DQ 2. Identify the anode and cathode in the Cu-Pb voltaic cell. Provide the balanced half-reactions that describe the reaction taking place at each electrode. Combine the two half-reactions to provide the net ionic equation that represents the overall reaction within the Cu-Pb voltaic cell.

Identify the anode and cathode in the Cu-Pb voltaic cell -

                                       Cu (s) ---> Cu 2+ (aq) + 2 e-

                          Cu (s) + Pb 2+ (aq) ---> Cu 2+ (aq) + Pb (s)

A galvanic cell or voltaic cell reaction is a redox reaction that takes place in two half-reactions oxidation and reduction. At anode oxidation reaction and at cathode reduction reaction takes place.  

anode:

                              Cu (s) ---> Cu 2+ (aq) + 2 e-

cathode:                

                                Pb 2+ (aq) + 2 e- ---> Pb (s)

combined:

                          Cu (s) + Pb 2+ (aq) ---> Cu 2+ (aq) + Pb (s)

Thus, Identify the anode and cathode in the Cu-Pb voltaic cell -

                                       Cu (s) ---> Cu 2+ (aq) + 2 e-

                          Cu (s) + Pb 2+ (aq) ---> Cu 2+ (aq) + Pb (s)

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Anode: Pb

Cathode: Cu

Overall cell reaction:

Pb(s) + Cu^2+(aq) > Pb^2+(aq) + Cu(s)  

Explanation:

Anode half reaction;

Pb(s) > Pb^2+(aq) + 2e                -0.13 V

Cathode half reaction;

Cu^2+(aq) + 2e > Cu(s)             +0.34 V

Overall cell reaction;

Pb(s) + Cu^2+(aq) > Pb^2+(aq) + Cu(s)    

Pb is the anode because it has a more negative reduction potential. This implies that it is above copper in the electrochemical series. In any voltaic cell comprising of a metal cathode and anode; the metal that is higher in the electrochemical series serves as the anode while the other metal serves as the cathode.

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