If hydrogen electrodes dipped in two solutions of `pH=3` and `pH=6` are connected by a salt bridge, the `EMF_(cell)` is

Options

A. `0.052V`
B. `0.104V`
C. `0.177V`
D. `0.3V`

Explanation

To find the EMF of the cell formed by hydrogen electrodes dipped in two solutions of different pH values, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the pH Values**:
- pH of solution 1 (pH1) = 3
- pH of solution 2 (pH2) = 6

2. **Calculate the Concentration of H⁺ Ions**:
- The concentration of H⁺ ions can be calculated using the formula:
\[ [H^+] = 10^{-\text{pH}} \]
- For pH1 (3):
\[ [H^+]_1 = 10^{-3} \, \text{M} \]
- For pH2 (6):
\[ [H^+]_2 = 10^{-6} \, \text{M} \]

3. **Use the Nernst Equation**:
- The Nernst equation for the hydrogen electrode is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[H^+]_1}{[H^+]_2} \right) \]
- Here, \(E^\circ_{\text{cell}} = 0\) V for hydrogen electrodes, and \(n = 1\) (since 1 mole of electrons is involved in the half-reaction).

4. **Substitute the Values into the Nernst Equation**:
- Substitute the concentrations into the equation:
\[ E_{\text{cell}} = 0 - \frac{0.0591}{1} \log \left( \frac{10^{-3}}{10^{-6}} \right) \]

5. **Simplify the Logarithm**:
- The logarithm simplifies as follows:
\[ \log \left( \frac{10^{-3}}{10^{-6}} \right) = \log(10^{3}) = 3 \]

6. **Calculate the EMF**:
- Now substitute back into the equation:
\[ E_{\text{cell}} = -0.0591 \times 3 \]
- Calculate the value:
\[ E_{\text{cell}} = -0.1773 \, \text{V} \]
- Since we are interested in the magnitude, we take the positive value:
\[ E_{\text{cell}} = 0.177 \, \text{V} \]

### Final Answer:
The EMF of the cell is \(0.177 \, \text{V}\).

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