The half reactions that occur in a lead acid battery are:
`PbSO_(3)(s)+2e^(-)toPb(s)+SO_(4)^(2-)(aq)E^(@)=-0.36V`
`PbO_(2)(s)+4H^(+)(aq)+SO_(4)^(2-)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(l)E^(@)=+1.69V`
Calculate the overall potential for the cell in discharging reaction, `E_(cell)^(@)` Give answer in nearest single digit integer.

Explanation

To calculate the overall potential for the cell in the discharging reaction, we need to follow these steps:

### Step 1: Identify the Half Reactions
We have two half-reactions provided:
1. **Reduction Reaction (Cathode):**
\[ \text{PbO}_2(s) + 4\text{H}^+(aq) + \text{SO}_4^{2-}(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) \quad E^\circ = +1.69 \, \text{V} \]

2. **Oxidation Reaction (Anode):**
\[ \text{PbSO}_4(s) + 2e^- \rightarrow \text{Pb}(s) + \text{SO}_4^{2-}(aq) \quad E^\circ = -0.36 \, \text{V} \]

### Step 2: Determine Which Reaction is Reduction and Which is Oxidation
- The reaction with the higher standard reduction potential (E°) will occur at the cathode (reduction).
- The reaction with the lower E° will occur at the anode (oxidation).

Here, \( E^\circ = +1.69 \, \text{V} \) (for PbO2) is higher, so it is the reduction reaction. The \( E^\circ = -0.36 \, \text{V} \) (for PbSO4) is lower, so it is the oxidation reaction.

### Step 3: Calculate the Overall Cell Potential
The overall cell potential \( E^\circ_{\text{cell}} \) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = (+1.69 \, \text{V}) - (-0.36 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 1.69 \, \text{V} + 0.36 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 2.05 \, \text{V} \]

### Step 4: Round to the Nearest Single Digit Integer
The final step is to round \( E^\circ_{\text{cell}} \) to the nearest single digit integer:
\[ E^\circ_{\text{cell}} \approx 2 \]

### Final Answer
The overall potential for the cell in the discharging reaction is **2**.

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