For the redox change;

\(Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s)\), taking place

                 0.1 M       0.1 M

In a cell \(E^0_{cell}\) is 1.10 volt. Ecell for the cell would be

1.07 V

To determine the cell potential (\(E_{\text{cell}}\)) for the given redox change, we can use the Nernst equation:

\[ E_{\text{cell}} = E^0_{\text{cell}} - \left(\frac{0.0592}{n}\right)\log\left(\frac{{[Zn^{2+}]}}{{[Cu^{2+}]}}\right) \]

Given that \(E^0_{\text{cell}}\) is 1.10 V, and the concentrations of \([Zn^{2+}]\) and \([Cu^{2+}]\) are both 0.1 M, we can substitute these values into the equation. Since the number of electrons transferred in this reaction (\(n\)) is 2, we can calculate the value of \(E_{\text{cell}}\):

\[ E_{\text{cell}} = 1.10 \, \text{V} - \left(\frac{0.0592}{2}\right)\log\left(\frac{{0.1}}{{0.1}}\right) \]

Simplifying the equation, we get:

\[ E_{\text{cell}} = 1.10 \, \text{V} - 0.0296 \log(1) = 1.07 \, \text{V} \]

Therefore, the correct answer is (1) 1.07 V.