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Using standard electrode potentials to calculate electrochemical cell voltages

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

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Using standard electrode potentials to calculate electrochemical cell voltages

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

  2. 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.

  3. 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.

  4. Anode Cathode

  5. 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

  6. Tableof 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

  7. Eonet= Eooxidation + Eoreduction • Eonet = 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.

  8. Eonet= Eooxidation + Eoreduction • Eooxidation = 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

  9. Eonet= Eooxidation + Eoreduction • Eoreduction = 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.

  10. 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+. • Identify the anode and cathode • Write the oxidation and reduction half reactions • Calculate the cell voltage

  11. 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) Eonet = Eooxidation + Eoreduction = (+0.76V) + ( +0.80V) = +1.56V

  12. http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/volticCell.htmlhttp://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/volticCell.html • Look at this electrochemical cell demonstration. Do the same as our example with Zinc and Silver

  13. Exercise: • An electrochemical cell was designed using the metals Mg and Al and solutions containing Mg+2 and Al+3. • Identify the anode and cathode • Write the oxidation and reduction half reactions • Calculate the cell voltage • 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. • Identify the anode and cathode • Write the oxidation and reduction half reactions • Calculate the cell voltage

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