Question

Write the reactions taking place at anode and cathode in a lead storage battery, when it is in use.

Hint:

On discharge, lead is oxidised at the anode and lead dioxide is reduced at the cathode, and both form lead sulphate.

Answer 1

Step 1: Understanding the Concept:
A lead storage battery is a secondary cell used in cars. When it supplies current, it works as a galvanic cell, so it is in the discharging state.
The anode is the electrode where oxidation happens, and the cathode is the electrode where reduction happens.
The electrolyte is about 38 percent by mass sulphuric acid, \(\mathrm{H_2SO_4}\).

Step 2: Key Formula or Approach:
Find which electrode loses electrons and which one gains them.
The anode is a grid of lead, Pb. The cathode is a grid packed with lead dioxide, \(\mathrm{PbO_2}\).

Step 3: Reaction at the anode (oxidation):
Lead metal has oxidation state 0. It loses two electrons and goes to \(\mathrm{Pb^{2+}}\).
This \(\mathrm{Pb^{2+}}\) combines with sulphate ions of the electrolyte and settles as insoluble lead sulphate.
\[ \mathrm{Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-} \]

Step 4: Reaction at the cathode (reduction):
In \(\mathrm{PbO_2}\), lead is in the +4 state. It gains two electrons and goes to the +2 state.
The oxide ions take up hydrogen ions from the acid to form water. The \(\mathrm{Pb^{2+}}\) again forms \(\mathrm{PbSO_4}\).
\[ \mathrm{PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)} \]

Step 5: Check the overall reaction:
Add the two half reactions. The electrons cancel and we get the cell reaction.
\[ \mathrm{Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)} \]
Sulphuric acid is used up, so its density falls as the battery discharges.

Final Answer:
Anode: lead is oxidised to lead sulphate. Cathode: lead dioxide is reduced to lead sulphate.
\[ \boxed{\text{Anode: } \mathrm{Pb + SO_4^{2-} \rightarrow PbSO_4 + 2e^-}} \] \[ \boxed{\text{Cathode: } \mathrm{PbO_2 + SO_4^{2-} + 4H^+ + 2e^- \rightarrow PbSO_4 + 2H_2O}} \]