For the cell:
`Cu(10g)|underset((C_(1)))(CuSO_(4))"solution""||"underset((C_(2)))(ZnSO_(4))"solution""|"Zn(10g),E_("cell")=EV`
`E_("cell")` for the cell : `Cu(20g)|underset((C_(1)))(CuSO_(4))"solution""||"underset((C_(2)))(ZnSO_(4))"solution""|"Zn(20g)`, is

Options

A. E volt
B. 2E volt
C. `E/2` volt
D. 1.1 volt

Answer

To solve the problem, we need to analyze the electrochemical cell represented and understand how the changes in the weight of the electrodes affect the cell potential (E_cell).

### Step-by-Step Solution:

1. **Identify the Cell Components**:
The cell is represented as:
\[ \text{Cu(10g)} | \text{CuSO}_4 \text{(C}_1\text{)} || \text{ZnSO}_4 \text{(C}_2\text{)} | \text{Zn(10g)} \]
Here, we have copper (Cu) and zinc (Zn) electrodes with their respective weights and concentrations of their sulfate solutions.

2. **Understand the Effect of Weight on E_cell**:
The problem states that the weight of the copper and zinc electrodes is doubled in the second cell:
\[ \text{Cu(20g)} | \text{CuSO}_4 \text{(C}_1\text{)} || \text{ZnSO}_4 \text{(C}_2\text{)} | \text{Zn(20g)} \]
However, the concentrations \(C_1\) and \(C_2\) remain unchanged.

3. **Use the Nernst Equation**:
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
In this case, the concentrations of the solutions are the same for both cells, which means that the logarithmic term will not change.

4. **Conclusion on E_cell**:
Since the concentrations \(C_1\) and \(C_2\) are the same in both scenarios, the E_cell will remain unchanged despite the increase in the weight of the electrodes. Therefore, we can conclude:
\[ E_{\text{cell}} \text{ for the second cell} = E_{\text{cell}} \text{ for the first cell} \]
Hence, if the E_cell for the first cell is \(E_V\), the E_cell for the second cell will also be \(E_V\).

### Final Answer:
The E_cell for the cell with \(Cu(20g)\) and \(Zn(20g)\) is \(E_V\).