To explain the chemical action that occurs in battery consider a voltaic cell. (a). In an insulating container there is a solution of sulphuric acid H2 SO4. In this solution electrodes of copper and zinc are immersed. In solution molecules of H2 SO4 separate into H+ and SO4 — ions. The action of this cell generating an e.m.f. is illustrated in various step as shown in Fig. (b), (c), (d), (e).
Now suppose a piece of metallic 7inc (Zn) inserted into the solution as indicated in (c) some of the Zn atoms from Zn++ Ions and enter the solution, In the ionization Zn joins the SO4 — ions and forms zinc sulphate ZnSO4 Each atom of Zn ionized therefore losses two electrons. These electrons are left on the zinc electrode as excess making that electrode negative w.r.t. the solution.
Now suppose a copper (Cu) electrode separated from zinc electrode is added. (d). Some of the H ions pick up one electron each from the copper become neutral gaseous hydrogen, which forms bubbles on the copper electrode. The loss of these electrons to the H ions causes copper electrode to have positive charge. Now when the circuit is completed as shown. (e) electrons flow from the ionized zinc through the zinc electrode to the external circuit, the copper plate and back to charge more H ions to gaseous hydrogen.
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