I want to ask why battery open circuit voltage drops when battery is discharged.
For alkaline battery – chemical reactions theoretically give 1.43V voltage.
(source: wikipedia)
The half-reactions are:
$$Zn(s) + 2OH^−(aq) \rightarrow ZnO(s) + H_2O(l) + 2e^−\ [e° = 1.28\,\mathrm{V}] $$
$$2MnO_2(s) + H_2O(l) + 2e^− \rightarrow Mn_2O_3(s) + 2OH^−(aq)\ [e° = +0.15 \,\mathrm{V}] $$Overall reaction:
$$Zn(s) + 2MnO_2(s) \Leftrightarrow ZnO(s) + Mn_2O_3(s)\ [e° = 1.43\,\mathrm{V}]$$
So – why new alkaline battery have voltage above 1.6V and it drops to 0.85-1V when battery is empty?
Voltage from chemical reaction (theoretical 1.43V in alkaline) is not constant?
Best Answer
While the equation you cited is indeed the basic reaction by which an alkaline cell works, the actual picture in a commercial cell is much more complicated. As the cell discharges, other reactions come into play that compete with the original reaction, and the external terminal voltage you can measure is a complex function of these various reactions combined with the effects of increasing internal current leakage and rising internal resistance as the concentration and physical distribution of reactants changes.