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.
In a dry cell (Leclanché cell), which of the following substances is reduced at the cathode?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Manganese dioxide
In the dry cell the zinc container is the anode and a graphite rod surrounded by powdered \( \ce{MnO2} \) and carbon is the cathode; the space between them holds a moist paste of \( \ce{NH4Cl} \) and \( \ce{ZnCl2} \). Reduction (gain of electrons) takes place at the cathode:
\( \displaystyle \ce{MnO2 + NH4^+ + e^- -> MnO(OH) + NH3} \)
Mn in \( \ce{MnO2} \): x + 2(−2) = 0, so x = +4.
Mn in \( \ce{MnO(OH)} \): x + 2(−2) + 1 = 0, so x = +3.
Manganese gains one electron, so \( \ce{MnO2} \) is the substance reduced.
Why the other options are incorrect
Option (1) Ammonium chloride — Incorrect. \( \ce{NH4^+} \) takes part in the cathode reaction, but it only gives up \( \ce{H^+} \) and leaves as \( \ce{NH3} \); nitrogen stays at −3, so it is not reduced.
Option (3) Zinc chloride — Incorrect. It is only part of the electrolyte paste and does not appear in the cathode reaction.
Option (4) Zinc — Incorrect. Zinc is the anode and is oxidised: \( \ce{Zn(s) -> Zn^2+ + 2e^-} \).
NCERT (2026-27 edition):
2.6.1 Primary Batteries – “… The electrode reactions are complex, but they can be written approximately as follows :
Anode: \( \displaystyle \ce{Zn(s) -> Zn^2+ + 2e^-} \)
Cathode: \( \displaystyle \ce{MnO2 + NH4^+ + e^- -> MnO(OH) + NH3} \)
In the reaction at cathode, manganese is reduced from the + 4 oxidation state to the +3 state.”