For the cell Zn|Zn²⁺(0.01M)||Cu²⁺(1M)|Cu, if E°cell = 1.1V, the EMF at 298K is (R = 8.314, F = 96500):

Using the Nernst equation:

$$E = E^{\circ} - \frac{RT}{nF}\ln Q = E^{\circ} - \frac{0.0592}{n}\log Q$$

For $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, $n = 2$

$$Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.01}{1} = 0.01$$

$$E = 1.1 - \frac{0.0592}{2}\log(0.01) = 1.1 - 0.0296 \times (-2)$$

$$= 1.1 + 0.0592 = \boxed{1.1592 \text{ V}}$$