During discharging of a Pb storage battery, the concentration of H₂SO₄ decreases with time. Calculate the time for which a current of 1 A can be drawn as the acid concentration falls from 1 M (initial) to 0.5 M (final). (Assume the solution volume remains 5 L and stays constant.) 24 h 55 h 67 h 83 h

1. The cell reaction

In discharge, the overall balanced reaction for one lead-acid cell is:

$\mathrm{Pb + PbO_2 + 2\;H_2SO_4 \;\longrightarrow\; 2\;PbSO_4 + 2\;H_2O}$

Notice 2 moles of $H_2SO_4$ disappear when the cell sends 2 moles of electrons (i.e. $2 \times 96\,500\,\text{C}$). Hence 1 mole $H_2SO_4 \;\leftrightarrow\; 1$ Faraday.

2. Link between concentration change and charge

• Concentration is moles per litre.
• Volume is kept constant (5 L), so any change in molarity translates directly to change in moles:

$\Delta n_{\text{acid}} = (M_\text{initial} - M_\text{final}) \times V$

Each mole lost → $96\,500$ C.

3. Current & time

Current is the rate of charge flow: $I = \dfrac{Q}{t}$. Rearranged, $t = \dfrac{Q}{I}$.

Put everything together to get the required time.

What’s happening here?

Think of a lead-acid battery like a water tank that empties while giving water to a tap.
• The “water” inside is the acid $H_2SO_4$.
• As the battery gives electric current, the amount of acid slowly goes down – like the tank level dropping.
• We are told how much the level drops (from 1 M to 0.5 M in a 5-litre tank).
• We want to know how long a 1-ampere ‘tap’ can keep running before the level falls that much.

Big idea

In a lead battery, every mole of acid that disappears moves exactly 1 Faraday (96 500 C) of charge through the wires.
So:

1. Find moles of acid lost.
2. Change those moles into charge (Coulombs).
3. Time = Charge / Current.
   That’s really all!

Step-by-step calculation

1. Change in acid concentration
   $\Delta M = 1.0\,\text{M} - 0.50\,\text{M} = 0.50\,\text{M}$

2. Moles of $H_2SO_4$ consumed (volume $V = 5$ L):

   $\Delta n = \Delta M \times V = 0.50\,\text{mol\,L}^{-1} \times 5\,\text{L} = 2.5\,\text{mol}$

3. Charge corresponding to this loss
   Each mole → $1$ F $= 96\,500$ C, so

   $Q = 2.5\,\text{mol} \times 96\,500\,\text{C\,mol}^{-1} = 241\,250\,\text{C}$

4. Time for a current of 1 A
   $t = \frac{Q}{I} = \frac{241\,250\,\text{C}}{1\,\text{C\,s}^{-1}} = 241\,250\,\text{s}$

5. Convert seconds to hours
   $t_{\text{hours}} = \frac{241\,250}{3600} \approx 66.98\,\text{h} \;\approx\; 67\,\text{h}$

Hence, the battery can supply 1 A for roughly 67 hours.

Correct option: 67 h