Acetic acid has `K_a=1.8xx10^(-5)` while formic acid has `K_(a)=2.1xx10^(-4)`.What would be the magnitude of the emf of the cell
`Pt(H_(2))|0.1 M` acetic acid+
0.1 M sodium acetate ||0.1 M formic acid +
0.1 M sodium formate|`Pt(H_(2))` at `25^(@)C` ?

Options

A. 0.032 volt
B. 0.063 volt
C. 0.0456 volt
D. 0.055 volt

Explanation

To determine the magnitude of the EMF of the cell described in the question, we can follow these steps:

### Step 1: Identify the components of the cell
The cell is composed of two half-cells:
1. **Anode**: Platinum (Pt) electrode in a solution of 0.1 M acetic acid and 0.1 M sodium acetate.
2. **Cathode**: Platinum (Pt) electrode in a solution of 0.1 M formic acid and 0.1 M sodium formate.

### Step 2: Write the relevant equilibrium expressions
For acetic acid (weak acid):
- The dissociation can be represented as:
\[ \text{CH}_3\text{COOH} \leftrightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \]
- The equilibrium constant \( K_a \) for acetic acid is given as \( 1.8 \times 10^{-5} \).

For formic acid (weak acid):
- The dissociation can be represented as:
\[ \text{HCOOH} \leftrightarrow \text{HCOO}^- + \text{H}^+ \]
- The equilibrium constant \( K_a \) for formic acid is given as \( 2.1 \times 10^{-4} \).

### Step 3: Calculate the cell potential using the Nernst equation
The Nernst equation for calculating the EMF of the cell is:
\[ E_{cell} = E^\circ - \frac{0.059}{n} \log \frac{[H^+]_{cathode}}{[H^+]_{anode}} \]
Where:
- \( E^\circ \) is the standard electrode potential (which we can assume to be 0 for the hydrogen electrode).
- \( n \) is the number of electrons transferred (which is 1 in this case).

### Step 4: Calculate the concentrations of \( [H^+] \)
Using the \( K_a \) values, we can find the \( [H^+] \) concentrations for both acids:
1. For acetic acid:
\[ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \implies 1.8 \times 10^{-5} = \frac{x \cdot 0.1}{0.1 - x} \approx \frac{x \cdot 0.1}{0.1} \implies x \approx 1.8 \times 10^{-5} \text{ M} \]
Thus, \( [H^+]_{anode} = 1.8 \times 10^{-5} \text{ M} \).

2. For formic acid:
\[ K_a = \frac{[H^+][HCOO^-]}{[HCOOH]} \implies 2.1 \times 10^{-4} = \frac{y \cdot 0.1}{0.1 - y} \approx \frac{y \cdot 0.1}{0.1} \implies y \approx 2.1 \times 10^{-4} \text{ M} \]
Thus, \( [H^+]_{cathode} = 2.1 \times 10^{-4} \text{ M} \).

### Step 5: Substitute values into the Nernst equation
Now substituting the values into the Nernst equation:
\[ E_{cell} = 0 - \frac{0.059}{1} \log \left( \frac{2.1 \times 10^{-4}}{1.8 \times 10^{-5}} \right) \]
Calculating the logarithm:
\[ \frac{2.1 \times 10^{-4}}{1.8 \times 10^{-5}} = 11.67 \implies \log(11.67) \approx 1.067 \]
Thus,
\[ E_{cell} = -0.059 \cdot 1.067 \approx -0.063 \text{ V} \]
The magnitude of the EMF is:
\[ |E_{cell}| \approx 0.063 \text{ V} \]

### Final Answer
The magnitude of the EMF of the cell is approximately **0.063 V**.

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