Question

lowering the pH of the solution inside the battery will: a. make E larger than E b. make E smaller than E c. have no effect on E vs. E d. make the battery last longer

Answer 1

Consider the lead acid battery. The reactions taking place at the cathode and the anode are given as

Cathode: PbO₂ (s) + HSO₄^- (aq) + 3 H⁺ (aq) + 2 e^- ---------> PbSO₄ (s) + 2 H₂O (l)

Anode: Pb (s) + HSO₄^- (aq) --------> PbSO₄ (s) + H⁺ (aq) + 2 e^-

The overall reaction in a lead-acid battery is the sum of the reactions at the cathode and the anode and is denoted as

PbO₂ (s) + Pb (s) + 2 HSO₄^- (aq) + 2 H⁺ (aq) -------> 2 PbSO₄ (s) + 2 H₂O (l)

The Nernst equation for the cell is written as

E = E⁰ – 2.303RT/nF*log 1/[HSO₄^-]²[H⁺]²

where E⁰ is the standard cell potential and n = 2 (number of electrons transferred) and F = 96485 J/V.mol is the Faraday’s constant.

Therefore,

E = E⁰ – 2.303RT/nF*(log 1 – log [HSO₄^-]² – log [H⁺]²)

= E⁰ + 2*2.303RT/nF*log [HSO₄^-] + 2*2.303RT/nF*log [H⁺]

= E⁰ + 4.606RT/nF*log [HSO₄^-] + 4.606RT/nF*(-pH) (pH = -log [H⁺])

= E⁰ + 4.606RT/nF*log [HSO₄^-] – 4.606RT/nF*Ph

As the pH of the solution decreases, the value of E increases and becomes greater than E⁰ (all other factors remaining constant). Therefore, (a) is the correct statement.