Calculate the e.m.f. of the cell,
`Zn(s)//Zn^(2+)(0.1 M) || Pb^(2+)(0.02 M)//Pb(s)`
`E_(Zn^(2+)//Zn)^(@)=-0.76" V " and E_(Pb^(2+)//Pb)^(@)=-0.13" V "`

Explanation

To calculate the e.m.f. (electromotive force) of the cell represented by the reaction:

\[ \text{Zn(s)} // \text{Zn}^{2+}(0.1 \, M) || \text{Pb}^{2+}(0.02 \, M) // \text{Pb(s)} \]

we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials
The half-reactions and their standard electrode potentials are given as:
- For Zinc:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad E^\circ = -0.76 \, V \]
- For Lead:
\[ \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \quad E^\circ = -0.13 \, V \]

### Step 2: Determine the anode and cathode
In this electrochemical cell:
- The anode (oxidation) is where zinc is oxidized:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- The cathode (reduction) is where lead is reduced:
\[ \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \]

### Step 3: Calculate the standard cell potential (E°cell)
The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.13 \, V) - (-0.76 \, V) = -0.13 + 0.76 = 0.63 \, V \]

### Step 4: Apply the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Pb}^{2+}]} \right) \]

Where:
- \( n = 2 \) (the number of electrons transferred)
- \( [\text{Zn}^{2+}] = 0.1 \, M \)
- \( [\text{Pb}^{2+}] = 0.02 \, M \)

### Step 5: Substitute the values into the Nernst equation
Substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 0.63 \, V - \frac{0.0591}{2} \log \left( \frac{0.1}{0.02} \right) \]

### Step 6: Calculate the logarithm
Calculate the concentration ratio:
\[ \frac{0.1}{0.02} = 5 \]
Now, calculate the logarithm:
\[ \log(5) \approx 0.699 \]

### Step 7: Substitute back into the equation
Now substitute back into the Nernst equation:
\[ E_{\text{cell}} = 0.63 \, V - \frac{0.0591}{2} \times 0.699 \]

### Step 8: Calculate the final value
Calculating the second term:
\[ \frac{0.0591}{2} \times 0.699 \approx 0.0206 \]
Now calculate:
\[ E_{\text{cell}} = 0.63 \, V - 0.0206 \approx 0.6094 \, V \]

### Final Answer
Thus, the e.m.f. of the cell is approximately:
\[ E_{\text{cell}} \approx 0.61 \, V \]

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