Step 1: Introduction to Lead Storage Battery - The lead storage battery is a secondary battery that can be recharged.
- It consists of lead (Pb) as the anode and lead dioxide (PbO\(_2\)) as the cathode, immersed in sulfuric acid (H\(_2\)SO\(_4\)) electrolyte.
Step 2: Cell Reactions During Discharge
Anode Reaction (Oxidation): \[ \text{Pb (s)} + \text{SO}_4^{2-} (aq) \rightarrow \text{PbSO}_4 (s) + 2e^- \]
Cathode Reaction (Reduction): \[ \text{PbO}_2 (s) + 4H^+ (aq) + \text{SO}_4^{2-} (aq) + 2e^- \rightarrow \text{PbSO}_4 (s) + 2H_2O (l) \]
Overall Cell Reaction: \[ \text{Pb} (s) + \text{PbO}_2 (s) + 4H^+ (aq) + 2\text{SO}_4^{2-} (aq) \rightarrow 2\text{PbSO}_4 (s) + 2H_2O (l) \]
Step 3: Working Principle - During discharge, Pb and PbO\(_2\) get converted into PbSO\(_4\), releasing electrical energy.
- During charging, the reaction is reversed, regenerating Pb and PbO\(_2\).
(i) Explain Kohlrausch's law. Write its applications.
Calculate the e.m.f. of the following cell: \[ \text{Cu(s)} | \text{Cu}^{2+} (1M) || \text{Ag}^{+} (0.01M) | \text{Ag(s)} \] Given: \[ E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34V, \quad E^0_{\text{Ag}^{+}/\text{Ag}} = +0.80V \]