Calculate the EMF of the electrode concentration cell Hg-Zn `(c_(1)M)|(Zn^(2+)(c M)|` Hg-Zn `(c_(2)M)` at `25^(@)C`, if the concentration of the zinc amalgam are 2 g per 100 g of mercury and 1g per 100 g of mercury in anode and cathode half-cell respectively.

Options

A. `6.8xx10^(-2)V`
B. `8.8xx10^(-3)V`
C. `5.7 xx 10^(-2)V`
D. `7.8xx10^(-3)V`

Explanation

To calculate the EMF of the electrode concentration cell \( \text{Hg-Zn} \) at \( 25^\circ C \), we will follow these steps:

### Step 1: Identify the half-cell reactions
For the concentration cell, we have two half-cells:
- **Anode (oxidation)**: \( \text{Zn} \) (from \( C_1 \)) is oxidized to \( \text{Zn}^{2+} \).
- **Cathode (reduction)**: \( \text{Zn}^{2+} \) (from \( C_2 \)) is reduced to \( \text{Zn} \).

The half-cell reactions can be written as:
- Anode: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \) (oxidation)
- Cathode: \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) (reduction)

### Step 2: Determine the concentrations
Given:
- For the anode, the concentration of zinc amalgam is \( 2 \, \text{g} \) per \( 100 \, \text{g} \) of mercury.
- For the cathode, the concentration of zinc amalgam is \( 1 \, \text{g} \) per \( 100 \, \text{g} \) of mercury.

To find the molarity, we need to convert grams to moles using the molar mass of zinc (\( 65.4 \, \text{g/mol} \)):
- Moles of zinc in anode:
\[ C_1 = \frac{2 \, \text{g}}{65.4 \, \text{g/mol}} = 0.0306 \, \text{mol} \]
- Moles of zinc in cathode:
\[ C_2 = \frac{1 \, \text{g}}{65.4 \, \text{g/mol}} = 0.0153 \, \text{mol} \]

### Step 3: Apply the Nernst equation
The Nernst equation for the cell is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[C_2]}{[C_1]} \right) \]
Where:
- \( n = 2 \) (number of electrons transferred)
- \( E^\circ_{\text{cell}} = 0 \, \text{V} \) (for concentration cells)

Substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 0 - \frac{0.0591}{2} \log \left( \frac{C_2}{C_1} \right) \]
\[ E_{\text{cell}} = -0.02955 \log \left( \frac{0.0153}{0.0306} \right) \]

### Step 4: Calculate the logarithm
Calculating the ratio:
\[ \frac{C_2}{C_1} = \frac{0.0153}{0.0306} = 0.5 \]
Now, calculate the logarithm:
\[ \log(0.5) \approx -0.301 \]

### Step 5: Substitute back into the equation
Substituting the logarithm back into the Nernst equation:
\[ E_{\text{cell}} = -0.02955 \times (-0.301) \approx 0.00888 \, \text{V} \, \text{or} \, 8.8 \, \text{mV} \]

### Final Answer
The EMF of the electrode concentration cell is approximately \( 8.8 \, \text{mV} \).

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