The EMF of a Daniel cell at 298K is `E_(1)`. The cell is: `Zn(s)abs( ZnSO_(4) (0.01M)) | abs(CuSO_(4)(1.0M)) Cu(s).` When the concentration of `ZnSO_(4)` is 1.0 M and that of `CuSO_(4)` is `0.01` M the emf changes to `E_(2)`. What is the relationship between `E_(1) and E_(2)`?

Options

A. `E_(2) = 0 ne E_(1)`
B. `E_(1) gt E_(2)`
C. `E_(1) lt E_(2)`
D. `E_(1) = E_(2)`

Answer

To determine the relationship between the EMF of a Daniel cell at different concentrations of the electrolyte solutions, we can utilize the Nernst equation. The Nernst equation relates the cell potential (E) to the standard electrode potential (E°) and the concentrations of the reactants and products involved in the electrochemical reaction.

### Step-by-Step Solution:

1. **Identify the Standard Electrode Potential (E°)**:
For the Daniel cell, the standard electrode potential (E°) is given as 1.1 V.

2. **Write the Nernst Equation**:
The Nernst equation is given by:
\[ E = E° - \frac{0.0591}{n} \log \left( \frac{[\text{reduced form}]}{[\text{oxidized form}]} \right) \]
where \( n \) is the number of moles of electrons transferred in the half-reaction.

3. **Determine the Half-Reactions**:
- Oxidation at the anode:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- Reduction at the cathode:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
Here, \( n = 2 \).

4. **Calculate E₁** (when [ZnSO₄] = 0.01 M and [CuSO₄] = 1.0 M):
\[ E_1 = E° - \frac{0.0591}{2} \log \left( \frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]} \right) \]
Substituting the concentrations:
\[ E_1 = 1.1 - \frac{0.0591}{2} \log \left( \frac{1.0}{0.01} \right) \]
\[ E_1 = 1.1 - \frac{0.0591}{2} \log(100) \]
\[ E_1 = 1.1 - \frac{0.0591}{2} \times 2 \]
\[ E_1 = 1.1 - 0.0591 \]
\[ E_1 = 1.0409 \, \text{V} \approx 1.041 \, \text{V} \]

5. **Calculate E₂** (when [ZnSO₄] = 1.0 M and [CuSO₄] = 0.01 M):
\[ E_2 = E° - \frac{0.0591}{2} \log \left( \frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]} \right) \]
Substituting the concentrations:
\[ E_2 = 1.1 - \frac{0.0591}{2} \log \left( \frac{0.01}{1.0} \right) \]
\[ E_2 = 1.1 - \frac{0.0591}{2} \log(0.01) \]
\[ E_2 = 1.1 - \frac{0.0591}{2} \times (-2) \]
\[ E_2 = 1.1 + 0.0591 \]
\[ E_2 = 1.1591 \, \text{V} \approx 1.159 \, \text{V} \]

6. **Compare E₁ and E₂**:
From the calculations:
- \( E_1 \approx 1.041 \, \text{V} \)
- \( E_2 \approx 1.159 \, \text{V} \)

Therefore, we find that:
\[ E_1 < E_2 \]

### Conclusion:
The relationship between \( E_1 \) and \( E_2 \) is:
\[ E_1 < E_2 \]