# Battery EMF vs SOC (state of charge)

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?

• Sounds like a chemistry question. – Olin Lathrop May 25 '14 at 12:58
• @OlinLathrop: Not really -- it's more about the physical engineering of a commerically-viable cell. – Dave Tweed May 25 '14 at 13:56

## 1 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.

• Hmmm that makes sense. Simple lead acid batteries have very small "extra voltage" and after charging voltage drops without load to 2.1V per cell anyway. Modern NiMH have mix of metal hydrides and there are many diffrent reactions. – Kamil May 25 '14 at 18:11