At temperature of 298 K the emf of the following electrochemical cell `Ag(s) | Ag^(+) (0.1 M)|| Zn^(2+) (0.1 M) |Zn(s)` will be (given: `E_("cell")^(@)= -1.562 V)`)

Options

A. `-1.532 V`
B. `-1.503 V`
C. `1.532 V`
D. `-3.06 V`

Explanation

To find the emf of the electrochemical cell `Ag(s) | Ag^(+) (0.1 M) || Zn^(2+) (0.1 M) | Zn(s)` at a temperature of 298 K, we will use the Nernst equation. The Nernst equation is given by:

\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]

Where:
- \( E_{\text{cell}} \) is the emf of the cell.
- \( E^\circ_{\text{cell}} \) is the standard cell potential.
- \( n \) is the number of moles of electrons transferred in the reaction.
- \( Q \) is the reaction quotient.

### Step 1: Identify the half-reactions
The half-reactions for the cell are:
1. For the silver electrode (reduction):
\[ Ag^+ + e^- \rightarrow Ag \quad (E^\circ = +0.80 \, V) \]
2. For the zinc electrode (oxidation):
\[ Zn \rightarrow Zn^{2+} + 2e^- \quad (E^\circ = -0.76 \, V) \]

### Step 2: Calculate the standard cell potential
The overall standard cell potential \( E^\circ_{\text{cell}} \) is calculated as follows:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 \, V - (-0.76 \, V) = 1.56 \, V \]

### Step 3: Determine the number of electrons transferred
From the half-reactions, we see that 2 electrons are transferred in the overall reaction. Therefore, \( n = 2 \).

### Step 4: Calculate the reaction quotient \( Q \)
The reaction quotient \( Q \) for the cell can be expressed as:
\[ Q = \frac{[Zn^{2+}]}{[Ag^+]^2} \]
Given that both concentrations are 0.1 M:
\[ Q = \frac{0.1}{(0.1)^2} = \frac{0.1}{0.01} = 10 \]

### Step 5: Apply the Nernst equation
Now we can substitute the values into the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
Substituting \( E^\circ_{\text{cell}} = 1.56 \, V \), \( n = 2 \), and \( Q = 10 \):
\[ E_{\text{cell}} = 1.56 \, V - \frac{0.0591}{2} \log(10) \]
Since \( \log(10) = 1 \):
\[ E_{\text{cell}} = 1.56 \, V - \frac{0.0591}{2} \cdot 1 \]
Calculating:
\[ E_{\text{cell}} = 1.56 \, V - 0.02955 \, V = 1.53045 \, V \]

### Step 6: Final answer
Thus, the emf of the cell at 298 K is approximately:
\[ E_{\text{cell}} \approx 1.53 \, V \]