Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal. The rusting of iron, tarnishing of silver, development of green coating on copper and bronze are some of the examples of corrosion. In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of iron (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex but it may be considered essentially as an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode:
Anode: 2Fe(s) → 2Fe2+ + 4e−; E°(Fe2+/Fe) = −0.44 V
Electrons released at the anodic spot move through the metal and go to another spot on the metal and reduce oxygen in the presence of H+ (which is believed to be available from H2CO3 formed due to dissolution of carbon dioxide from air into water). This spot behaves as cathode:
Cathode: O2(g) + 4H+(aq) + 4e− → 2H2O(l); E° = 1.23 V
The overall reaction is:
2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l); E°cell = 1.67 V
The ferrous ions are further oxidised by atmospheric oxygen to ferric ions, which come out as rust in the form of hydrated ferric oxide (Fe2O3.xH2O), with further production of hydrogen ions.
Which of the following takes place when a lead storage battery is discharged (in use)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Sulphuric acid is consumed
The lead storage battery has a lead anode, a grid of lead packed with \( \ce{PbO2} \) as cathode and 38% sulphuric acid as electrolyte. When it is in use (discharging), both electrodes turn into lead sulphate and the net cell reaction is:
\( \displaystyle \ce{Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)} \)
\( \ce{H2SO4} \) is on the left side, so it is used up while water is formed, and the acid becomes more dilute as the battery discharges.
Why the other options are incorrect
Option (1) SO2 is evolved — Incorrect. The sulphate ions end up in solid \( \ce{PbSO4} \); no gas is given off in the cell reaction.
Option (2) Lead is formed — Incorrect. Lead is consumed at the anode on discharge; Pb is formed again only when the battery is charged.
Option (3) Lead sulphate is consumed — Incorrect. \( \ce{PbSO4} \) is formed at both electrodes on discharge; it is converted back into Pb and \( \ce{PbO2} \) on charging.
NCERT (2026-27 edition):
2.6.2 Secondary Batteries – “The cell reactions when the battery is in use are given below:
Anode: \( \displaystyle \ce{Pb(s) + SO4^2-(aq) -> PbSO4(s) + 2e^-} \)
Cathode: \( \displaystyle \ce{PbO2(s) + SO4^2-(aq) + 4H^+(aq) + 2e^- -> PbSO4(s) + 2H2O(l)} \)
i.e., overall cell reaction consisting of cathode and anode reactions is:
\( \displaystyle \ce{Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)} \)”