Calculate emf of cell at `25^(@)c`
Cell notation.
`M|underset(0.01)(M^(2+))||underset(0.0001)(M^(2+))|M`
if value of `E_(cell)^(0)` is 4 volt (given `(RT)/(F)` in 10=0.06)

Options

A. 3.94 volt
B. 4.06 volt
C. 2.03 volt
D. 8.18 volt

Explanation

To calculate the EMF of the cell at 25°C, we will use the Nernst equation. Let's break down the steps.

### Step 1: Identify the components of the cell notation
The cell notation given is:
\[ M | \underset{0.01}{M^{2+}} || \underset{0.0001}{M^{2+}} | M \]

This indicates:
- Anode (oxidation): \( M \) is oxidized to \( M^{2+} \) with a concentration of \( 0.01 \, M \).
- Cathode (reduction): \( M^{2+} \) is reduced to \( M \) with a concentration of \( 0.0001 \, M \).

### Step 2: Write the Nernst equation
The Nernst equation is given by:
\[ E = E^{\circ}_{cell} - \frac{0.059}{n} \log \left( \frac{[M^{2+}]_{cathode}}{[M^{2+}]_{anode}} \right) \]

Where:
- \( E^{\circ}_{cell} = 4 \, V \) (given)
- \( n = 2 \) (since 2 electrons are involved in the half-reaction)

### Step 3: Substitute the values into the Nernst equation
Substituting the values into the Nernst equation:
- \( [M^{2+}]_{cathode} = 0.0001 \, M \)
- \( [M^{2+}]_{anode} = 0.01 \, M \)

Thus, we have:
\[ E = 4 - \frac{0.059}{2} \log \left( \frac{0.0001}{0.01} \right) \]

### Step 4: Simplify the logarithmic term
Calculating the logarithm:
\[ \frac{0.0001}{0.01} = 10^{-4} / 10^{-2} = 10^{-2} \]
So,
\[ \log(10^{-2}) = -2 \]

### Step 5: Substitute back into the equation
Now substituting back:
\[ E = 4 - \frac{0.059}{2} \times (-2) \]
\[ E = 4 + 0.059 \]

### Step 6: Calculate the final EMF
Calculating the final value:
\[ E = 4 + 0.059 = 4.059 \, V \]

### Final Answer
The EMF of the cell at 25°C is approximately:
\[ \boxed{4.059 \, V} \]

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