`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 act as the cathode and which as the anode.
- The half-reaction with the higher standard reduction potential will occur at the cathode (reduction), while the one with the lower standard reduction potential will occur at the anode (oxidation).
- Here, since \(E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V}\) is greater than \(E^\circ_{\text{Al}^{3+}/\text{Al}} = -1.66 \, \text{V}\), copper will be reduced at the cathode and aluminum will be oxidized at the anode.

### Step 3: Write the oxidation half-reaction for aluminum.
- The oxidation half-reaction for aluminum is the reverse of the reduction reaction:
\[ \text{Al}(s) \rightarrow \text{Al}^{3+} + 3e^- \quad E^\circ = +1.66 \, \text{V} \, (\text{reverse sign}) \]

### Step 4: Calculate the standard cell potential \(E^\circ_{\text{cell}}\).
- The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Al}^{3+}/\text{Al}} = 0.34 \, \text{V} - (-1.66 \, \text{V}) \]
- This simplifies to:
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} + 1.66 \, \text{V} = 2.00 \, \text{V} \]

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