The standard electrode potential for the two electrode `A^(+)//A and B^(+)//B` are respectively 0.5V and 0.75V. The emf of the given cell `A|A^(+)(a=1)||B^(+)(a=1)|` B will be

Options

A. 1.25V
B. `-1.25V`
C. `-0.25V`
D. `0.25V`

Explanation

To calculate the EMF (Electromotive Force) of the given electrochemical cell `A|A^(+)(a=1)||B^(+)(a=1)|B`, we can follow these steps:

### Step 1: Identify the standard electrode potentials
The standard electrode potentials for the two electrodes are given as:
- For electrode A: \( E^\circ_A = 0.5 \, \text{V} \)
- For electrode B: \( E^\circ_B = 0.75 \, \text{V} \)

### Step 2: Determine the anode and cathode
In an electrochemical cell:
- The anode is where oxidation occurs.
- The cathode is where reduction occurs.

In this cell, since A is on the left and B is on the right, we can assign:
- Anode: A (oxidation)
- Cathode: B (reduction)

### Step 3: Write the formula for EMF
The EMF of the cell can be calculated using the formula:
\[ \text{EMF} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

### Step 4: Substitute the values into the formula
Substituting the values we have:
\[ \text{EMF} = E^\circ_B - E^\circ_A \]
\[ \text{EMF} = 0.75 \, \text{V} - 0.5 \, \text{V} \]

### Step 5: Calculate the EMF
Now, perform the subtraction:
\[ \text{EMF} = 0.25 \, \text{V} \]

### Final Answer
The EMF of the given cell is \( 0.25 \, \text{V} \).

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