The standard reduction potential of `Pb` and `Zn` electrodes are `-0.126` and `-0.763 V` respectively. The cell equation will be

Options

A. `Pb^(2+)+Zn rarr Pb+Zn^(2+)`
B. `Pb^(4+)+2 Zn rarr Pb+2 Zn^(2+)`
C. `Zn^(2+)+Pb rarr Zn+Pb^(2+)`
D. None of the above

Answer

To determine the cell equation involving the lead (Pb) and zinc (Zn) electrodes based on their standard reduction potentials, we follow these steps:

### Step 1: Identify the half-reactions
The standard reduction potentials given are:
- For Pb: \( E^\circ = -0.126 \, V \)
- For Zn: \( E^\circ = -0.763 \, V \)

The half-reactions for the reduction of Pb and Zn are:
- Pb: \( \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \)
- Zn: \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \)

### Step 2: Determine which species is oxidized and which is reduced
The species with the higher standard reduction potential will be reduced, while the species with the lower standard reduction potential will be oxidized.

- Pb has a higher reduction potential (-0.126 V) compared to Zn (-0.763 V). Therefore, Pb will be reduced and Zn will be oxidized.

### Step 3: Write the oxidation half-reaction for Zn
Since Zn is being oxidized, we write the oxidation half-reaction:
- \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)

### Step 4: Write the reduction half-reaction for Pb
The reduction half-reaction for Pb is:
- \( \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \)

### Step 5: Combine the half-reactions to form the overall cell equation
Now, we combine the oxidation and reduction half-reactions:
- Oxidation: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
- Reduction: \( \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \)

Combining these, we get:
\[ \text{Zn} + \text{Pb}^{2+} \rightarrow \text{Zn}^{2+} + \text{Pb} \]

### Final Cell Equation
The overall cell equation is:
\[ \text{Zn} + \text{Pb}^{2+} \rightarrow \text{Zn}^{2+} + \text{Pb} \]