The cell , `Zn | Zn^(2+) (1M) || Cu^(2+) (1M) Cu (E_("cell")^@ = 1. 10 V)`,
Was allowed to be completely discharfed at `298 K `. The relative concentration of `2+` to `Cu^(2+) [(Zn^(2=))/(Cu^(2+))]` is :

Options

A. (1:1 )
B. (3:1)
C. (3:2)
D. (5:3)

Explanation

To solve the problem, we need to determine the relative concentration of \( \text{Zn}^{2+} \) to \( \text{Cu}^{2+} \) after the cell has been completely discharged. We will use the Nernst equation and the information provided in the question.

### Step-by-Step Solution:

1. **Identify the Components of the Cell:**
The cell is given as \( \text{Zn} | \text{Zn}^{2+} (1M) || \text{Cu}^{2+} (1M) | \text{Cu} \). Here, zinc is oxidized and copper is reduced.

2. **Write the Half-Reactions:**
- Oxidation (Anode):
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- Reduction (Cathode):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

3. **Overall Cell Reaction:**
Combining the half-reactions gives:
\[ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \]

4. **Nernst Equation:**
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
Where:
- \( E^{\circ}_{\text{cell}} = 1.10 \, \text{V} \)
- \( n = 2 \) (number of electrons transferred)

5. **Condition for Complete Discharge:**
When the cell is completely discharged, \( E_{\text{cell}} = 0 \). Therefore, we can set the Nernst equation to zero:
\[ 0 = 1.10 - \frac{0.0591}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]

6. **Rearranging the Equation:**
Rearranging gives:
\[ \frac{0.0591}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) = 1.10 \]
\[ \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) = \frac{2 \times 1.10}{0.0591} \]

7. **Calculating the Logarithm:**
\[ \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) = \frac{2.20}{0.0591} \approx 37.24 \]

8. **Finding the Ratio:**
To find the ratio, we take the antilog:
\[ \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 10^{37.24} \]

### Final Result:
The relative concentration of \( \text{Zn}^{2+} \) to \( \text{Cu}^{2+} \) is approximately:
\[ \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \approx 10^{37.24} \]