Galvanic cells generate electrical energy at the expense of a spontaneous redox reaction. In an electrode concentration cell two like electrodes having different concentrations, either because they are gas electrodes operating at different pressures or because they are amalgams (solutions in mercury) with different concentrations are dipped into the same solution.
Eg. An example is a cell composed of two chlorine electrodes with different pressure of `Cl_(2)`:
`Pt_(L)|Cl_(2)(P_(L))|HCl(aq)|Cl_(2)(P_(R))|Pt_(R)`
Where `P_(L)` and `P_(R)` are the `Cl_(2)` pressure at the left and right electrodes.
Calculate the EMF of the electrode concentration cell represented by :
`Hg-Zn(c_(1))|Zn^(+2)(aq)|Hg-Zn(c_(2))`
At `25^(@)C.c_(1)=2g` of `Zn` per `100g` of `Hg` and `c_(2)=1g` of `Zn` per `50g` of `Hg`

Options

A. `0.059V`
B. `3`
C. `0`
D. data insufficient

Explanation

To calculate the EMF of the electrode concentration cell represented by `Hg-Zn(c_(1))|Zn^(+2)(aq)|Hg-Zn(c_(2))`, we will follow these steps:

### Step 1: Understand the Concentration Cell
In a concentration cell, the EMF is generated due to the difference in concentration of the same species on both sides of the cell. Here, we have zinc in mercury amalgam at two different concentrations.

### Step 2: Identify the Given Concentrations
- For the left electrode (anode), `c_(1) = 2g` of `Zn` per `100g` of `Hg`.
- For the right electrode (cathode), `c_(2) = 1g` of `Zn` per `50g` of `Hg`.

### Step 3: Convert Concentrations to Molarity
To compare the concentrations, we need to express them in the same unit. We can convert the given concentrations into molarity (moles per liter).

1. **Calculate the molar mass of zinc (Zn)**:
- Molar mass of Zn = 65.38 g/mol.

2. **Calculate the concentration at the anode (c1)**:
- `c_(1) = (2 g Zn / 100 g Hg) * (1 mol Zn / 65.38 g Zn) = (2/100) * (1/65.38) mol Zn per g Hg`.
- Since we need to express it in terms of moles per liter, we can assume the density of mercury is approximately 13.6 g/cm³, which is 13600 g/L.
- Therefore, the concentration in molarity is:
\[ c_{1} = \frac{(2/100) \times (1/65.38)}{(100/13600)} = \frac{2 \times 13600}{100 \times 65.38} \text{ mol/L} \]

3. **Calculate the concentration at the cathode (c2)**:
- `c_(2) = (1 g Zn / 50 g Hg) * (1 mol Zn / 65.38 g Zn) = (1/50) * (1/65.38) mol Zn per g Hg`.
- Similarly, using the density of mercury:
\[ c_{2} = \frac{(1/50) \times (1/65.38)}{(50/13600)} = \frac{1 \times 13600}{50 \times 65.38} \text{ mol/L} \]

### Step 4: Apply the Nernst Equation
The Nernst equation for a concentration cell is given by:
\[ E_{cell} = E^{\circ} + \frac{RT}{nF} \ln \left( \frac{[Zn^{2+}]_{anode}}{[Zn^{2+}]_{cathode}} \right) \]
Where:
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( T = 298 \, K \) (25°C)
- \( n = 2 \) (for Zn)
- \( F = 96485 \, C/mol \)

Since both electrodes are zinc, \( E^{\circ} = 0 \).

### Step 5: Calculate the EMF
Substituting the values:
\[ E_{cell} = 0 + \frac{(8.314)(298)}{2(96485)} \ln \left( \frac{c_{1}}{c_{2}} \right) \]

### Step 6: Evaluate the Concentration Ratio
Since we have already calculated the concentrations:
- \( c_{1} = \frac{2 \times 13600}{100 \times 65.38} \)
- \( c_{2} = \frac{1 \times 13600}{50 \times 65.38} \)

Now, simplify the ratio \( \frac{c_{1}}{c_{2}} \):
\[ \frac{c_{1}}{c_{2}} = \frac{(2/100)}{(1/50)} = \frac{2 \times 50}{100 \times 1} = 1 \]

### Step 7: Conclusion
Since \( \ln(1) = 0 \), the EMF becomes:
\[ E_{cell} = 0 \]

### Final Answer
The EMF of the electrode concentration cell is **0 V**.

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