want to test ur iq...meh2 cni..heheh..

1. An electrochemical cell is created using gold and magnesium half-cells.
   
   
1.    Determine which half-cell will undergo oxidation and which will undergo reduction, identify anode and cathode, and calculate the voltage for the cell. You do not need to diagram the cell. 
   
   
2.      If the mass of the magnesium electrode changes by 5.0 g, what will be the change in mass of the gold electrode, and will its mass increase or decrease?

Part a.

balance for e- E° (V) cathode reduction × 2 Au3+ + 3e- → Au(s) +1.50 + electrode anode oxidation × 3 Mg(s) → Mg2+ + 2 e- +2.37 - electrode  balanced net equation   3Mg(s) + 2Au3+ → 2Au(s) + 3Mg2+ +3.87 total voltage

Part b.

The balanced equation is required to answer part b: 3Mg(s) + 2Au3+ → 2Au(s) + 3Mg2+ From this equation we see that 3 moles of Mg react with 2 moles of Au to give us the ratio: 3 Mg 2 Au Since the question concerns mass, we will need to know the molar masses of each element, in order to convert between moles and mass. Molar masses are equivalent to atomic masses, found in the periodic table. The unit for molar mass is g/mol. (Need a review? Check here) molar mass of Mg:  24.3 g ·mol-1      molar mass of Au: 197.0 g ·mol-1 The question gives us the mass of Mg and asks us to find mass of Au. Set up the equation so that all units will cancel except g Au (mass of Au): g Au = 5.0 g Mg × 1 mol Mg 24.3 g × 197.0 g mol Au × 2 Au 3 Mg = 27.0 g Au answer The mass of gold will change by 27.0 g. Our half-reactions tell us that gold ions combine with gold ions to produce solid gold. Thus our gold electrode will increase it’s mass by 27.0 g.

    Balance the following half-reactions for both atoms and electrons by adding the appropriate number of electrons to the correct side of the equation. Also identify each as either an oxidation or reduction.

a.	Pb2+ → Pb	Pb2+ + 2e- → Pb	reduction
b.	Cl2 → Cl-	Cl2 + 2 e- → 2 Cl-	reduction
c.	Fe3+ → Fe2+	Fe3+ + e- → Fe2+	reduction
d.	N2O + H2O → NO + H+
	N2O + H2O → 2NO + 2 H+ + 2e-		oxidation

   Break each equation into two half-reactions. Identify each half-reaction as oxidation or reduction.

a.	Cu + 2 H+ → Cu2+ + H2
	Cu → Cu2+ + 2 e-	oxidation
	2 H+ + 2 e- → H2	reduction
b.	2 Al + 3 S → Al2S3
	2 Al → 2Al3+ + 6 e-	oxidation
	3S + 6e- → 3 S2-	reduction