Suppose, the circuit is open.

I understood from @Dale's answer that the negative terminal of the battery is indeed electrostatically negatively charged. Suppose, it can have a charge of $-0.5C$.

However, I'm a bit confused about the positive terminal (Cu electrode). According to @Poutnik, in an open circuit, both these reactions are occuring at the two terminals,

$$\require{mhchem} \ce{Zn(s) <=> Zn^2+(aq) + 2 e-}$$ $$\ce{Cu(s) <=> Cu^2+(aq) + 2 e-}$$

However, Zn's tendency to dissolve is greater than that of Cu.

So, in an open circuit, if the electrostatic charge at the negative terminal (Zn electrode) is $-0.5C$, my hypothesis is that the electrostatic charge at the positive terminal (Cu electrode) will be say $-0.3C$.

In conclusion, is it appropriate for me to say that in an open circuit, the Cu electrode/positive terminal too is also negatively charged, but it is just less negatively charged than the Zn electrode/negative terminal?


1 Answer 1


While the electrodes do become charged, the amount of charge is not a fixed amount like -0.5 C. The electrodes become charged until the electrical potential difference between the electrode and the electrolyte is greater than the electrochemical potential of the reaction at the electrode surface. So the amount of charge depends not only on the battery, but also on the circuit.

For example, if it is in a circuit where the negative terminal is grounded, then the negative terminal will be uncharged and the positive terminal will be positively charged. Or if it is in a circuit where the positive terminal is grounded then the positive terminal will be uncharged and the negative terminal will be negatively charged. It is also possible to set it so that both terminals are positively charged (with the positive terminal being more charged), and it is possible to have both terminals negatively charged (with the negative terminal being more negatively charged).

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    $\begingroup$ @tryingtobeastoic oops, yes. Fixing it shortly $\endgroup$
    – Dale
    Apr 29, 2022 at 22:34
  • $\begingroup$ If the circuit is open (a wire isn't connecting the two terminals), and if there is no grounding, the positive side will be positively charged electrostatically and the negative side will be negatively charged electrostatically: is that fair to say, sir? $\endgroup$ Apr 30, 2022 at 6:00
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    $\begingroup$ @tryingtobeastoic if the battery is disconnected from any circuit (not just connected to an open circuit) including not being grounded and if the net charge on the battery is 0 and there is no external E field then yes it is fair to say that the positive side will be positively charged and the negative side will be negatively charged. $\endgroup$
    – Dale
    Apr 30, 2022 at 9:50
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    $\begingroup$ @tryingtobeastoic any specific statement about the charge on the electrode requires some more detail about the circuit. I think your confusion is that you are trying to make a general claim about something that requires detailed specifics. I never made any general claims about charge, just the electrochemical potential. I think it is a mistake to try to make general statements about the charge. Anyway, we are getting the “extended discussion” warning, so I am done. Sorry I couldn’t help better $\endgroup$
    – Dale
    Apr 30, 2022 at 10:30

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