Two concentration cells of Ag with Ag electrode in `AgNO_(3)` in first cell concentration of one electrode is 1M and other electrode is 0.1 M and emf is 0.065 V in second cell concentration of one electrode is 1 M and other electrode is 0.01 M calculate the emf of second cell

Options

A. 0.12 V
B. 0.06 V
C. 0.09 V
D. 0.16 V

Explanation

To calculate the EMF of the second concentration cell, we can follow these steps:

### Step 1: Understand the Concentration Cells
In a concentration cell, the electrodes are made of the same material (in this case, Ag) and the EMF is generated due to the difference in concentration of the electrolyte. The half-reaction for silver can be represented as:
- At the anode: \( \text{Ag} \rightarrow \text{Ag}^+ + e^- \)
- At the cathode: \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)

### Step 2: Use the Nernst Equation
The Nernst equation for a concentration cell is given by:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
Where:
- \( E^\circ \) is the standard EMF (which is 0 for a concentration cell since both electrodes are the same),
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin (assumed to be 298 K),
- \( n \) is the number of moles of electrons transferred (1 for silver),
- \( F \) is Faraday's constant (96485 C/mol),
- \( Q \) is the reaction quotient.

### Step 3: Calculate the Reaction Quotient \( Q \)
For the first cell with concentrations of 1 M and 0.1 M:
\[ Q_1 = \frac{[\text{Ag}^+]_{\text{anode}}}{[\text{Ag}^+]_{\text{cathode}}} = \frac{0.1}{1} = 0.1 \]
For the second cell with concentrations of 1 M and 0.01 M:
\[ Q_2 = \frac{[\text{Ag}^+]_{\text{anode}}}{[\text{Ag}^+]_{\text{cathode}}} = \frac{0.01}{1} = 0.01 \]

### Step 4: Calculate the EMF for the First Cell
Given that the EMF for the first cell is 0.065 V, we can use the Nernst equation:
\[ 0.065 = 0 - \frac{(8.314)(298)}{(1)(96485)} \ln(0.1) \]
Calculating the constant:
\[ \frac{(8.314)(298)}{(1)(96485)} \approx 0.025693 \]
Now, substituting into the equation:
\[ 0.065 = -0.025693 \ln(0.1) \]
Solving for \( \ln(0.1) \):
\[ \ln(0.1) \approx -2.3026 \]
Thus:
\[ 0.065 = 0.025693 \times 2.3026 \approx 0.059 \]
This confirms our first cell EMF.

### Step 5: Calculate the EMF for the Second Cell
Now, we apply the same Nernst equation for the second cell:
\[ E_2 = 0 - \frac{(8.314)(298)}{(1)(96485)} \ln(0.01) \]
Calculating:
\[ \ln(0.01) \approx -4.6052 \]
Substituting into the equation:
\[ E_2 = -0.025693 \times (-4.6052) \approx 0.118 \]
Thus, the EMF for the second cell is approximately:
\[ E_2 \approx 0.12 \, \text{V} \]

### Final Answer
The EMF of the second cell is approximately **0.12 V**.

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