The \(E_{\text {cell }}^{\circ}\) for the following cell is \(1.54 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) : $$\mathrm{U}(s) \mid \mathrm{U}^{3+}(a q)\left\|\mathrm{Ni}^{2+}(a q)\right\| \mathrm{Ni}(s)$$ Calculate the standard reduction potential for the \(\mathrm{U}^{3+} / \mathrm{U}\) half-cell.

Understand that the net potential of the cell, known as \(E_{\text {cell }}^{\circ}\), is equal to the sum of the standard potentials of the half-reactions. We'll denote the reduction potential of the \(\mathrm{Ni}^{2+}/\mathrm{Ni}\) half-cell as \(E_{\text {Ni }}^{\circ}\), and the standard reduction potential of the \(\mathrm{U}^{3+}/\mathrm{U}\) half-cell as \(E_{\text {U }}^{\circ}\). So, the cell potential \(E_{\text {cell }}^{\circ}\) is given by the equation \(E_{\text {cell }}^{\circ} = E_{\text {U }}^{\circ} + E_{\text {Ni }}^{\circ}\). Note that the standard reduction potential for the \(\mathrm{Ni}^{2+}/\mathrm{Ni}\) half-cell, \(E_{\text {Ni }}^{\circ}\), is \(-0.257 \, \mathrm{V}\) at 25°C and the known \(E_{\text {cell }}^{\circ}\) of the whole cell is \(1.54 \, \mathrm{V}\). The goal is to calculate \(E_{\text {U }}^{\circ}\).

Substitute the given values into the equation to solve for \(E_{\text {U }}^{\circ}\). Rearranging the cell potential equation, we get \(E_{\text {U }}^{\circ} = E_{\text {cell }}^{\circ} - E_{\text {Ni }}^{\circ}\). Substitute \(E_{\text {cell }}^{\circ} = 1.54 \, \mathrm{V}\) and \(E_{\text {Ni }}^{\circ} = -0.257 \, \mathrm{V}\) into the equation, we finally get the standard reduction potential for the \(\mathrm{U}^{3+}/ \mathrm{U}\) half-cell.