The standard `EMF` of decinormal calomel electrode is `0.268 V`. The `EMF` is

Options

A. `-0.298V`
B. `0.327V`
C. `-0.327V`
D. `0.298V`

Explanation

To solve the problem regarding the EMF of a decinormal calomel electrode, we will follow these steps:

### Step 1: Understand the Given Information
The standard EMF of the calomel electrode is given as 0.268 V. A decinormal solution means that the concentration of the solution is 0.1 N (10^-1 N).

### Step 2: Write the Nernst Equation
The Nernst equation relates the EMF of a cell to its standard EMF and the concentrations of the reactants and products:
\[ E_{cell} = E^0_{cell} - \frac{0.059}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]
where:
- \(E_{cell}\) is the EMF of the cell.
- \(E^0_{cell}\) is the standard EMF of the cell.
- \(n\) is the number of electrons transferred in the half-reaction.

### Step 3: Identify the Half-Reaction
For the calomel electrode, the half-reaction can be represented as:
\[ \text{Hg}_2\text{Cl}_2 + 2e^- \rightleftharpoons 2\text{Hg} + 2\text{Cl}^- \]
Here, \(n = 2\) because 2 electrons are involved in the half-reaction.

### Step 4: Substitute Values into the Nernst Equation
Given:
- \(E^0_{cell} = 0.268 \, V\)
- Concentration of Cl^- ions in a decinormal solution is \(0.1 \, M\).

Substituting into the Nernst equation:
\[ E_{cell} = 0.268 - \frac{0.059}{2} \log \frac{[\text{Cl}^-]^2}{1} \]
Since the concentration of Cl^- ions is \(0.1 \, M\), we can write:
\[ E_{cell} = 0.268 - \frac{0.059}{2} \log (0.1^2) \]

### Step 5: Calculate the Logarithm
Calculating the logarithm:
\[ \log (0.1^2) = \log (10^{-2}) = -2 \]
So, we have:
\[ E_{cell} = 0.268 - \frac{0.059}{2} \times (-2) \]

### Step 6: Simplify the Equation
Now, simplifying the equation:
\[ E_{cell} = 0.268 + \frac{0.059}{2} \times 2 \]
\[ E_{cell} = 0.268 + 0.059 \]
\[ E_{cell} = 0.327 \, V \]

### Final Answer
The EMF of the decinormal calomel electrode is \(0.327 \, V\).

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