The half cell reactions with their oxidation potentials are
(a) Pb(s)`-2e^(-)toPb^(2+)(aq),E_("oxi")^(@)=+0.13V` (b) `Ag(s)-e^(-)toAg^(+)(aq),E_("oxi")^(@)= -0.80V`
Write the cell reaction and calculate its emf.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Identify the Half-Reactions and Their Potentials
We have two half-cell reactions with their oxidation potentials:
1. \( \text{Pb(s)} \rightarrow \text{Pb}^{2+}(aq) + 2e^- \), \( E_{\text{oxi}}^\circ = +0.13 \, \text{V} \)
2. \( \text{Ag(s)} \rightarrow \text{Ag}^+(aq) + e^- \), \( E_{\text{oxi}}^\circ = -0.80 \, \text{V} \)

### Step 2: Determine Which Reaction is Oxidation and Which is Reduction
- The reaction with the higher oxidation potential will be the oxidation reaction, and the one with the lower oxidation potential will be the reduction reaction.
- Here, \( \text{Pb} \) has a higher oxidation potential than \( \text{Ag} \), so:
- Oxidation: \( \text{Pb(s)} \rightarrow \text{Pb}^{2+}(aq) + 2e^- \)
- Reduction: \( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag(s)} \)

### Step 3: Balance the Electrons
To balance the electrons, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. The oxidation of lead produces 2 electrons, while the reduction of silver consumes 1 electron. Therefore, we need to multiply the silver half-reaction by 2:

\[ 2 \text{Ag}^+(aq) + 2e^- \rightarrow 2 \text{Ag(s)} \]

### Step 4: Write the Overall Cell Reaction
Now we can combine the two half-reactions:

\[ \text{Pb(s)} + 2 \text{Ag}^+(aq) \rightarrow \text{Pb}^{2+}(aq) + 2 \text{Ag(s)} \]

### Step 5: Calculate the EMF of the Cell
The EMF (electromotive force) of the cell can be calculated using the formula:

\[ E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ \]

From our half-reactions:
- The cathode (reduction) is silver: \( E_{\text{cathode}}^\circ = -0.80 \, \text{V} \)
- The anode (oxidation) is lead: \( E_{\text{anode}}^\circ = +0.13 \, \text{V} \)

Substituting the values:

\[ E_{\text{cell}}^\circ = (-0.80 \, \text{V}) - (+0.13 \, \text{V}) = -0.80 \, \text{V} - 0.13 \, \text{V} = -0.93 \, \text{V} \]

### Step 6: Correct the Sign for Oxidation Potential
Since we are using the oxidation potential for lead, we take the negative of the oxidation potential:

\[ E_{\text{cell}}^\circ = 0.13 \, \text{V} - (-0.80 \, \text{V}) = 0.13 \, \text{V} + 0.80 \, \text{V} = 0.93 \, \text{V} \]

### Final Answer
The overall cell reaction is:
\[ \text{Pb(s)} + 2 \text{Ag}^+(aq) \rightarrow \text{Pb}^{2+}(aq) + 2 \text{Ag(s)} \]
The EMF of the cell is:
\[ E_{\text{cell}}^\circ = 0.93 \, \text{V} \]

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