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

Options

A. There were `2.00 mol ` of `Zn^(2+)` but only `1.00 mol ` of `Cu^(2+)`
B. The `Zn` electrode had twice the surface of the `Cu` electrode.
C. The `[Zn^(2+)]` was larger than the `[Cu^(2+)]`.
D. The volume of the `Zn^(2+)` solution was larger than the volume of the `Cu^(2+)` solution.

Explanation

To solve the problem regarding the electrochemical cell represented as `Zn(s)|Zn^(2+)||Cu^(2+)|Cu(s)`, we need to analyze the information given and apply the Nernst equation to understand the observed cell voltage.

### Step-by-Step Solution:

1. **Identify the Standard Cell Voltage (E°cell)**:
- The standard cell voltage for the given cell is provided as \( E^\circ_{cell} = 1.10 \, V \).

2. **Measured Cell Voltage (Ecell)**:
- The measured cell voltage when the cell was prepared in the lab is \( E_{cell} = 0.98 \, V \).

3. **Understanding the Nernst Equation**:
- The Nernst equation relates the cell potential under non-standard conditions to the standard cell potential:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
where \( n \) is the number of moles of electrons transferred in the reaction, and \( Q \) is the reaction quotient.

4. **Determine the Number of Electrons (n)**:
- In the given cell, the half-reaction for zinc is:
\[ Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \]
- The half-reaction for copper is:
\[ Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \]
- Therefore, \( n = 2 \) (2 electrons are transferred).

5. **Substituting Values into the Nernst Equation**:
- Rearranging the Nernst equation to find \( Q \):
\[ 0.98 = 1.10 - \frac{0.0591}{2} \log Q \]
- Rearranging gives:
\[ 0.12 = \frac{0.0591}{2} \log Q \]
- Multiplying both sides by 2:
\[ 0.24 = 0.0591 \log Q \]
- Dividing by 0.0591:
\[ \log Q = \frac{0.24}{0.0591} \approx 4.06 \]

6. **Calculating Q**:
- To find \( Q \):
\[ Q = 10^{4.06} \approx 11500 \]
- The reaction quotient \( Q \) is defined as:
\[ Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} \]

7. **Interpreting the Result**:
- Since \( Q \) is significantly greater than 1, it indicates that the concentration of \( Zn^{2+} \) is much higher than that of \( Cu^{2+} \). This could explain why the measured cell voltage is lower than the standard cell voltage.

8. **Conclusion**:
- One possible explanation for the observed voltage of 0.98 V (which is lower than the standard voltage of 1.10 V) is that the concentration of \( Zn^{2+} \) is significantly greater than the concentration of \( Cu^{2+} \).