For the cell, `Zn(s)abs(Zn^(2+))abs(Cu^(2+))Cu(s)`, the standard cell voltage, `E^(0)""_(cell)` is `1.10` V. When a cell using these reagents was prepared in the lab, the measured cell voltage was `0.98 ` One possible explanation for the observed voltage is :

Options

A. there were `2.00` mole of `Cu^(2+)` but only `1.00` mol of `Zn^(2+)`
B. the Zn electrode had twice the surface area as that of the Cu electrode
C. the `[Zn^(2+)]` concentration was larger than the `[Cu^(2+)]` concentration
D. the volume of the `Zn^(2+)` ion solution was greater than the volume of the `Cu^(2+)` ion solution.

Answer

To solve the problem, we need to analyze the electrochemical cell given and understand the relationship between the standard cell voltage, the measured cell voltage, and the concentrations of the ions involved.

### Step-by-Step Solution:

1. **Identify the Cell Reactions**:
- The cell is represented as `Zn(s) | Zn^(2+) || Cu^(2+) | Cu(s)`.
- In this cell, zinc (Zn) is oxidized to zinc ions (Zn²⁺), and copper ions (Cu²⁺) are reduced to copper (Cu).
- The half-reactions are:
- Oxidation: \( Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \)
- Reduction: \( Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \)

2. **Standard Cell Voltage**:
- The standard cell voltage \( E^\circ_{cell} \) is given as 1.10 V.

3. **Measured Cell Voltage**:
- The measured cell voltage \( E_{cell} \) is 0.98 V, which is less than the standard cell voltage.

4. **Nernst Equation**:
- The Nernst equation relates the cell potential to the concentrations of the reactants and products:
\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q \]
- Where \( Q \) is the reaction quotient, given by:
\[ Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} \]
- Here, \( n = 2 \) (the number of electrons transferred).

5. **Understanding the Decrease in Voltage**:
- Since \( E_{cell} < E^\circ_{cell} \), it implies that the term \( \frac{RT}{nF} \ln Q \) must be positive, leading to a decrease in the cell potential.
- This can occur if the concentration of \( Zn^{2+} \) is greater than that of \( Cu^{2+} \), which would make \( Q \) greater than 1, resulting in a positive \( \ln Q \).

6. **Conclusion**:
- The observed lower voltage can be explained by the concentration of \( Zn^{2+} \) being larger than that of \( Cu^{2+} \). Thus, the correct explanation for the observed voltage is that the concentration of zinc ions is greater than the concentration of copper ions.

### Final Answer:
The correct explanation for the observed voltage is that the concentration of zinc ions (\( Zn^{2+} \)) was larger than the concentration of copper ions (\( Cu^{2+} \)).