There are two principal types of electrochemical cells. A galvanic cell is an electrochemical cell that produces electricity as a result of spontaneous reaction occurring inside it. An electrolytic cell is an electrochemical cell in which a non-spontaneous reaction is driven by an external source of current. any redox reaction may be expressed in terms of two half reactions which are conceptual reactions showing the lowss and gain of electrons. each half reaction has a difinite value of standard electrode potential. the overall reaction is represented by a universally accepted method. knowing the standard electrode potential of the half reactions, the standard EMF of the cell can be calculted. the standard EMF further helps in the calculation of free energy change, equilibrium constant of the cell reaction as well as parameters like solublity products of a sparingly soluble salt. a cell can also be set up in which the two electrodes may be of the same (type, e.g., both may be hydrogen electrodes but the concentration of `H^(+)` ions in the two solutions may be different. Such cells are called concentration cells.
Q. If hydrogen electrodees dipped in two solutions of pH=3 and pH=6 are connected by a salt bridge, the emf of the resulting cell is

Options

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

Explanation

To solve the problem of calculating the EMF of a concentration cell made of hydrogen electrodes in solutions of pH 3 and pH 6, we can follow these steps:

### Step-by-Step Solution:

1. **Understand pH and H⁺ Concentration**:
- The pH of a solution is related to the concentration of hydrogen ions (H⁺) by the formula:
\[ \text{pH} = -\log[H^+] \]
- Therefore, we can find the concentration of H⁺ ions in each solution.

2. **Calculate H⁺ Concentration for pH 3**:
- For pH 3:
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \text{ M} \]

3. **Calculate H⁺ Concentration for pH 6**:
- For pH 6:
\[ [H^+] = 10^{-\text{pH}} = 10^{-6} \text{ M} \]

4. **Identify the Standard EMF Formula**:
- The EMF (E) of the concentration cell can be calculated using the Nernst equation:
\[ E = \frac{0.0591}{n} \log \left( \frac{C_2}{C_1} \right) \]
- Here, \(C_1\) is the concentration of H⁺ in the more concentrated solution (pH 3), and \(C_2\) is the concentration of H⁺ in the less concentrated solution (pH 6). Since both electrodes are hydrogen electrodes, \(n = 1\).

5. **Substitute Values into the Formula**:
- Substitute \(C_1 = 10^{-3}\) M and \(C_2 = 10^{-6}\) M into the equation:
\[ E = 0.0591 \log \left( \frac{10^{-6}}{10^{-3}} \right) \]

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

7. **Calculate the EMF**:
- Now substitute back into the equation:
\[ E = 0.0591 \times (-3) = -0.1773 \text{ V} \]
- Since EMF is expressed as a positive value, we take the absolute value:
\[ E = 0.1773 \text{ V} \]

### Final Answer:
The EMF of the resulting cell is approximately **0.177 V**.