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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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