$$ \ce{Al^{3+}(aq) + 3e^- -> Al(s)} \hskip{2em} E^\circ = \pu{-1.66 V} $$
$$ \ce{Ag^+(aq) + e^- -> Ag(s) } \hskip{2em} E^\circ = \pu{+0.80 V} $$
According to the standard reduction potentials given above, what is the standard cell potential for the reaction represented below?
$$ \ce{3Ag^+(aq) + Al(s) -> 3Ag(s) + Al^{3+}(aq)}$$
To obtain the reaction, we can reverse the first reaction.
We would also need to multiply each coefficient by 3 in the second reaction, but recall that this does not change the cell potential.
$$ \ce{Al^{3+}(aq) + 3e^- -> Al(s)} \hskip{2em} E^\circ = \pu{-1.66 V} $$
$$ \ce{ Al(s) -> Al^{3+}(aq) + 3e^- } \hskip{2em} E^\circ = \pu{+1.66 V} $$
$$ \ce{3Ag^+(aq) + 3e^- -> 3Ag(s) } \hskip{2em} E^\circ = \pu{+0.80 V} $$
$$ \ce{3Ag^+(aq) + Al(s) -> 3Ag(s) + Al^{3+}(aq)} \hskip{2em} E^\circ = 1.66 + 0.80 = 2.46$$