The `EMF` of concentration cell consisting of two zinc electrodes, one dipping into `M//4` solution of `ZnSO_(4)` and the other into `M//16` solution of the same salt at `25^(@)C` is

Options

A. `0.0125V`
B. `0.0250V`
C. `0.0178V`
D. `0.0356V`

Explanation

To find the EMF of a concentration cell consisting of two zinc electrodes, one dipping into a \( M/4 \) solution of \( ZnSO_4 \) and the other into a \( M/16 \) solution of the same salt at \( 25^\circ C \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Cell Type**:
A concentration cell consists of two electrodes made of the same material (in this case, zinc) but immersed in solutions of different concentrations of the same electrolyte.

2. **Standard Reduction Potential**:
Since both electrodes are zinc, the standard reduction potential (\( E^\circ \)) for both electrodes is the same. Therefore, the overall standard cell potential (\( E^\circ_{cell} \)) is zero:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0 \]

3. **Nernst Equation**:
The EMF of the cell can be calculated using the Nernst equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q \]
where \( n \) is the number of electrons transferred, and \( Q \) is the reaction quotient.

4. **Determine \( n \)**:
For the zinc half-reaction:
\[ Zn^{2+} + 2e^- \rightarrow Zn \]
The number of electrons transferred (\( n \)) is 2.

5. **Calculate Reaction Quotient \( Q \)**:
The reaction quotient \( Q \) for the concentration cell is given by:
\[ Q = \frac{[Zn^{2+}]_{anode}}{[Zn^{2+}]_{cathode}} \]
Here, the anode is the electrode in the \( M/16 \) solution and the cathode is in the \( M/4 \) solution:
\[ Q = \frac{M/16}{M/4} = \frac{1/16}{1/4} = \frac{1}{4} \]

6. **Substitute into Nernst Equation**:
Now substituting the values into the Nernst equation:
\[ E_{cell} = 0 - \frac{0.059}{2} \log \left(\frac{1}{4}\right) \]
Simplifying further:
\[ E_{cell} = -0.0295 \log \left(\frac{1}{4}\right) \]

7. **Calculate \( \log \left(\frac{1}{4}\right) \)**:
We know that:
\[ \log \left(\frac{1}{4}\right) = -\log(4) = -\log(2^2) = -2\log(2) \approx -2 \times 0.301 = -0.602 \]

8. **Final Calculation**:
Now substituting this value back:
\[ E_{cell} = -0.0295 \times (-0.602) \approx 0.0178 \, \text{V} \]

### Final Answer:
The EMF of the concentration cell is approximately \( 0.0178 \, \text{V} \).