The standard electrode potentials for the reactions,
`Ag^(+) (a.q) + e^(-) rarr Ag (s)`
`Sn^(2+) (a.q) + 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

Answer

To find the emf (electromotive force) 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.

The half-reactions given are:

1. For silver:
\[ \text{Ag}^{+} (aq) + e^{-} \rightarrow \text{Ag} (s) \quad E^\circ = +0.80 \, \text{V} \]

2. For tin:
\[ \text{Sn}^{2+} (aq) + 2e^{-} \rightarrow \text{Sn} (s) \quad E^\circ = -0.14 \, \text{V} \]

### Step 2: Determine the oxidation and reduction reactions.

In the cell, tin (Sn) will be oxidized and silver (Ag) will be reduced. Therefore, we need to reverse the tin half-reaction to represent oxidation:

\[ \text{Sn} (s) \rightarrow \text{Sn}^{2+} (aq) + 2e^{-} \quad E^\circ = +0.14 \, \text{V} \quad (\text{oxidation potential}) \]

### Step 3: Calculate the overall cell potential (emf).

The overall cell potential (emf) can be calculated using the formula:

\[ E_{cell} = E_{reduction} - E_{oxidation} \]

Here, the reduction potential for silver is \( +0.80 \, \text{V} \) and the oxidation potential for tin is \( +0.14 \, \text{V} \).

Substituting the values:

\[ E_{cell} = E_{Ag} - E_{Sn} = 0.80 \, \text{V} - 0.14 \, \text{V} \]

\[ E_{cell} = 0.80 \, \text{V} + 0.14 \, \text{V} = 0.94 \, \text{V} \]

### Final Answer:
The emf of the cell is \( 0.94 \, \text{V} \).

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