Calculate the emf of the following cell at `25^(@)C`
`Ag(s)|AgNO_(3) (0.01" mol "Kg^(-1))| AgNO_(3) ("0.05 mole "Kg^(-1))|Ag(s)`

Options

A. 0.046V
B. 0.828V
C. `-0.046V`
D. 0.46V

Explanation

To calculate the EMF of the given cell at 25°C, we will use the Nernst equation. The Nernst equation is given by:

\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \left( \frac{[Ox]}{[Red]} \right) \]

Where:
- \( E_{cell} \) is the cell potential (EMF).
- \( E^0_{cell} \) is the standard cell potential.
- \( n \) is the number of electrons transferred in the half-reaction.
- \([Ox]\) is the concentration of the oxidized species.
- \([Red]\) is the concentration of the reduced species.

### Step 1: Identify the half-reactions
In the given cell:
- At the anode (oxidation), silver (Ag) is oxidized to Ag⁺:
\[ \text{Ag(s)} \rightarrow \text{Ag}^+ + e^- \]

- At the cathode (reduction), Ag⁺ is reduced to silver (Ag):
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag(s)} \]

### Step 2: Determine the standard cell potential
For the silver half-reaction, the standard reduction potential \( E^0 \) for Ag⁺/Ag is approximately +0.80 V. Since both half-reactions involve the same species, the overall standard cell potential \( E^0_{cell} \) is:
\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.80 V - 0.80 V = 0 V \]

### Step 3: Determine the concentrations
From the problem, we have:
- Concentration of Ag⁺ at the anode: \( [Ag^+] = 0.01 \, \text{mol/kg} \)
- Concentration of Ag⁺ at the cathode: \( [Ag^+] = 0.05 \, \text{mol/kg} \)

### Step 4: Calculate the EMF using the Nernst equation
Substituting the values into the Nernst equation:
- \( n = 1 \) (one electron is transferred)
- \( E^0_{cell} = 0 \, V \)
- \([Ox] = 0.01 \, \text{mol/kg}\) (anode concentration)
- \([Red] = 0.05 \, \text{mol/kg}\) (cathode concentration)

Now, substituting these values into the Nernst equation:
\[ E_{cell} = 0 - \frac{0.0591}{1} \log \left( \frac{0.01}{0.05} \right) \]

Calculating the logarithm:
\[ \log \left( \frac{0.01}{0.05} \right) = \log(0.2) \approx -0.699 \]

Now substituting this back into the equation:
\[ E_{cell} = 0 - 0.0591 \times (-0.699) \approx 0.041309 \, V \]

### Step 5: Final answer
Thus, the EMF of the cell is approximately:
\[ E_{cell} \approx 0.042 \, V \]