The cell `Zn//Zn^(2+) (1 M) ||Cu^(2+)(1M)//Cu(E_("cell")^(0) = 1.10 V)`was allowed to be completely discharged at 298 K. The relative concentration of `Zn^(2+)` to `Cu^(2+) (([Zn^(2+)])/([Cu^(2+)]))` is `10^(x)`. The value of x is :
(take `(2.303 RT)/(F) = 0.059` round off your answer up to one decimal)

Answer

To solve the problem, we need to find the value of \( x \) in the expression \( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 10^x \) after the cell has been completely discharged.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The cell reaction can be written as:
\[ \text{Zn (s)} + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu (s)} \]

2. **Understand the Condition of Complete Discharge**:
When the cell is completely discharged, the cell potential \( E_{\text{cell}} \) becomes zero.

3. **Use the Nernst Equation**:
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{2.303RT}{nF} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
Here, \( E^{\circ}_{\text{cell}} = 1.10 \, \text{V} \), \( n = 2 \) (since 2 electrons are involved), and at 298 K, we are given \( \frac{2.303RT}{F} = 0.059 \).

4. **Set Up the Equation for Complete Discharge**:
Since \( E_{\text{cell}} = 0 \) at complete discharge:
\[ 0 = 1.10 - \frac{0.059}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]

5. **Rearrange the Equation**:
Rearranging gives:
\[ \frac{0.059}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) = 1.10 \]
Multiplying both sides by \( \frac{2}{0.059} \):
\[ \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) = \frac{1.10 \times 2}{0.059} \]

6. **Calculate the Right Side**:
Calculate \( \frac{1.10 \times 2}{0.059} \):
\[ \frac{2.20}{0.059} \approx 37.288 \]

7. **Find the Concentration Ratio**:
Thus,
\[ \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \approx 37.288 \]
Taking the antilog gives:
\[ \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \approx 10^{37.288} \]

8. **Determine \( x \)**:
Since \( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 10^x \), we have \( x \approx 37.288 \).

9. **Round Off**:
Rounding \( x \) to one decimal place gives:
\[ x \approx 37.3 \]

### Final Answer:
The value of \( x \) is \( 37.3 \).