A conventional method of representing Daniell cell is:
`Zn(s), Zn^(2+) (aq)(1M)"||"Cu^(2+) (aq)(1M), Cu(s)`
(i) Draw a diagram of the cell and mark anode and cathode.
`Cu^(2+)//Cu=0.34V, Zn^(2+)//Zn = -0.76V`
(ii) Give the net reaction as current is drawn.
(iii) What is the cell potential at 298 K?

Explanation

### Step-by-Step Solution:

#### (i) Draw a diagram of the cell and mark anode and cathode.

1. **Draw two containers**: Start by sketching two beakers or containers. Label the left container as "Zinc half-cell" and the right container as "Copper half-cell".

2. **Add solutions**: In the left container, indicate that it contains a solution of Zinc sulfate (ZnSO₄) and in the right container, indicate that it contains Copper sulfate (CuSO₄).

3. **Insert electrodes**: Draw a zinc electrode (Zn) in the left container and a copper electrode (Cu) in the right container.

4. **Connect with a salt bridge**: Draw a salt bridge connecting the two containers. This bridge allows ions to flow between the two half-cells.

5. **Mark the anode and cathode**:
- Label the zinc electrode as the **anode** (where oxidation occurs).
- Label the copper electrode as the **cathode** (where reduction occurs).

6. **Indicate electron flow**: Draw arrows to show that electrons flow from the anode (Zn) to the cathode (Cu).

7. **Label the terminals**: Mark the left side (anode) as negative (-) and the right side (cathode) as positive (+).

#### Diagram:
```
Zn(s) | Zn²⁺(aq) (1M) || Cu²⁺(aq) (1M) | Cu(s)
[Anode] [Cathode]
```

#### (ii) Give the net reaction as current is drawn.

1. **Write the oxidation half-reaction** at the anode:
\[ \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^{-} \]

2. **Write the reduction half-reaction** at the cathode:
\[ \text{Cu}^{2+}(aq) + 2e^{-} \rightarrow \text{Cu(s)} \]

3. **Combine the two half-reactions** to get the net reaction:
- The electrons cancel out:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)} \]

#### (iii) What is the cell potential at 298 K?

1. **Identify the standard reduction potentials**:
- For Cu²⁺/Cu: \( E^\circ = +0.34 \, V \)
- For Zn²⁺/Zn: \( E^\circ = -0.76 \, V \)

2. **Calculate the standard cell potential** \( E^\circ_{cell} \):
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
\[ E^\circ_{cell} = 0.34 \, V - (-0.76 \, V) = 0.34 \, V + 0.76 \, V = 1.10 \, V \]

3. **Use the Nernst equation** to find the cell potential at 298 K:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
- Here, \( n = 2 \) (number of electrons transferred).
- Since both concentrations are 1 M, the log term becomes:
\[ \log \left( \frac{1}{1} \right) = 0 \]

4. **Substituting values**:
\[ E_{cell} = 1.10 \, V - \frac{0.0591}{2} \cdot 0 = 1.10 \, V \]

### Final Answers:
- **(i)** Diagram of the Daniell cell with anode and cathode labeled.
- **(ii)** Net reaction:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)} \]
- **(iii)** Cell potential at 298 K: \( E_{cell} = 1.10 \, V \)