\begin{align*} \text{Given: } E_{\text{Ni}^0} &= -0.25 \, \text{V}, \quad E_{\text{Al}^0} = -1.66 \, \text{V} \\ \text{To find: } \text{Standard cell potential} \\ \text{Formula: } E_{\text{cell}^0} &= E_{\text{cathode}^0} - E_{\text{anode}^0} \\ \text{Calculation: } \text{Electrode reactions are} \\ \text{At anode: } \ce{Al_{(s)} &-> Al^{3+}_{(aq)} + 3e^-} \\ \text{At cathode: } \ce{Ni^{2+}_{(aq)} + 2e^- &-> Ni_{(s)}} \\ \text{The standard electrode potential is given by} \\ E_{\text{cell}^0} &= E_{\text{cathode}^0} - E_{\text{anode}^0} \\ E_{\text{cell}^0} &= E_{\text{Ni}^0} - E_{\text{Al}^0} \\ &= (-0.25 \, \text{V}) - (-1.66 \, \text{V}) \\ &= 1.41 \, \text{V} \\ \text{The standard cell potential for the reaction is } 1.41 \, \text{V.} \end{align*}