# Using standard electrode potentials to calculate electrochemical cell voltages Chapter 19 Pages 768-773.

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Using standard electrode potentials to calculate electrochemical cell voltages Chapter 19 Pages 768-773

An electrochemical (voltaic) cell consists of an oxidation reaction and a reduction reaction to produce a voltage for the cell. Ex: Batteries Need to know: which electrode is the cathode, which is the anode, and whether the chemical reaction is spontaneous.

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

When choosing your anode and cathode follow this rule: Look at the reduction potential for the ions, the cathode will be the more positive reduction potential (p.11) Reason: higher reduction potential means more easily reduced, therefore reduction takes place there, therefore the cathode

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. Page 11 of our data booklet

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.

E o net = E o oxidation + E o reduction E o oxidation = oxidation half reaction taking place in the cell Oxidation half reactions are the reverse reaction from the table on p.11, so their sign changes! If the reduction potential of Ag+ = 0.80V, then the oxidation potential of Ag(s) = - 0.80V

E o net = E o oxidation + E o reduction 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 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 +2 + 2e-  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 +2 + 2e-+0.76V c) E o net = E o oxidation + E o reduction = (+0.76V) + ( +0.80V) = +1.56V

http://www.chem.iastate.edu/group/Green bowe/sections/projectfolder/flashfiles/elect roChem/volticCell.htmlhttp://www.chem.iastate.edu/group/Green bowe/sections/projectfolder/flashfiles/elect roChem/volticCell.html Look at this electrochemical cell demonstration. Do the same as our example with Zinc and Silver

Exercise: 1.An electrochemical cell was designed using the metals Mg and Al and solutions containing Mg +2 and Al +3. a)Identify the anode and cathode b)Write the oxidation and reduction half reactions c)Calculate the cell voltage 2.An electrochemical cell was designed using the metals Zn and Pb and solutions containing the positive ions of these metals, Zn +2 and Pb +2. a)Identify the anode and cathode b)Write the oxidation and reduction half reactions c)Calculate the cell voltage

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