Question

Calculate the theoretical cell potential (E°) of a galvanic cell under standard conditions made up of copper and magnesium (see Part II and Table 1 for more information).

PARTIL Creating and Testing Voltaic Cells Introduction and Background for the Voltaic Cells A galvanic cell (sometimes more appropriately called a voltaic cell) consists of two half-cells joined by a salt bridge that allow ions to pass between the two sides in order to maintain electroneutrality. Each half-cell contains the Components of a half-reaction. If we were going to make a voltaic cell from zinc and magnesium metals, we would create one half-cell with zinc solid metal dipped in a zinc nitrate solution (to provide aqueous Zn?). The other half-cell would contain magnesium metal dipped in a magnesium nitrate solution. Looking at Table 1 we can find a list of standard reduction potentials and corresponding reduction reactions. For zinc and magnesium the important reactions are: Eo Zn2+(aq) 2e Zn (s) 0.76 V v (s) Mg2+(aq) 2e M -2.37 The more positive (or less negative) the reduction potential (EO), the greater the tendency for a substance to be reduced (or act as an oxidizing agent) in order to create a spontaneous reaction. Thus, in this example, zinc would be reduced, which leaves manganese to be oxidized. The reactions then become Reduction: In 2 aq 2e Zn(s) MgCs) Mg2+(aq) 2e Oxidation: 2+ Mg(s) 2+ Zn(s) (ag) (ag) NET The resulting cell potential will thus be 1.61 V Table 1: Standard reduction potentials at 25°C Half-reactions Eo (V) Cu 2e Cu(s) -+0.34 V 2+ 0.76 V 2+ 2e-- Mg(s) 2.37 V

Answer 1

Calculate the theoretical cell potential (E°) of a galvanic cell under standard conditions made up of copper and magnesium (see Part II and Table 1 for more information).

For copper:

Cu+2 + 2e- --> Cu(s) E = 0.34 V

For Magnesium

Mg2+ + 2e- --> Mg(s) E = -2.37 V

so

Mg will oxidize, since it has much lower potential

then

Cu+2 + 2e- --> Cu(s) E = 0.34 V

Mg(s) --> Mg2+ + 2e-   E =+-2.37 V

Ecell = Ered + Eox = 0.34 +2.37 = 2.71 V