The emf of the cell, `Ni|Ni^(2+)(1.0M)||Ag^(+)(1.0M)|Ag [E^(@)` for `Ni^(2+)//Ni =- 0.25` volt, `E^(@)` for `Ag^(+)//Ag = 0.80` volt] is given by : `[E^(@)` for `Ag^(+)//Ag = 0.80` volt]

Options

A. `-0.25 +0.80 = 0.55` volt
B. `-0.25 -(+0.80) =- 1.05` volt
C. `0 +0.80 -(-0.25)=+ 1.05` volt
D. `-0.80 -(+0.25) =- 0.55` volt

Explanation

To find the EMF of the cell composed of Nickel and Silver, we will follow these steps:

### Step 1: Identify the half-reactions
- **Oxidation half-reaction**: Nickel (Ni) is oxidized to Nickel ions (Ni²⁺).
\[ \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \]
- **Reduction half-reaction**: Silver ions (Ag⁺) are reduced to Silver (Ag).
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

### Step 2: Determine the standard reduction potentials
- Given:
- \( E^\circ \) for \( \text{Ni}^{2+} / \text{Ni} = -0.25 \, \text{V} \)
- \( E^\circ \) for \( \text{Ag}^+ / \text{Ag} = +0.80 \, \text{V} \)

### Step 3: Identify the anode and cathode
- The anode is where oxidation occurs, which is Nickel in this case.
- The cathode is where reduction occurs, which is Silver.

### Step 4: Write the equation for the EMF of the cell
The EMF of the cell can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}} - E^\circ_{\text{Ni}} = 0.80 \, \text{V} - (-0.25 \, \text{V}) \]

### Step 5: Calculate the EMF
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} + 0.25 \, \text{V} = 1.05 \, \text{V} \]

### Step 6: Conclusion
The EMF of the cell is \( 1.05 \, \text{V} \). This positive value indicates that the reaction is spontaneous.