Calculate the cell potential for the cell,
`Ni(s)//Ni^(2+)(0.1 M) || Ag^(+)(0.1 M)//Ag(s)`
`["Given",E_(Ag^(+)//Ag)^(@)=+0.80" V ", E_(Ni^(2+)//Ni)^(@)=-0.25" V ", R=8.314 JK^(-1) mol^(-1),F=96500 C mol^(-1)]`

Explanation

To calculate the cell potential for the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials.
- The cell is composed of two half-cells:
- Nickel half-cell: \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}(s) \) with \( E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.25 \, \text{V} \)
- Silver half-cell: \( \text{Ag}^+ + e^- \rightarrow \text{Ag}(s) \) with \( E^\circ_{\text{Ag}^+/Ag} = +0.80 \, \text{V} \)

### Step 2: Determine which half-reaction is the anode and which is the cathode.
- The half-reaction with the higher standard reduction potential will occur at the cathode, while the one with the lower standard reduction potential will occur at the anode.
- In this case:
- Cathode: Silver half-cell (higher potential, \( +0.80 \, \text{V} \))
- Anode: Nickel half-cell (lower potential, \( -0.25 \, \text{V} \))

### Step 3: Use the formula for cell potential.
- The standard cell potential \( E^\circ_{\text{cell}} \) is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

### Step 4: Substitute the values into the formula.
- Substitute the values of the standard reduction potentials into the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^+/Ag} - E^\circ_{\text{Ni}^{2+}/Ni} \]
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} - (-0.25 \, \text{V}) \]

### Step 5: Perform the calculation.
- Simplifying the equation:
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} + 0.25 \, \text{V} = 1.05 \, \text{V} \]

### Final Answer:
- The cell potential for the given electrochemical cell is \( E^\circ_{\text{cell}} = 1.05 \, \text{V} \).

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