$$\ce{Al^{3+}(aq) + 3e^- -> Al(s)} \hskip{2em} E^\circ = \pu{-1.66 V}$$ $$\ce{Ag^+(aq) + e^- -> Ag(s) } \hskip{2em} E^\circ = \pu{+0.80 V}$$

According to the standard reduction potentials given above, what is the standard cell potential for the reaction represented below?

$$\ce{3Ag^+(aq) + Al(s) -> 3Ag(s) + Al^{3+}(aq)}$$

Summary Submit

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.

$$\ce{Al^{3+}(aq) + 3e^- -> Al(s)} \hskip{2em} E^\circ = \pu{-1.66 V}$$ $$\ce{ Al(s) -> Al^{3+}(aq) + 3e^- } \hskip{2em} E^\circ = \pu{+1.66 V}$$ $$\ce{3Ag^+(aq) + 3e^- -> 3Ag(s) } \hskip{2em} E^\circ = \pu{+0.80 V}$$

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$$\ce{3Ag^+(aq) + Al(s) -> 3Ag(s) + Al^{3+}(aq)} \hskip{2em} E^\circ = 1.66 + 0.80 = 2.46$$