`underset(("anode"))(Zn_((s))|Zn_((aq))^(2+)||underset(("cathode"))(Cu_((aq))^(2+)|Cu_((s))` is

Options

A. Weston cell
B. Daniel cell
C. Calomel cell
D. Faraday cell

Explanation

To solve the question regarding the electrochemical cell represented by the notation \( \text{Zn(s)} | \text{Zn}^{2+}(aq) || \text{Cu}^{2+}(aq) | \text{Cu(s)} \), we will follow these steps:

### Step 1: Identify the Anode and Cathode
In the given cell notation:
- The anode is where oxidation occurs. Here, zinc (Zn) is oxidized to zinc ions (\( \text{Zn}^{2+} \)).
- The cathode is where reduction occurs. Here, copper ions (\( \text{Cu}^{2+} \)) are reduced to copper (Cu).

### Step 2: Write the Half-Reactions
- **Oxidation half-reaction (Anode):**
\[ \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^- \]

- **Reduction half-reaction (Cathode):**
\[ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu(s)} \]

### Step 3: Write the Overall Cell Reaction
Combining the oxidation and reduction half-reactions gives the overall cell reaction:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)} \]

### Step 4: Calculate the Standard Cell Potential (\( E^\circ \))
To find the standard cell potential, we need the standard reduction potentials for both half-reactions:
- For \( \text{Zn}^{2+}/\text{Zn} \): \( E^\circ = -0.76 \, \text{V} \)
- For \( \text{Cu}^{2+}/\text{Cu} \): \( E^\circ = +0.34 \, \text{V} \)

Using the formula for the cell potential:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} - (-0.76 \, \text{V}) = 0.34 \, \text{V} + 0.76 \, \text{V} = 1.10 \, \text{V} \]

### Step 5: Conclusion
The electrochemical cell represented by the notation is a **Daniel cell** (or galvanic cell) with a standard cell potential of \( 1.10 \, \text{V} \).

### Final Answer
The answer is **Daniel cell**.

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