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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?

tryingtobeastoic
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1 Answers1

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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).

Dale
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    @tryingtobeastoic oops, yes. Fixing it shortly – Dale Apr 29 '22 at 22:34
  • A video that asserts that the battery's terminals are positively and negatively charged electrostatically – tryingtobeastoic Apr 30 '22 at 05:09
  • 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? – tryingtobeastoic Apr 30 '22 at 06:00
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    @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. – Dale Apr 30 '22 at 09:50
  • Ok, I understand. But I'm struggling to reconcile between what you just said and your previous answer. Let us consider a battery disconnected from a circuit. According to your previous answer, both Cu and Zn metal are getting dissolved and turning to Zn2+ and Cu2+ although Zn's dissolving tendency is greater. (continued) – tryingtobeastoic Apr 30 '22 at 09:57
  • Due to this, there will be excess charges (free electrons) present in the Zn electrode/terminal and electrostatic negative charge will develop. Now, the part that I'm struggling with will come up. Cu metal is also dissolving and forming Cu2+ ions and free electrons. Shouldn't there be a negative electrostatic charge in the Cu electrode as well? Granted, the dissolving tendency of Cu is lower, so the magnitude of negative charge will be lower in the Cu electrode than that in the Zn electrode. So, both electrodes/terminals have negative charge. (continued) – tryingtobeastoic Apr 30 '22 at 10:03
  • @tryingtobeastoic I am not sure what you think requires reconciliation. My previous answer made no claim that the battery met any of those conditions and therefore I never claimed there was a specific amount of charge. If it meets all those conditions then there will be a specific charge, otherwise not. What will be consistent is the potential difference, not the charge. – Dale Apr 30 '22 at 10:04
  • It's just that the magnitude of the negative charge in the Cu is lower. This is what I hypothesized after reading your first answer. However, now you are saying that the Cu terminal is positively charged, so I'm a bit confused :-) – tryingtobeastoic Apr 30 '22 at 10:05
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    @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 – Dale Apr 30 '22 at 10:30