What is the cell potential of the following cell ?
`Zn(s)|Zn^(2+)(1.0M)||Pb^(2+)(1.0M)|Pb(s)`
Given :`E_(Pb^(2+)//Pb)^(@)=-0.12" V and " E_(Zn^(2+)//Zn)^(@)=-0.76" V"`

Answer

To determine the cell potential of the given electrochemical cell, we can follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials
The cell consists of the following half-reactions:
- For zinc: \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) with \( E^\circ = -0.76 \, \text{V} \)
- For lead: \( \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \) with \( E^\circ = -0.12 \, \text{V} \)

### Step 2: Calculate the standard cell potential (\( E^\circ_{\text{cell}} \))
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 where reduction occurs (lead) and the anode is where oxidation occurs (zinc).

Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.12 \, \text{V}) - (-0.76 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = -0.12 + 0.76 = 0.64 \, \text{V} \]

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

### Step 4: Calculate the reaction quotient (\( Q \))
In this case, the concentrations are given as 1.0 M for both \( \text{Zn}^{2+} \) and \( \text{Pb}^{2+} \):
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Pb}^{2+}]} = \frac{1.0}{1.0} = 1 \]

### Step 5: Substitute the values into the Nernst equation
Now substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 0.64 \, \text{V} - \frac{0.0591}{2} \log(1) \]
Since \( \log(1) = 0 \):
\[ E_{\text{cell}} = 0.64 \, \text{V} - 0 = 0.64 \, \text{V} \]

### Final Answer
The cell potential of the given cell is:
\[ \boxed{0.64 \, \text{V}} \]