Chapter 19 Last Unit Electrochemistry: Voltaic Cells and Reduction Potentials

Electrochemical Processes: All electrochemical processes involve redox reactions. Electric current = a flow of electrons in a circuit Cells: Chemical Energy  Electrical Energy (spontaneous) Electrolysis: Electrical Energy  Chemical Energy (nonspontaneous) We know that oxidation and reduction occurs in the interaction of a metal with aqueous ions. Oxidation: Cu(s)  Cu2+(aq)+ 2e- Reduction: Cu2+(aq) + 2e-  Cu(s)

Voltaic Cells: Aka Galvanic cell An electrochemical (voltaic) cell consists of an oxidation reaction and a reduction reaction to produce a voltage for the cell. Prevents reactants for ox and reduc rxns coming in contact with each other, e - flow through an external circuit Ex: Batteries Need to know: which electrode is the cathode, which is the anode, and whether the chemical reaction is spontaneous.

More…. Electrodes = conductors that carry e - into and out of a cell, there are two: Anode = negative electrode = where oxidation occurs Cathode = positive electrode = where reduction occurs Electrons travel from the anode to the cathode, along a conductive wire = voltage = electricity.

Anode Cathode

Galvanic Cell Notation: Short hand that shows anode on left, cathode on right, double vertical line represents the porous barrier or salt bridge Example: Zn(s) | Zn +2 (aq) || Cu +2 (aq) | Cu(s)

Dry Cell Battery:

Try it Pg 761 # 2(a), 3

Cell Potentials: Chemical changes in voltaic cells are accompanied by changes in potential energy Electrons flow spontaneously from a position of higher potential energy (at the anode) to a position of lower potential energy (at the cathode) The moving electron can do work! (ie. light a light blub) The difference in potential energy between the anode and the cathode is called electric potential, E. Unit = volt (V)

Choosing the anode and cathode rule: Look at the reduction potential for the ions, the cathode will be the more positive reduction potential Reason: higher reduction potential means more easily reduced, therefore reduction takes place there, therefore the cathode In data booklet see Table of standard reduction potentials (shows the relative ability for a substance to be reduced (gain electrons). It expresses this potential to gain electrons by assigning a voltage for each reduction half reaction in the table). Or page 848

E o net = E o oxidation + E o reduction E o net = total net voltage of the cell (electric potential energy of the cell) Must be greater than zero for the reaction to be spontaneous If less than zero, than non-spontaneous reaction and no electricity produced.

Continued…. E o oxidation = oxidation half reaction taking place in the cell Oxidation half reactions are the reverse reaction So take the value from Table of standard reduction potentials and change the sign! Example: If the reduction potential of Ag+ = 0.80V, then the oxidation potential of Ag(s) = V

More…. E o reduction = reduction half reaction taking place in the cell The larger the reduction potential (the more positive), the greater the tendency for the reaction to occur. The smaller the reduction potential (the most negative), the least likely for a reaction to occur.

Example: Calculate the standard cell potential for the galvanic cell in which the following reaction occurs: 2I - (aq) + Br 2 (l)  I 2 (s) + 2Br Write the half reactions and decide on nodes: 2I - (aq)  I 2 (s) + 2e - (oxidation = anode) Br 2 (l) + 2e -  2Br 1- (reduction = cathode) 2. Use table to find reduction potentials: 2I - (aq)  I 2 (s) + 2e - E ° red = 0.54 V Br 2 (l) + 2e -  2Br 1- E ° red = V 3. Change sign on the oxidation reaction: 2I - (aq)  I 2 (s) + 2e - E ° ox = V 4. E o net = E o ox + E o red =1.066 V V = v -

Try it Pg 773 # 5, 6, 7, 8

Example Problem 1: Design an electrochemical cell that uses the reactions of the metals Zn and Ag, and solutions of their ions, Zn +2 and Ag +. a) Identify the anode and cathode b) Write the oxidation and reduction half reactions c) Calculate the cell voltage

Solution: a), b) Zn e-  Zn(s)-0.76V Ag +1 +1e-  Ag(s) +0.80V Ag+ more easily reduced, therefore Ag(s) is the cathode, so oxidation takes place at the anode (Zn) and the oxidation potential half reaction is: Zn(s)  Zn e-+0.76V c) E o net = E o oxidation + E o reduction = (+0.76V) + ( +0.80V) = +1.56V