The standard electrode potentials for two metals are given below.

\[\begin{array}{*{20}{l}} {{\text{A}}{{\text{l}}^{3 + }}{\text{(aq)}} + {\text{3}}{{\text{e}}^ - } \rightleftharpoons {\text{Al(s)}}}&{{E^\Theta } = - 1.66{\text{ V}}} \\ {{\text{N}}{{\text{i}}^{2 + }}{\text{(aq)}} + {\text{2}}{{\text{e}}^ - } \rightleftharpoons {\text{Ni(s)}}}&{{E^\Theta } = - 0.23{\text{ V}}} \end{array}\]

What is the equation and cell potential for the spontaneous reaction that occurs?

\(\begin{array}{*{20}{l}} {{\text{A.}}}&{2{\text{A}}{{\text{l}}^{3 + }}({\text{aq)}} + 3{\text{Ni(s)}} \to {\text{2Al(s)}} + 3{\text{N}}{{\text{i}}^{2 + }}({\text{aq)}}}&{{E^\Theta } = 1.89{\text{ V}}} \\ {{\text{B.}}}&{2{\text{Al(s)}} + 3{\text{N}}{{\text{i}}^{2 + }}({\text{aq)}} \to {\text{2A}}{{\text{l}}^{3 + }}({\text{aq)}} + 3{\text{Ni(s)}}}&{{E^\Theta } = 1.89{\text{ V}}} \\ {{\text{C.}}}&{2{\text{A}}{{\text{l}}^{3 + }}({\text{aq)}} + 3{\text{Ni(s)}} \to {\text{2Al(s)}} + 3{\text{N}}{{\text{i}}^{2 + }}({\text{aq)}}}&{{E^\Theta } = 1.43{\text{ V}}} \\ {{\text{D.}}}&{2{\text{Al(s)}} + 3{\text{N}}{{\text{i}}^{2 + }}({\text{aq)}} \to {\text{2A}}{{\text{l}}^{3 + }}({\text{aq)}} + 3{\text{Ni(s)}}}&{{E^\Theta } = 1.43{\text{ V}}} \end{array}\)

D