Question

Which of the following option(s) is/are correct regarding the discharging of lead storage battery?

$\square$ Pb(s) on anode is converted into $PbSO_4(s)$

$\square$ Oxidation number of lead at cathode changes from +2 to 0

$\square$ The number of moles of $H_2SO_4$ consumed when it is used up at 9.65 ampere for 1000 seconds is 0.05 [1F = 96500 C]

$\square$ The mass of $H_2SO_4$ consumed when it is used up at 9.65 ampere for 1000 seconds is 9.8 g [1F = 96500 C]

• A. Pb(s) on anode is converted into $PbSO_4(s)$
• B. Oxidation number of lead at cathode changes from +2 to 0
• C. The number of moles of $H_2SO_4$ consumed when it is used up at 9.65 ampere for 1000 seconds is 0.05 [1F = 96500 C]
• D. The mass of $H_2SO_4$ consumed when it is used up at 9.65 ampere for 1000 seconds is 9.8 g [1F = 96500 C]

Answer 1

Answer: (A), (D)

Pb(s) on anode is converted into $PbSO_4(s)$, The mass of $H_2SO_4$ consumed when it is used up at 9.65 ampere for 1000 seconds is 9.8 g [1F = 96500 C]

Answer 2

The lead storage battery undergoes the following reactions during discharging:

1. Anode (Negative Electrode - Oxidation):

Lead metal (Pb) is oxidized to lead sulfate ($PbSO_4$).

$Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$

The oxidation state of lead changes from 0 to +2.

2. Cathode (Positive Electrode - Reduction):

Lead dioxide ($PbO_2$) is reduced to lead sulfate ($PbSO_4$).

$PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$

The oxidation state of lead in $PbO_2$ is +4. The oxidation state of lead in $PbSO_4$ is +2. So, the oxidation state of lead changes from +4 to +2.

3. Overall Reaction:

Combining the anode and cathode reactions:

$Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$

During discharging, sulfuric acid ($H_2SO_4$) is consumed, and water is produced, leading to a decrease in the density of the electrolyte.

Now let's evaluate each option:

$\square$ Pb(s) on anode is converted into $PbSO_4(s)$

This statement is correct. As per the anode reaction, lead metal is oxidized to lead sulfate.

$\square$ Oxidation number of lead at cathode changes from +2 to 0

This statement is incorrect. At the cathode, lead in $PbO_2$ (oxidation state +4) is reduced to lead in $PbSO_4$ (oxidation state +2). So, the oxidation number changes from +4 to +2.

$\square$ The number of moles of $H_2SO_4$ consumed when it is used up at 9.65 ampere for 1000 seconds is 0.05 [1F = 96500 C]

First, calculate the total charge (Q) passed:

$Q = I \times t = 9.65 \text{ A} \times 1000 \text{ s} = 9650 \text{ C}$

Next, convert the charge to Faradays:

Number of Faradays ($n_F$) = $\frac{Q}{1F} = \frac{9650 \text{ C}}{96500 \text{ C/F}} = 0.1 \text{ F}$

From the overall reaction, $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$, it is clear that 2 moles of electrons (2F) are transferred for every 2 moles of $H_2SO_4$ consumed.

Therefore, 1 mole of $H_2SO_4$ is consumed for every 1 Faraday of charge.

Since 0.1 F of charge is passed, 0.1 moles of $H_2SO_4$ are consumed.

This statement is incorrect as it states 0.05 moles.

$\square$ The mass of $H_2SO_4$ consumed when it is used up at 9.65 ampere for 1000 seconds is 9.8 g [1F = 96500 C]

From the previous calculation, the number of moles of $H_2SO_4$ consumed is 0.1 mol.

The molar mass of $H_2SO_4$ is $2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 \text{ g/mol}$.

Mass of $H_2SO_4$ consumed = moles $\times$ molar mass = $0.1 \text{ mol} \times 98 \text{ g/mol} = 9.8 \text{ g}$.

This statement is correct.