1. Identify the Half-Reactions
* Aluminum (Al) will act as the anode (oxidation occurs):
Al(s) → Al³⁺(aq) + 3e⁻
* Copper (Cu) will act as the cathode (reduction occurs):
Cu²⁺(aq) + 2e⁻ → Cu(s)
2. Look Up Standard Reduction Potentials
You'll need a table of standard reduction potentials (E°) to find the values for each half-reaction. Here are the typical values:
* Al³⁺(aq) + 3e⁻ → Al(s) E° = -1.66 V
* Cu²⁺(aq) + 2e⁻ → Cu(s) E° = +0.34 V
3. Determine the Overall Cell Reaction
* Since aluminum has a more negative reduction potential, it will be oxidized (the reverse of the reduction reaction).
* To balance the electrons, multiply the aluminum half-reaction by 2 and the copper half-reaction by 3:
* 2Al(s) → 2Al³⁺(aq) + 6e⁻
* 3Cu²⁺(aq) + 6e⁻ → 3Cu(s)
* The overall cell reaction is:
2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)
4. Calculate the Standard Cell Potential (Emf)
The emf of the cell is the difference between the standard reduction potentials of the cathode and the anode. Remember that the anode's potential is reversed:
Emf = E°(cathode) - E°(anode)
Emf = +0.34 V - (-1.66 V)
Emf = +2.00 V
Therefore, the standard emf of this voltaic cell is +2.00 V.
Important Note: This calculation assumes standard conditions (25°C, 1 atm pressure, 1 M concentration of ions). The actual emf may vary depending on the actual concentrations of the ions in solution.
