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 (electromotive force) of the cell formed by hydrogen electrodes dipped in solutions of pH 3 and pH 6, we can follow these steps:

### Step 1: Understand the pH values
The pH of a solution is related to the concentration of hydrogen ions \([H^+]\) in the solution. The relationship is given by:
\[ [H^+] = 10^{-\text{pH}} \]
For pH 3:
\[ [H^+] = 10^{-3} \, \text{M} \]
For pH 6:
\[ [H^+] = 10^{-6} \, \text{M} \]

### Step 2: Identify the electrodes
In this setup:
- The electrode in the pH 3 solution will act as the anode (oxidation occurs here).
- The electrode in the pH 6 solution will act as the cathode (reduction occurs here).

### Step 3: Use the Nernst equation
The Nernst equation for a hydrogen electrode can be expressed as:
\[ E = E^\circ - \frac{0.059}{n} \log \frac{[H^+]_{\text{anode}}}{[H^+]_{\text{cathode}}} \]
For hydrogen electrodes, \(E^\circ = 0\) V and \(n = 2\) (since 2 electrons are involved in the half-reaction).

### Step 4: Calculate the EMF of the cell
Substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 0 - \frac{0.059}{2} \log \frac{[H^+]_{\text{anode}}}{[H^+]_{\text{cathode}}} \]
Substituting the concentrations:
\[ E_{\text{cell}} = -\frac{0.059}{2} \log \frac{10^{-3}}{10^{-6}} \]
Calculating the log term:
\[ \log \frac{10^{-3}}{10^{-6}} = \log(10^{3}) = 3 \]
Now substituting this back into the equation:
\[ E_{\text{cell}} = -\frac{0.059}{2} \cdot 3 \]
\[ E_{\text{cell}} = -0.0885 \, \text{V} \]

### Step 5: Finalize the EMF value
Since the EMF is negative, we take the absolute value to express it as a positive potential:
\[ E_{\text{cell}} = 0.0885 \, \text{V} \]

### Conclusion
The EMF of the cell is approximately **0.0885 V**.

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