Calculate the `EMF` of a Daniel cell when the concentartion of `ZnSO_(4)` and `CuSO_(4)` are `0.001M` and `0.1M` respectively. The standard potential of the cell is `1.1V`.

Explanation

To calculate the EMF of a Daniel cell with given concentrations of ZnSO₄ and CuSO₄, we will use the Nernst equation. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Standard Potential (E°)**:
The standard potential of the Daniel cell is given as:
\[ E° = 1.1 \, \text{V} \]

2. **Write the Nernst Equation**:
The Nernst equation is given by:
\[ E = E° - \frac{0.059}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]
For the Daniel cell, the half-reactions are:
- Oxidation: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
- Reduction: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)

Here, \( n = 2 \) (the number of electrons transferred).

3. **Identify the Concentrations**:
The concentrations are given as:
- \([\text{Zn}^{2+}] = 0.001 \, \text{M} = 10^{-3} \, \text{M}\)
- \([\text{Cu}^{2+}] = 0.1 \, \text{M} = 10^{-1} \, \text{M}\)

4. **Substitute into the Nernst Equation**:
Substitute the values into the Nernst equation:
\[ E = 1.1 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} \]
\[ E = 1.1 - \frac{0.059}{2} \log \frac{10^{-3}}{10^{-1}} \]

5. **Calculate the Logarithm**:
Simplify the logarithm:
\[ \log \frac{10^{-3}}{10^{-1}} = \log (10^{-3} \times 10^{1}) = \log (10^{-2}) = -2 \]

6. **Substitute the Logarithm Value**:
Now substitute this value back into the equation:
\[ E = 1.1 - \frac{0.059}{2} \times (-2) \]
\[ E = 1.1 + 0.059 \]

7. **Final Calculation**:
\[ E = 1.159 \, \text{V} \]

### Final Answer:
The EMF of the Daniel cell is:
\[ E = 1.159 \, \text{V} \]