The standard electrode potential for the reactions,
`Ag^(+)(aq)+e^(-) rarr Ag(s)`
`Sn^(2+)(aq)+2e^(-) rarr Sn(s)`
at `25^(@)C` are `0.80 ` volt and `-0.14 ` volt, respectively. The `emf`of the cell `Sn|Sn^(2+)(1M)||Ag^(+)(1M)Ag` is `:`

Options

A. `0.66 `volt
B. `0.80 ` volt
C. `1.08` volt
D. `0.94` volt

Explanation

To calculate the emf of the cell represented by the reaction \( \text{Sn} | \text{Sn}^{2+}(1M) || \text{Ag}^{+}(1M) | \text{Ag} \), we will follow these steps:

### Step 1: Identify the Half-Reactions and Their Standard Electrode Potentials
We have two half-reactions:
1. \( \text{Ag}^{+}(aq) + e^{-} \rightarrow \text{Ag}(s) \) with a standard electrode potential \( E^\circ = +0.80 \, \text{V} \)
2. \( \text{Sn}^{2+}(aq) + 2e^{-} \rightarrow \text{Sn}(s) \) with a standard electrode potential \( E^\circ = -0.14 \, \text{V} \)

### Step 2: Determine the Oxidation and Reduction Reactions
In the cell, the Sn electrode is oxidized (loses electrons) and the Ag ion is reduced (gains electrons). Therefore, we need to reverse the Sn half-reaction:
- Oxidation: \( \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq) + 2e^{-} \)
- Reduction: \( \text{Ag}^{+}(aq) + e^{-} \rightarrow \text{Ag}(s) \)

### Step 3: Adjust the Half-Reactions for Electron Transfer
Since the oxidation of Sn involves 2 electrons and the reduction of Ag involves 1 electron, we need to multiply the Ag half-reaction by 2 to balance the number of electrons:
- Oxidation: \( \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq) + 2e^{-} \)
- Reduction: \( 2\text{Ag}^{+}(aq) + 2e^{-} \rightarrow 2\text{Ag}(s) \)

### Step 4: Write the Overall Cell Reaction
Combining the two half-reactions gives us:
\[ \text{Sn}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Sn}^{2+}(aq) + 2\text{Ag}(s) \]

### Step 5: Calculate the Standard Cell Potential (emf)
The standard cell potential \( E^\circ_{cell} \) can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} \]

From our half-reactions:
- \( E^\circ_{reduction} \) (for Ag) = \( +0.80 \, \text{V} \)
- \( E^\circ_{oxidation} \) (for Sn, reversed) = \( -(-0.14 \, \text{V}) = +0.14 \, \text{V} \)

Now substituting these values:
\[ E^\circ_{cell} = 0.80 \, \text{V} + 0.14 \, \text{V} = 0.94 \, \text{V} \]

### Final Answer
The emf of the cell \( \text{Sn} | \text{Sn}^{2+}(1M) || \text{Ag}^{+}(1M) | \text{Ag} \) is \( 0.94 \, \text{V} \).

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