Describe the components and reactions of a Zinc-copper galvanic cell, including standard electrode potentials and the overall cell potential.

Here's the solution for the Zinc-copper cell:

i)
The standard electrode potential of Zinc ($-0.76 \text{ V}$) is more negative than that of Copper ($+0.34 \text{ V}$). This means Zinc will be oxidized (anode) and Copper ions will be reduced (cathode).
W is the electrode where oxidation occurs.
The identity of W is the Zinc electrode ($\text{Zn}$).

ii)
X is the electrolyte in the half-cell where reduction occurs. Copper ions are reduced to copper metal.
The identity of X is a Copper(II) salt solution (e.g., $\text{CuSO}_4(aq)$ or $\text{Cu(NO}_3)_2(aq)$).

iii)
The equation for the overall cell reaction:
Anode (oxidation): $\text{Zn}(s) \to \text{Zn}^{2+}(aq) + 2\text{e}^-$
Cathode (reduction): $\text{Cu}^{2+}(aq) + 2\text{e}^- \to \text{Cu}(s)$
Overall cell reaction:

$$\boxed{\text{Zn}(s) + \text{Cu}^{2+}(aq) \to \text{Zn}^{2+}(aq) + \text{Cu}(s)}$$

iv)
The reading on the voltmeter is the standard cell potential ($E^\circ_{\text{cell}}$).

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}}$$ $$E^\circ_{\text{cell}} = (+0.34 \text{ V}) - (-0.76 \text{ V})$$ $$E^\circ_{\text{cell}} = 0.34 \text{ V} + 0.76 \text{ V}$$ $$E^\circ_{\text{cell}} = \boxed{1.10 \text{ V}}$$