Given that (at T = 298 K)
`Cu(s) | Cu^(2+)(1.0M) || Ag^(+)(1.0M)| Ag(s) underset(E_(cell)^(@) = 0.46V`
`Zn(s) | Zn^(2+) (1.0M) || Cu^(2+) (1.0M) | Cu(s) underset(E_(cell)^(@) = 1.10V`
Then `E_(cell)` for,
`Zn | Zn^(2+)(0.1M) || Ag^(+)(0.1M) | Ag` at 298 K will be:

Options

A. `1.59V`
B. `1.53V`
C. `2.53V`
D. Cannot be calculated due to the insuficient data

Explanation

To find the cell potential \( E_{cell} \) for the cell reaction \( \text{Zn | Zn}^{2+}(0.1M) || \text{Ag}^{+}(0.1M) | \text{Ag} \) at 298 K, we will follow these steps:

### Step 1: Determine the Standard Cell Potential \( E^\circ_{cell} \)

We have two half-cell reactions with their standard potentials given:

1. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with \( E^\circ = 0.46 \, \text{V} \)
2. \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) with \( E^\circ = 1.10 \, \text{V} \)

The overall standard cell potential for the reaction involving Zn and Ag can be calculated as follows:

\[ E^\circ_{cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

In our case, the cathode reaction is the reduction of \( \text{Ag}^{+} \) to \( \text{Ag} \) and the anode reaction is the oxidation of \( \text{Zn} \) to \( \text{Zn}^{2+} \).

From the given data:
- \( E^\circ_{\text{Ag}^{+}/\text{Ag}} = 0.46 \, \text{V} \) (reduction)
- \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = 1.10 \, \text{V} \) (oxidation)

Thus, we can express \( E^\circ_{cell} \) as:

\[ E^\circ_{cell} = 0.46 \, \text{V} + 1.10 \, \text{V} = 1.56 \, \text{V} \]

### Step 2: Calculate the Reaction Quotient \( Q_c \)

The reaction quotient \( Q_c \) for the cell reaction is given by:

\[ Q_c = \frac{[\text{Zn}^{2+}]}{[\text{Ag}^{+}]^2} \]

Given that both concentrations are \( 0.1 \, M \):

\[ Q_c = \frac{0.1}{(0.1)^2} = \frac{0.1}{0.01} = 10 \]

### Step 3: Use the Nernst Equation

The Nernst equation relates the cell potential under non-standard conditions to the standard cell potential:

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

Where:
- \( n = 2 \) (number of moles of electrons transferred)
- \( Q_c = 10 \)

Substituting the values into the Nernst equation:

\[ E_{cell} = 1.56 \, \text{V} - \frac{0.0591}{2} \log(10) \]

Since \( \log(10) = 1 \):

\[ E_{cell} = 1.56 \, \text{V} - \frac{0.0591}{2} \cdot 1 \]

Calculating the second term:

\[ E_{cell} = 1.56 \, \text{V} - 0.02955 \, \text{V} \]

\[ E_{cell} = 1.53045 \, \text{V} \]

### Step 4: Final Answer

Rounding to three significant figures, we get:

\[ E_{cell} \approx 1.53 \, \text{V} \]