`A1^(3+) + 3e^(-) to Al(s)`, `E^(@) = -1.66V`
`Cu^(2+)+ 2e^(-) rightarrow CU(s)`, `E^(@)=+0.34V`
What voltage is produced under standard conditions by combining the half reactions with these standard electrode potentials?

Explanation

To find the voltage produced under standard conditions by combining the given half-reactions, we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
- The first half-reaction is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al}(s) \quad E^\circ = -1.66 \, \text{V} \]
- The second half-reaction is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s) \quad E^\circ = +0.34 \, \text{V} \]

### Step 2: Determine which half-reaction will undergo oxidation and which will undergo reduction.
- The half-reaction with the more negative standard electrode potential will act as the anode (oxidation). Here, aluminum (Al) is oxidized:
\[ \text{Al}(s) \rightarrow \text{Al}^{3+} + 3e^- \]
- The half-reaction with the more positive standard electrode potential will act as the cathode (reduction). Here, copper (Cu) is reduced:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s) \]

### Step 3: Balance the number of electrons transferred in the half-reactions.
- The oxidation of aluminum produces 3 electrons, while the reduction of copper consumes 2 electrons. To balance the electrons, we can multiply the aluminum half-reaction by 2 and the copper half-reaction by 3:
\[ 2 \times \left(\text{Al}(s) \rightarrow \text{Al}^{3+} + 3e^-\right) \quad \text{(oxidation)} \]
\[ 3 \times \left(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)\right) \quad \text{(reduction)} \]

### Step 4: Write the balanced overall reaction.
- The balanced overall reaction will be:
\[ 2\text{Al}(s) + 3\text{Cu}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Cu}(s) \]

### Step 5: Calculate the standard cell potential \(E^\circ_{\text{cell}}\).
- The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Here, \(E^\circ_{\text{cathode}} = 0.34 \, \text{V}\) (for Cu) and \(E^\circ_{\text{anode}} = -1.66 \, \text{V}\) (for Al):
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} - (-1.66 \, \text{V}) = 0.34 \, \text{V} + 1.66 \, \text{V} = 2.00 \, \text{V} \]

### Conclusion:
The voltage produced under standard conditions by combining the half-reactions is:
\[ \boxed{2.00 \, \text{V}} \]