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. Calculate the emf of the cell.

Options

A. 6.01 V
B. 0.412 V
C. 0.609 V
D. 4.12 V

Explanation

To calculate the emf of the galvanic cell consisting of a metallic zinc plate immersed in 0.1 M Zn(NO₃)₂ solution and a metallic lead plate in 0.02 M Pb(NO₃)₂ solution, we can follow these steps:

### Step 1: Identify the half-reactions and standard reduction potentials
- **Zinc (Zn)** is oxidized:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-} \quad (E^\circ = -0.76 \, \text{V}) \]
- **Lead (Pb)** is reduced:
\[ \text{Pb}^{2+} + 2e^{-} \rightarrow \text{Pb} \quad (E^\circ = -0.13 \, \text{V}) \]

### Step 2: Determine the standard cell potential (E°cell)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is lead and the anode is zinc:
\[ E^\circ_{\text{cell}} = (-0.13) - (-0.76) = 0.63 \, \text{V} \]

### Step 3: Use the Nernst equation to calculate the cell potential (Ecell)
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]
where \(n\) is the number of moles of electrons transferred (which is 2 in this case).

### Step 4: Substitute the concentrations into the Nernst equation
- Concentration of Zn²⁺ = 0.1 M (from Zn(NO₃)₂)
- Concentration of Pb²⁺ = 0.02 M (from Pb(NO₃)₂)

The Nernst equation becomes:
\[ E_{\text{cell}} = 0.63 - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Pb^{2+}]} \]
Substituting the concentrations:
\[ E_{\text{cell}} = 0.63 - \frac{0.0591}{2} \log \frac{0.1}{0.02} \]

### Step 5: Calculate the logarithm
\[ \log \frac{0.1}{0.02} = \log 5 = 0.699 \]

### Step 6: Substitute back into the equation
\[ E_{\text{cell}} = 0.63 - \frac{0.0591}{2} \times 0.699 \]
Calculating the second term:
\[ \frac{0.0591}{2} \times 0.699 \approx 0.0206 \]
Thus,
\[ E_{\text{cell}} = 0.63 - 0.0206 = 0.6094 \, \text{V} \]

### Final Answer
The emf of the cell is approximately **0.6094 V**.

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