site of reduction). This force is known as the potential difference or electromotive force (emf or E). Potential difference is measured in volts (V) x

STANDARD ELECTRODE POTENTIALS Tables of Standard Reduction Potentials for Half-Reactions allow us to determine the voltage of electrochemical cells. These tables compare the ability of different half-reactions to become reduced.Standard Reduction Potentials for Half-Reactions Since half-reactions cannot occur on their own, all values in the table are determined by comparing a half-reaction with a hydrogen half-cell: 2H + (aq) + 2e - → H 2 (g) E° = 0.00 V the degree symbol following the E (E°) indicates standard conditions: temperature = 25°C; pressure = 100 kPa; concentration of aqueous solutions = 1 mol·L -1 (Find this half-reaction, and other the other half-reactions described below, in the Table.) This hydrogen half-cell has been assigned a voltage of 0.00 V. If a half-reaction is better at competing for electrons than this half-cell, that half-reaction will undergo reduction and the hydrogen will be oxidized. That other half-reaction will then be assigned a positive voltage.

ELECTRODE POTENTIALS – EXAMPLE 1 If copper and hydrogen half-cells are joined together we find that the copper half-cell will gain electrons from the hydrogen half-cell. Thus the copper half-cell is given a positive voltage and given a relative value of V: Cu 2+ (aq) + 2e - → Cu (s) E° = 0.34 V Since both half-reactions cannot undergo reduction, we must reverse the equation of the reaction that will undergo oxidation. This will give us an electrochemical cell voltage of 0.34 V Half-Reactions E° Cu 2+ (aq) + 2e - → Cu (s) 0.34 V H 2 (g) → 2H + (aq) + 2e V Cu 2+ (aq) + H 2 (g) → 2H + (aq) + Cu (s) 0.34 V

ELECTRODE POTENTIALS – EXAMPLE 2 We see in the Table of Standard Reduction Potentials that zinc has a negative E° indicating that it is not as good at competing for electrons as hydrogen. Zn 2+ (aq) + 2e - → Zn (s) E° = V Therefore if zinc and hydrogen are paired together in an electrochemical cell, the hydrogen would be reduced (gain the electrons) and zinc would be oxidized (losing electrons). To determine the net redox reaction as well as the voltage of the electrochemical cell we reverse the zinc equation, and also reverse it's sign before adding the equations and E° together: Half-Reactions E° Zn (s) → Zn 2+ (aq) + 2e V 2H + (aq) + 2e - → H 2 (g) V Zn (s) + 2H + (aq) → Zn 2+ (aq) + H 2 (g) 0.76 V