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 represented by the notation `Ni|Ni^(2+)(1.0M)||Ag^(+)(1.0M)|Ag`, we can follow these steps:

### Step 1: Identify the Anode and Cathode
- In the given cell notation, the left side represents the anode and the right side represents the cathode.
- Anode: `Ni|Ni^(2+)`
- Cathode: `Ag^(+)|Ag`

### Step 2: Write the Half-Reactions
- **Anode (Oxidation)**: The oxidation reaction at the anode involves nickel (Ni) losing electrons:
\[ Ni \rightarrow Ni^{2+} + 2e^- \]
- **Cathode (Reduction)**: The reduction reaction at the cathode involves silver ions (Ag^+) gaining electrons:
\[ Ag^+ + e^- \rightarrow Ag \]

### Step 3: Identify Standard Reduction Potentials
- From the question, we have the following standard reduction potentials:
- For `Ni^(2+)//Ni`: \( E^\circ = -0.25 \, \text{V} \)
- For `Ag^(+)//Ag`: \( E^\circ = 0.80 \, \text{V} \)

### Step 4: Calculate the Standard EMF of the Cell
- The standard EMF of the cell can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
- Plugging in the values:
\[ E^\circ_{cell} = E^\circ_{Ag} - E^\circ_{Ni} = 0.80 \, \text{V} - (-0.25 \, \text{V}) \]
- Simplifying this gives:
\[ E^\circ_{cell} = 0.80 \, \text{V} + 0.25 \, \text{V} = 1.05 \, \text{V} \]

### Step 5: Conclusion
- The EMF of the cell is \( 1.05 \, \text{V} \).

### Final Answer
The standard EMF of the cell is \( 1.05 \, \text{V} \).

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