The standard reduction potential of `Pb` and `Zn` electrodes are `-0.126` and `-0.763` volts respectively . The e.m.f of the cell
`Zn|Zn^(2+)(0.1M)||Pb^(2+)|(1M)Pb` is

Options

A. `0.637 V`
B. `lt0.637V`
C. `gt0.637V`
D. `0.889`

Explanation

To find the e.m.f. of the cell `Zn|Zn^(2+)(0.1M)||Pb^(2+)|(1M)Pb`, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials
- The standard reduction potential for `Pb^(2+) + 2e^- → Pb` is `E°(Pb) = -0.126 V`.
- The standard reduction potential for `Zn^(2+) + 2e^- → Zn` is `E°(Zn) = -0.763 V`.

### Step 2: Determine the oxidation and reduction reactions
- In this cell, zinc is oxidized and lead is reduced.
- The oxidation half-reaction for zinc is:
\[ Zn → Zn^{2+} + 2e^- \]
- The reduction half-reaction for lead is:
\[ Pb^{2+} + 2e^- → Pb \]

### Step 3: Calculate the standard cell potential (E°cell)
The standard cell potential can be calculated using the formula:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
Here, the cathode is lead (Pb) and the anode is zinc (Zn):
\[ E°_{cell} = E°(Pb) - E°(Zn) = (-0.126) - (-0.763) \]
Calculating this gives:
\[ E°_{cell} = -0.126 + 0.763 = 0.637 V \]

### Step 4: Use the Nernst equation to find the e.m.f. of the cell
The Nernst equation is given by:
\[ E = E°_{cell} - \frac{0.0591}{n} \log \frac{[Ox]}{[Red]} \]
Where:
- \( n = 2 \) (the number of moles of electrons transferred)
- \( [Ox] = [Zn^{2+}] = 0.1 \, M \)
- \( [Red] = [Pb^{2+}] = 1 \, M \)

Substituting these values into the Nernst equation:
\[ E = 0.637 - \frac{0.0591}{2} \log \frac{0.1}{1} \]
Calculating the logarithm:
\[ \log \frac{0.1}{1} = \log 0.1 = -1 \]
Now substituting this back into the equation:
\[ E = 0.637 - \frac{0.0591}{2} \times (-1) \]
\[ E = 0.637 + \frac{0.0591}{2} \]
Calculating \( \frac{0.0591}{2} \):
\[ \frac{0.0591}{2} = 0.02955 \]
Now substituting this value:
\[ E = 0.637 + 0.02955 = 0.66655 V \]

### Final Answer
The e.m.f. of the cell is approximately **0.667 V**.

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