Read the passage given below and answer the question.
A galvanic cell consists of a metallic zinc plate immersed in 0.1 M `Zn(NO_(3))_(2)` solution and metallic plate of lean in 0.02 M Pb`(NO_(3))_(2)` solution.
Q. How will the cell be represented?

Options

A. `Zn(s)|Zn^(2+)(aq)||Pb^(2+)(aq)|Pb(s)`
B. `Zn^(2+)(s)|Zn(aq)||Pb^(2+)(aq)|Pb(s)`
C. `Pb^(2+)(aq)|Pb(s)||Zn^(2+)(s)|Zn(aq)`
D. `Pb(s)|Pb^(2+)(aq)||Zn^(2+)(s)|Zn(aq)`

Explanation

To represent the galvanic cell described in the passage, we need to follow the standard notation for electrochemical cells. Here’s a step-by-step breakdown of how to represent the cell:

### Step 1: Identify the Anode and Cathode
- The anode is where oxidation occurs, and in this case, it is the zinc plate (Zn).
- The cathode is where reduction occurs, which is the lead plate (Pb).

### Step 2: Write the Half-Reactions
- For the anode (oxidation of zinc):
\[ \text{Zn (s)} \rightarrow \text{Zn}^{2+} (aq) + 2e^- \]
- For the cathode (reduction of lead):
\[ \text{Pb}^{2+} (aq) + 2e^- \rightarrow \text{Pb (s)} \]

### Step 3: Determine the Concentrations
- The zinc solution has a concentration of 0.1 M Zn(NO3)2, which means the concentration of Zn²⁺ ions is 0.1 M.
- The lead solution has a concentration of 0.02 M Pb(NO3)2, which means the concentration of Pb²⁺ ions is 0.02 M.

### Step 4: Write the Cell Notation
- The cell notation follows the format:
\[ \text{Anode} | \text{Anode Solution} || \text{Cathode Solution} | \text{Cathode} \]
- Based on the information:
- Anode: Zn (s) in 0.1 M Zn²⁺ (aq)
- Cathode: 0.02 M Pb²⁺ (aq) in Pb (s)

### Step 5: Combine the Information
- The complete cell representation is:
\[ \text{Zn (s)} | \text{Zn}^{2+} (0.1 M) || \text{Pb}^{2+} (0.02 M) | \text{Pb (s)} \]

### Final Representation
Thus, the cell can be represented as:
\[ \text{Zn (s)} | \text{Zn}^{2+} (0.1 M) || \text{Pb}^{2+} (0.02 M) | \text{Pb (s)} \]

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