Q.(a) A lead storage battery has been used for 20 days at the rate of two hours per day by drawing a constant current of 2 amperes. What would be the quantity of $H_2SO_4$ consumed by the battery ? (Two electrons involved in the reaction) OR

Lead Acid Battery Recharging: Why the Key Formulas Hold

Let's start with the core chemical reaction during recharging — understanding why the formulas work requires knowing what happens inside the battery.

1. The Recharging Reaction (The "Why" Behind the Voltage)

During discharge, lead sulfate ($PbSO_4$) forms on both electrodes. Recharging reverses this:

At the negative terminal (cathode during recharge):

At the positive terminal (anode during recharge):

Overall recharge reaction:

Key insight: Recharging regenerates sulfuric acid ($H_2SO_4$) — this is why the electrolyte's specific gravity increases during charging.

2. The Voltage Formula: $V_{\text{cell}} = 2.1 + \frac{0.059}{2} \log \frac{[H^+]^4}{[SO_4^{2-}]^2}$

Why this formula holds — step by step:

Step 1: Nernst equation for each half-cell

For the negative half-cell (Pb/PbSO₄):

For the positive half-cell (PbO₂/PbSO₄):

Step 2: Cell voltage = $E_+ - E_-$

Step 3: Simplify using standard potentials

$E_+^\circ - E_-^\circ = 2.1\ \text{V}$ (standard cell voltage). Combining logs:

Why this matters for recharging:

• As $H_2SO_4$ concentration increases during charging, $[H^+]$ rises and $[SO_4^{2-}]$ rises → the log term becomes more positive → voltage rises.
• This is why a fully charged battery shows ~2.6–2.7 V per cell during charging (higher than the 2.1 V open-circuit voltage).

3. The Charging Current Formula: $I_{\text{charge}} = \frac{C}{10}$ (for constant current)

Why "C/10" is the standard rule:

Reasoning:

• C = battery capacity in Ah (e.g., 100 Ah)
• C/10 = 10 A for a 100 Ah battery

Why this rate?

• At higher rates (e.g., C/5), the internal resistance causes excessive heat and gassing (water electrolysis into $H_2$ and $O_2$).
• At lower rates (e.g., C/20), charging takes too long and may not fully reverse sulfation.
• C/10 balances:
  • Reaction kinetics: The reverse reaction rate matches the diffusion of ions.
  • Thermal management: Heat generated = $I^2R$ — kept low enough to avoid damage.
  • Gas evolution: At C/10, most current goes to the desired reaction, not water splitting.

Exam tip: For lead-acid, the charging voltage is typically 2.4–2.45 V per cell (not 2.1 V) — this overpotential is needed to overcome the activation energy of the reverse reaction.

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