EMF of the given cell
`A_((s))|A_((aq))^(2+)||B_((aq))^(2+)|B_(s)`
Given `E_(A//A^(2+))^(@):+1.4V " and "E_(B//B^(2+))^(@):-1.4V`.

Options

A. 2.8 V
B. 1.8 V
C. 0 V
D. `-1.8` V

Answer

To calculate the EMF of the given cell represented as `A_(s)|A_(aq)^(2+)||B_(aq)^(2+)|B_(s)`, we will follow these steps:

### Step 1: Identify the half-reactions
For the cell notation `A_(s)|A_(aq)^(2+)||B_(aq)^(2+)|B_(s)`, we can identify the half-reactions:
- **Oxidation half-reaction**: \( A \rightarrow A^{2+} + 2e^- \)
- **Reduction half-reaction**: \( B^{2+} + 2e^- \rightarrow B \)

### Step 2: Determine the standard reduction potentials
We are given the standard reduction potentials:
- For A: \( E^{\circ}_{A/A^{2+}} = +1.4 \, V \) (this is the reduction potential)
- For B: \( E^{\circ}_{B/B^{2+}} = -1.4 \, V \) (this is the reduction potential)

### Step 3: Convert oxidation potential to reduction potential
Since A is being oxidized, we need to convert its oxidation potential to a reduction potential:
- Oxidation potential of A is \( -E^{\circ}_{A/A^{2+}} = -1.4 \, V \)

### Step 4: Identify the anode and cathode
- **Anode** (oxidation occurs): A
- **Cathode** (reduction occurs): B

### Step 5: Calculate the EMF of the cell
Using the formula for EMF of the cell:
\[ E_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]
Substituting the values:
\[ E_{cell} = E^{\circ}_{B/B^{2+}} - (-E^{\circ}_{A/A^{2+}}) \]
\[ E_{cell} = (-1.4 \, V) - (-1.4 \, V) \]
This simplifies to:
\[ E_{cell} = 1.4 \, V - (-1.4 \, V) = 1.4 \, V + 1.4 \, V = 2.8 \, V \]

### Final Answer
The EMF of the cell is \( 2.8 \, V \).

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