The electrochemical cell is set up as pt,
`H_(2)(1 atm)| HCl (0.1 M)|| CH_(3)COOH (0.1 M) | H_(2) (1 atm)`, pt. The emf of the cell is (Ka `=10^(-3) xx 1.8`)

Options

A. 0
B. Positive
C. Negative
D. Cannot be determined

Explanation

To determine the EMF of the electrochemical cell given by the notation:

\[ \text{Pt, } H_2(1 \text{ atm}) | \text{HCl (0.1 M)} || \text{CH}_3\text{COOH (0.1 M)} | H_2(1 \text{ atm}), \text{ Pt} \]

we will follow these steps:

### Step 1: Identify the Anode and Cathode Reactions

In the given cell, we have hydrogen gas (H₂) at both the anode and cathode. The half-reactions can be identified as follows:

- **Anode Reaction (Oxidation)**:
\[ H_2 \rightarrow 2H^+ + 2e^- \]

- **Cathode Reaction (Reduction)**:
\[ 2H^+ + 2e^- \rightarrow H_2 \]

### Step 2: Write the Overall Cell Reaction

Combining the half-reactions, we get the overall cell reaction:

\[ H_2 + 2H^+ \rightarrow 2H^+ + H_2 \]

This simplifies to:

\[ H_2 + 2H^+ \rightarrow H_2 + 2H^+ \]

### Step 3: Determine the Concentration of \( H^+ \)

- **HCl** is a strong acid and completely dissociates in solution. Therefore, the concentration of \( H^+ \) from HCl will be 0.1 M.

- **Acetic Acid (CH₃COOH)** is a weak acid and does not fully dissociate. Given \( K_a = 1.8 \times 10^{-3} \), we can find the concentration of \( H^+ \) produced from acetic acid using the formula:

\[ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \]

Assuming \( x \) is the concentration of \( H^+ \) produced, we have:

\[ K_a = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} \quad (\text{since } x \text{ is small}) \]

Thus,

\[ 1.8 \times 10^{-3} = \frac{x^2}{0.1} \]

Solving for \( x \):

\[ x^2 = 1.8 \times 10^{-4} \implies x = \sqrt{1.8 \times 10^{-4}} \approx 0.0134 \text{ M} \]

### Step 4: Apply the Nernst Equation

The Nernst equation is given by:

\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[H^+]_{\text{cathode}}^2}{[H^+]_{\text{anode}}^2} \right) \]

Here, \( n = 2 \) (number of electrons transferred).

Substituting the concentrations:

\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{2} \log \left( \frac{(0.0134)^2}{(0.1)^2} \right) \]

### Step 5: Calculate the Logarithmic Term

Calculating the logarithmic term:

\[ \log \left( \frac{(0.0134)^2}{(0.1)^2} \right) = \log \left( \frac{0.00017956}{0.01} \right) = \log(0.017956) \approx -1.72 \]

### Step 6: Substitute Back into the Nernst Equation

Substituting back into the Nernst equation:

\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{2} \times (-1.72) \]

Assuming \( E^0_{\text{cell}} \) is 0 (standard hydrogen electrode), we get:

\[ E_{\text{cell}} = 0 + 0.02955 \times 1.72 \approx 0.0509 \text{ V} \]

### Conclusion

The EMF of the cell is approximately **0.0509 V**.

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