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^(+)(1.0 M) | Ag` at 298 K will be:

Options

A. 1.59V
B. 1.53V
C. 2.53V
D. cannot be calcualted due to insufficient data

Answer

To find the standard cell potential \( E_{cell} \) for the cell \( Zn | Zn^{2+}(0.1M) || Ag^{+}(1.0M) | Ag \) at 298 K, we can follow these steps:

### Step 1: Identify the Half-Reactions
We have two half-cells:
1. **Zinc Half-Cell**:
\[ Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-} \quad (Oxidation) \]
The standard reduction potential for \( Zn^{2+} + 2e^{-} \rightarrow Zn \) is \( -0.76 \, V \).

2. **Silver Half-Cell**:
\[ Ag^{+}(aq) + e^{-} \rightarrow Ag(s) \quad (Reduction) \]
The standard reduction potential for \( Ag^{+} + e^{-} \rightarrow Ag \) is \( +0.80 \, V \).

### Step 2: Calculate the Standard Cell Potential \( E^\circ_{cell} \)
The overall cell reaction can be written as:
\[ Zn(s) + 2Ag^{+}(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s) \]
The standard cell potential is calculated as:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Where:
- \( E^\circ_{cathode} = +0.80 \, V \) (for Ag)
- \( E^\circ_{anode} = -0.76 \, V \) (for Zn)

Thus,
\[ E^\circ_{cell} = 0.80 - (-0.76) = 0.80 + 0.76 = 1.56 \, V \]

### Step 3: Use the Nernst Equation
Now, we will use the Nernst equation to calculate the cell potential under non-standard conditions:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[Zn^{2+}]}{[Ag^{+}]^2} \right) \]
Where:
- \( n = 2 \) (number of electrons transferred)
- \( [Zn^{2+}] = 0.1 \, M \)
- \( [Ag^{+}] = 1.0 \, M \)

Substituting the values:
\[ E_{cell} = 1.56 - \frac{0.0591}{2} \log \left( \frac{0.1}{(1.0)^2} \right) \]
\[ = 1.56 - 0.02955 \log(0.1) \]
Since \( \log(0.1) = -1 \):
\[ E_{cell} = 1.56 - 0.02955 \times (-1) \]
\[ E_{cell} = 1.56 + 0.02955 \]
\[ E_{cell} \approx 1.58955 \, V \approx 1.59 \, V \]

### Final Answer
The cell potential \( E_{cell} \) for the cell \( Zn | Zn^{2+}(0.1M) || Ag^{+}(1.0M) | Ag \) at 298 K is approximately **1.59 V**.

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