A voltmeter of very high resistance reads 1.03 V across the cell Zn(s) mid Zn²⁺(aq, 1.0 mol/L) ∥…

Step 1. The equation. $E = E^\circ - \dfrac{0.0591\ \text{V}}{2}\log Q$, with $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$.

Step 2. Solve for log Q. $\log Q = \dfrac{(1.10\ \text{V} - 1.03\ \text{V}) \times 2}{0.0591\ \text{V}} = 2.369$.

Step 3. Solve for the copper ion. $Q = 10^{2.369} = 234$. So $[\text{Cu}^{2+}] = \dfrac{1.0\ \text{mol/L}}{234} = 0.00428\ \text{mol/L}$.