The `E_(Cell)^(@)=1.18V` for `Zn(s)||Zn^(+2)(1M)||Cu^(+2)(1M)|Cu(s)`. The value of `x` if when excess granulated zinc is added to `1M Cu^(+2)` solution the `[Cu^(+2)]_(eq)` becomes `10^(-x)M` is `(T=298 K ,(2.303RT)/(F)=0.059)`

Options

A. 40
B. 30
C. 20
D. 10

Explanation

To solve the given problem, we need to analyze the electrochemical cell and apply the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write the half-reactions
The half-reactions for the cell are:
- At the anode (oxidation):
\[ \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \]
- At the cathode (reduction):
\[ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \]

### Step 2: Write the overall cell reaction
Combining the two half-reactions gives us the overall cell reaction:
\[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \]

### Step 3: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E_{cell}^0 - \frac{0.059}{n} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]
Where:
- \(E_{cell}^0 = 1.18 \, V\)
- \(n = 2\) (number of electrons transferred)

### Step 4: Set up the equation at equilibrium
At equilibrium, \(E_{cell} = 0\) (since the cell is at equilibrium). Thus, we can set up the equation:
\[ 0 = 1.18 - \frac{0.059}{2} \log \frac{[1]}{[\text{Cu}^{2+}]} \]
Given that \([\text{Cu}^{2+}] = 10^{-x}\), we can substitute this into the equation:
\[ 0 = 1.18 - \frac{0.059}{2} \log \frac{1}{10^{-x}} \]

### Step 5: Simplify the logarithm
The logarithm simplifies as follows:
\[ \log \frac{1}{10^{-x}} = \log(10^x) = x \]
Thus, our equation becomes:
\[ 0 = 1.18 - \frac{0.059}{2} x \]

### Step 6: Solve for \(x\)
Rearranging the equation gives:
\[ \frac{0.059}{2} x = 1.18 \]
\[ x = \frac{1.18 \times 2}{0.059} \]
Calculating this gives:
\[ x = \frac{2.36}{0.059} \approx 40 \]

### Final Answer
Thus, the value of \(x\) is:
\[ \boxed{40} \]