We study that when we connect a wire between the terminals of a battery ,an electric field in established between terminals due to potential difference .Is the potential difference result of accumulated charges on terminals. In other words when an electrode is put in the solution like Zn in ZnSO4 will Zn electrode will become -vely charged ? If not then how does electric field established into wire at exact moment when we connect wire between terminals (electrodes) which essential for electric current .


You can imagine the cell electrodes as capacitors, that are charged by chemical way.

But the capacitance of these capacitors

$$C=\frac {\mathrm{d}q}{\mathrm{d}E}$$

is very small, so does the accumulated charge $q$.

The charge loses due pushing electrons through the wires of a closed circuit are balanced by simultaneous recharging by ongoing redox reactions.

If we consider the zinc electrode of the Daniell cell, the reaction :

$$\ce{ Zn -> Zn^2+ + 2 e-}$$

is charging the electrode negatively, while the opposite reaction:

$$\ce{ Zn^2+ + 2 e- -> Zn}$$

is charging the electrode positively.

The electrode gains such a potential, where both reactions are ongoing at the same rate.

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  • $\begingroup$ "You can imagine the cell electrodes as capacitors, that are charged by chemical way." So it means "Before" we connect wire between electrodes , there is charge (may be small) on the each electrode due to their chemical reaction to the solution and when we talk about open circuit "Potential difference" between battery terminal the Potential difference is due to "these charges" . Right ? (Important point is here that I am talking about scenario when there is no wire connected between terminals .) $\endgroup$ – Nanda Sinha Oct 6 '19 at 21:51
  • $\begingroup$ Yes, it means that. But the positive/negative electrode does not have necessarily positive/negative charge . Rather more positive/,more negative charge than the other. $\endgroup$ – Poutnik Oct 7 '19 at 4:38

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