The EMF of a concentration cell consisting of two zinc electrodes, one dipping into `M/4` sol. Of zinc sulphate and the other into `M/6` sol. Of the same salt at `25^(@)`C. is:

Options

A. `0.0125V`
B. `0.0250V`
C. `0.0052V`
D. `0.9178V`

Answer

To calculate the EMF of a concentration cell consisting of two zinc electrodes, we can follow these steps:

### Step 1: Identify the concentrations
We have two zinc electrodes:
- Anode: Dipping into M/4 solution of zinc sulfate
- Cathode: Dipping into M/6 solution of zinc sulfate

### Step 2: Write the half-reactions
The oxidation and reduction half-reactions for zinc are:
- Oxidation (at the anode):
\[ \text{Zn (s)} \rightarrow \text{Zn}^{2+} + 2e^{-} \]
- Reduction (at the cathode):
\[ \text{Zn}^{2+} + 2e^{-} \rightarrow \text{Zn (s)} \]

### Step 3: Determine the reaction quotient (Q)
The reaction quotient \( Q \) for the concentration cell can be expressed as:
\[ Q = \frac{[\text{Zn}^{2+}]_{\text{anode}}}{[\text{Zn}^{2+}]_{\text{cathode}}} \]
Substituting the concentrations:
\[ Q = \frac{M/4}{M/6} = \frac{6}{4} = \frac{3}{2} \]

### Step 4: Use the Nernst equation
The Nernst equation for the cell is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log Q \]
For a concentration cell, \( E^{\circ}_{\text{cell}} = 0 \) (since both electrodes are the same material).

### Step 5: Substitute values into the Nernst equation
Here, \( n = 2 \) (the number of electrons transferred in the half-reaction):
\[ E_{\text{cell}} = 0 - \frac{0.0591}{2} \log \left(\frac{3}{2}\right) \]

### Step 6: Calculate the logarithm
Calculate \( \log \left(\frac{3}{2}\right) \):
\[ \log \left(\frac{3}{2}\right) \approx 0.1761 \]

### Step 7: Substitute the logarithm back into the equation
Now substituting back:
\[ E_{\text{cell}} = -\frac{0.0591}{2} \times 0.1761 \]
\[ E_{\text{cell}} = -0.0052 \, \text{V} \]

### Step 8: Determine the final EMF
Since the EMF is typically expressed as a positive value, we take the absolute value:
\[ E_{\text{cell}} = 0.0052 \, \text{V} \]

### Final Answer
The EMF of the concentration cell is \( 0.0052 \, \text{V} \).

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