Consider the cell `:`
`Cd(s)|Cd^(2+)(1.0M)||Cu^(2+)(1.0M)|Cu(s)`
If we wish to make a cell with a more positive voltage using the same substances, we should

Options

A. Increase both `[Cd^(2+)]` and `[Cu^(2+)]` to `2.0M`
B. Decrease the `[Cd^(2+]` to `0.1 M
C. Increase the `[Cu^(2+)]` to `2.0M`
D. Decrease both the `[Cd^(2+)]` and `[Cu^(2+)]` to `0.01M`

Explanation

To solve the problem, we need to analyze the electrochemical cell given and understand how the concentrations of the ions affect the cell potential (E_cell). The cell in question is:

\[ \text{Cd(s)} | \text{Cd}^{2+}(1.0M) || \text{Cu}^{2+}(1.0M) | \text{Cu(s)} \]

### Step 1: Identify the half-reactions
In this electrochemical cell:
- At the anode (oxidation), cadmium (Cd) is oxidized to cadmium ions (Cd²⁺):
\[ \text{Cd(s)} \rightarrow \text{Cd}^{2+} + 2e^- \]
- At the cathode (reduction), copper ions (Cu²⁺) are reduced to copper (Cu):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)} \]

### Step 2: Write the overall cell reaction
The overall cell reaction can be obtained by combining the half-reactions:
\[ \text{Cd(s)} + \text{Cu}^{2+} \rightarrow \text{Cd}^{2+} + \text{Cu(s)} \]

### Step 3: Use the Nernst equation
The Nernst equation relates the cell potential (E_cell) to the standard cell potential (E°_cell) and the concentrations of the reactants and products:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \left( \frac{[\text{Cd}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
Where:
- \( n \) is the number of moles of electrons transferred (which is 2 in this case).

### Step 4: Analyze how to increase E_cell
To increase the cell potential (E_cell), we can manipulate the concentrations of the ions involved:
1. **Increase the concentration of Cu²⁺**: This will decrease the value of the logarithmic term since Cu²⁺ is in the denominator.
2. **Decrease the concentration of Cd²⁺**: This will also decrease the value of the logarithmic term since Cd²⁺ is in the numerator.

### Step 5: Evaluate the options
Now let's evaluate the provided options:
- **Increase both Cd²⁺ and Cu²⁺ concentrations to 2M**: This will have no net effect on the logarithmic term since both concentrations increase equally.
- **Decrease Cd²⁺ concentration to 0.1M**: This will decrease the logarithmic term, increasing E_cell.
- **Increase Cu²⁺ concentration to 2M**: This will decrease the logarithmic term, increasing E_cell.
- **Decrease both Cd²⁺ and Cu²⁺ concentrations to 0.01M**: This will have no net effect on the logarithmic term since both concentrations decrease equally.

### Conclusion
The options that will result in a more positive voltage (higher E_cell) are:
- Decrease Cd²⁺ concentration to 0.1M
- Increase Cu²⁺ concentration to 2M

### Final Answer
To achieve a more positive voltage using the same substances, we should:
- Decrease the Cd²⁺ concentration to 0.1M or increase the Cu²⁺ concentration to 2M.