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$\text{E}^{\circ}_{\text{Al}^{3+}|\text{Al}}=-1.66\text{V, E}^{\circ}_{\text{Ag}^+|\text{Ag}}=+0.80\text{V}$
→$\text{E}^{\circ}_{\text{Al}^{3+}|\text{Al}}=-1.66\text{V, E}^{\circ}_{\text{Ag}^+|\text{Ag}}=+0.80\text{V}$