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 solution step by step.

### Step 1: Identify the half-cell reactions
From the cell notation `M|underset(0.01)(M^(2+))||underset(0.0001)(M^(2+))|M`, we can identify the half-reactions:

- **Anode (oxidation)**:
\[ M \rightarrow M^{2+} + 2e^- \]

- **Cathode (reduction)**:
\[ M^{2+} + 2e^- \rightarrow M \]

### Step 2: Write the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q \]
Where:
- \(E_{cell}^0\) = standard emf of the cell (given as 4 V)
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin (25°C = 298 K)
- \(n\) = number of moles of electrons transferred (2 in this case)
- \(F\) = Faraday's constant
- \(Q\) = reaction quotient

### Step 3: Substitute the values into the Nernst equation
We are given that \(\frac{RT}{F} \ln 10 = 0.06\). Therefore, we can rewrite the Nernst equation as:
\[ E_{cell} = E_{cell}^0 - \frac{0.059}{n} \log_{10} Q \]

### Step 4: Calculate the reaction quotient \(Q\)
The reaction quotient \(Q\) for the cell reaction is given by:
\[ Q = \frac{[M^{2+}]_{cathode}}{[M^{2+}]_{anode}} = \frac{0.0001}{0.01} = 0.01 \]

### Step 5: Substitute values into the Nernst equation
Now substituting the known values into the Nernst equation:
\[ E_{cell} = 4 - \frac{0.059}{2} \log_{10}(0.01) \]

### Step 6: Calculate \(\log_{10}(0.01)\)
\[ \log_{10}(0.01) = -2 \]

### Step 7: Substitute \(\log_{10}(0.01)\) into the equation
Now substituting this back:
\[ E_{cell} = 4 - \frac{0.059}{2} \times (-2) \]

### Step 8: Simplify the equation
\[ E_{cell} = 4 + 0.059 \]
\[ E_{cell} = 4 - 0.059 = 3.94 \text{ V} \]

### Final Answer
The emf of the cell at 25°C is:
\[ \boxed{3.94 \text{ V}} \]