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?