Calculating the emf from standard potentials:
A galvanic cell consists of an `Al` electrode in a `1.0M Al(NO_(3))_(3)` solution and an `Fe` electrode in a `1.0M Fe(NO_(3))_(2)` solution. Calculate the standard emf of this electrohemical cell at `25^(@)C`. Also write the cell reaction.
Strategy: From the table of standard electrode potentials, write the two reduction half reaction and standard electrode potentials for the cell. Change the direction of the half-cell reaction corresponding tol the smaller (or more negative) electrode potential. Multiply the when the half-reactions are added the electrons cancel. The sum of the half-reactions is the cell reaction. Add the electrode potentials to get the cell emf.

Explanation

According to Table 3.1, the readuction half-reactons and standard electrode pottentials are `{:(Al^(3+)(Aq.,1M)+3e^(-)hArrAl(s)E_(Al^(3+)+//Al)^(@)=1.6V),(Fe^(2+)(aq.,1M)+2e^(-)hArrFe(s)E_(Fe^(2+)//Fe)=-0.41V):}`
Since `Al^(3+)//Al` couple has the smaller (or more negative) elecrode potential, it undergose oxidation while the `Fe^(2+)//Fe` couple undergoes reduction. We reverse the first half-cell reaction and its half-cell potential to obtain
`{:(Al(s)overset(rarr)(larr)Al^(3+)(aq., 1M)+3e^(-)E_(Al//Al^(3+))^(@)=+1.66V),(Fe^(2+)(aq.)+2e^(-)hArrFe(s)E_(Fe^(2+)//Fe)^(@)=-0.41V):}`
We obtain the cell emgf by adding the half-cell potenitals, Because we alos want the cell reaction, we multiply the first half reaction by `2` and the second half-reacton by `3`, so that when the half-reations are added, the electrons cancel. We can do so because as an intensive property, `E^(@)` is not affeacted by this procedure. The addition of half-reaction is displayed below. The cell emf is `1.25 V`.
`{:(2Al(s)hArr2Al^(3+)(Aq.,1M)+6e^(-)" "E_(Al//Al^(3+))^(@)=+1.66V),(3Fe^(2+)(aq., 1M)+6e^(-)hArr 3Fe(s)" "E_(Fe^(2+)//Fe)^(@)=0.41V),(bar(2Al(s)+3Fe^(2+)(Aq., 1M)hArr 2Al^(3+)(aq.,1M)+3Fe(s)E_("cell")^(@)=E_(R)^(@)-E_(L)^(@)=1.25V)):}`
Note sthat the cell emf is also the standard electrode potential for the cathode (right electrode) minus the standard electrode potential for the (left electrode). Thus, we could have calculated the cell emf from the formula:
`E^(@) = E_(R)^(@) - E_(L)^(@)`
`= (-0.41 V) - (-1.66 V)`
`= 1.25 V`