Solveeit Logo
Login

Question

Question: A lead storage battery is considered having 2 kg of Pb and PbO₂ each, in the presence of excess H₂SO...

A lead storage battery is considered having 2 kg of Pb and PbO₂ each, in the presence of excess H₂SO₄.

Number of hours of current delivery having current strength 96 amperes is _______, if the battery operated till the reaction goes to completion. (Mw of Pb = 207)

Answer

4.67

Explanation

Solution

1. Determine the balanced chemical reaction for the discharge of a lead storage battery:

The overall reaction during discharge is: Pb(s) + PbO2(s) + 2H2SO4(aq)→2PbSO4(s) + 2H2O(l)\text{Pb(s) + PbO}_2\text{(s) + 2H}_2\text{SO}_4\text{(aq)} \rightarrow \text{2PbSO}_4\text{(s) + 2H}_2\text{O(l)}

2. Calculate the moles of each reactant (Pb and PbO2_2):

Given masses: Mass of Pb = 2 kg = 2000 g Mass of PbO2_2 = 2 kg = 2000 g

Molar masses: Molar mass of Pb = 207 g/mol (given) Molar mass of O = 16 g/mol Molar mass of PbO2_2 = M(Pb) + 2 * M(O) = 207 + 2 * 16 = 207 + 32 = 239 g/mol

Moles of Pb = Mass of PbMolar mass of Pb=2000 g207 g/mol≈9.662 mol\frac{\text{Mass of Pb}}{\text{Molar mass of Pb}} = \frac{2000 \text{ g}}{207 \text{ g/mol}} \approx 9.662 \text{ mol} Moles of PbO2_2 = Mass of PbO2Molar mass of PbO2=2000 g239 g/mol≈8.368 mol\frac{\text{Mass of PbO}_2}{\text{Molar mass of PbO}_2} = \frac{2000 \text{ g}}{239 \text{ g/mol}} \approx 8.368 \text{ mol}

3. Identify the limiting reactant:

From the balanced overall reaction, Pb and PbO2_2 react in a 1:1 molar ratio. Since 8.368 mol of PbO2_2 is less than 9.662 mol of Pb, PbO2_2 is the limiting reactant. The reaction will stop once all 8.368 moles of PbO2_2 are consumed.

4. Calculate the total charge (Q) transferred:

The half-reaction at the cathode (where PbO2_2 is consumed) is: PbO2(s) + SO42−(aq) + 4H+(aq) + 2e−→PbSO4(s) + 2H2O(l)\text{PbO}_2\text{(s) + SO}_4^{2-}\text{(aq) + 4H}^+\text{(aq) + 2e}^- \rightarrow \text{PbSO}_4\text{(s) + 2H}_2\text{O(l)} This reaction shows that 2 moles of electrons are transferred for every 1 mole of PbO2_2 consumed.

Moles of electrons transferred = 2 * Moles of PbO2_2 Moles of electrons transferred = 2 * 8.368 mol = 16.736 mol

Now, calculate the total charge Q using Faraday's constant (F = 96500 C/mol): Q = Moles of electrons * F Q = 16.736 mol * 96500 C/mol = 1615064 C

5. Calculate the time (t) for current delivery:

The relationship between charge (Q), current (I), and time (t) is Q = I * t. Given current (I) = 96 Amperes.

t = QI=1615064 C96 A≈16823.58 seconds\frac{\text{Q}}{\text{I}} = \frac{1615064 \text{ C}}{96 \text{ A}} \approx 16823.58 \text{ seconds}

6. Convert the time from seconds to hours:

t (hours) = t (seconds)3600 s/hour=16823.583600≈4.673 hours\frac{\text{t (seconds)}}{3600 \text{ s/hour}} = \frac{16823.58}{3600} \approx 4.673 \text{ hours}

The number of hours of current delivery is approximately 4.67.