Calculate EMF of following cell at 298 K
`Zn|ZnSO_(4) (0.1 M) ||CuSO_(4) (1.0 M)|Cu (s)` if `E_(cell) = 2.0` V

Options

A. `2.0296 V `
B. `2.0592 V`
C. `1.0508 V `
D. `2.0` V

Explanation

To calculate the EMF of the given electrochemical cell at 298 K, we will use the Nernst equation. Let's break down the steps:

### Step 1: Identify the half-reactions
In the given cell, we have:
- Anode (oxidation):
\[ \text{Zn (s)} \rightarrow \text{Zn}^{2+} (aq) + 2e^- \]
- Cathode (reduction):
\[ \text{Cu}^{2+} (aq) + 2e^- \rightarrow \text{Cu (s)} \]

### Step 2: Write the overall cell reaction
Combining the two half-reactions, we get the overall reaction:
\[ \text{Zn (s)} + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu (s)} \]

### Step 3: Write the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
Where:
- \( E^{\circ}_{\text{cell}} = 2.0 \, \text{V} \) (given)
- \( n = 2 \) (number of electrons transferred)

### Step 4: Substitute the concentrations
We have:
- \([\text{Zn}^{2+}] = 0.1 \, \text{M}\)
- \([\text{Cu}^{2+}] = 1.0 \, \text{M}\)

Substituting these values into the Nernst equation:
\[ E_{\text{cell}} = 2.0 - \frac{0.0591}{2} \log \left( \frac{0.1}{1.0} \right) \]

### Step 5: Calculate the logarithm
Calculate the logarithm:
\[ \log \left( \frac{0.1}{1.0} \right) = \log(0.1) = -1 \]

### Step 6: Substitute the logarithm value
Now substitute this value back into the equation:
\[ E_{\text{cell}} = 2.0 - \frac{0.0591}{2} \times (-1) \]
\[ E_{\text{cell}} = 2.0 + \frac{0.0591}{2} \]
\[ E_{\text{cell}} = 2.0 + 0.02955 \]
\[ E_{\text{cell}} = 2.02955 \, \text{V} \]

### Step 7: Final answer
Thus, the EMF of the cell at 298 K is approximately:
\[ E_{\text{cell}} \approx 2.03 \, \text{V} \]

---