The cell reaction of the galvanic cell : `Cu_((s))|Cu_((aq))^(2+) ||Hg_((aq))^(2+) |Hg_((l))` is

Options

A. `Hg+Cu^(2+) to Hg^(2+) +Cu`
B. `Hg+Cu^(2+) to Cu^(+) +Hg^+`
C. `Cu+Hg to CuHg`
D. `Cu+ Hg^(2+) to Cu^(2+) + Hg`

Answer

To derive the cell reaction of the galvanic cell represented by the notation `Cu(s)|Cu^(2+)(aq) || Hg^(2+)(aq)|Hg(l)`, we need to identify the oxidation and reduction half-reactions occurring at the anode and cathode, respectively.

### Step-by-Step Solution:

1. **Identify the Anode and Cathode:**
- In a galvanic cell, oxidation occurs at the anode and reduction occurs at the cathode.
- In this cell, copper (Cu) is the anode and mercury (Hg) is the cathode.

2. **Write the Oxidation Half-Reaction:**
- At the anode, copper is oxidized from its solid state to its ionic state:
\[ \text{Cu(s)} \rightarrow \text{Cu}^{2+}(aq) + 2e^- \]
- This reaction shows that solid copper loses two electrons to form copper ions.

3. **Write the Reduction Half-Reaction:**
- At the cathode, mercury ions are reduced to liquid mercury:
\[ \text{Hg}^{2+}(aq) + 2e^- \rightarrow \text{Hg(l)} \]
- This reaction shows that mercury ions gain two electrons to form liquid mercury.

4. **Combine the Half-Reactions:**
- To find the overall cell reaction, we combine the oxidation and reduction half-reactions. The electrons lost in the oxidation half-reaction must equal the electrons gained in the reduction half-reaction:
\[ \text{Cu(s)} + \text{Hg}^{2+}(aq) \rightarrow \text{Cu}^{2+}(aq) + \text{Hg(l)} \]

5. **Final Cell Reaction:**
- The overall cell reaction for the galvanic cell is:
\[ \text{Cu(s)} + \text{Hg}^{2+}(aq) \rightarrow \text{Cu}^{2+}(aq) + \text{Hg(l)} \]

### Summary:
The cell reaction of the galvanic cell `Cu(s)|Cu^(2+)(aq) || Hg^(2+)(aq)|Hg(l)` is:
\[ \text{Cu(s)} + \text{Hg}^{2+}(aq) \rightarrow \text{Cu}^{2+}(aq) + \text{Hg(l)} \]