Question 20.8: Calculating the emf from Standard Potentials Calculate the s...
Calculating the emf from Standard Potentials
Calculate the standard emf of the following voltaic cell at 25^{\circ} \mathrm{C} using standard electrode potentials.
\operatorname{Al}(s)\left|\mathrm{Al}^{3+}(a q) \| \mathrm{Fe}^{2+}(a q)\right| \mathrm{Fe}(s)
What is the cell reaction?
PROBLEM STRATEGY
From a table of electrode potentials, write the two reduction half-reactions and standard electrode potentials for the cell. The cell notation assumes that the anode (the oxidation half-cell) is on the left. Change the direction of this half-reaction and the sign of its electrode potential. (Assuming that the cell notation was written correctly, you change the direction of the half-reaction corresponding to the smaller, or more negative, electrode potential.) Multiply the half-reactions (but not the electrode potentials) by factors so that 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.
An alternative strategy is to apply the equation
E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}
The reduction half-reactions and standard electrode potentials are
(BOTH APPROACHES)
\begin{aligned} & \mathrm{Al}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(s) ; E_{\mathrm{Al}}^{\circ}=-1.66 \mathrm{~V} \\ & \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) ; E_{\mathrm{Fe}}^{\circ}=-0.41 \mathrm{~V} \end{aligned}
You reverse the first half-reaction and its half-cell potential to obtain
\begin{aligned} & \mathrm{Al}(s) \longrightarrow \mathrm{Al}^{3+}(a q)+3 \mathrm{e}^{-} ;-E_{\mathrm{A} 1}^{\circ}=1.66 \mathrm{~V} \\ & \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) ; E_{\mathrm{Fe}}^{\circ}=-0.41 \mathrm{~V} \end{aligned}
You obtain the cell emf by adding the half-cell potentials. Because you also want the cell reaction, you multiply the first half-reaction by 2 and the second halfreaction by 3 , so that when the half-reactions are added, the electrons cancel. The addition of half-reactions is displayed below. The cell emf is \mathbf{1 . 2 5} \mathbf{V}.
\begin{array}{rlr} 2 \mathrm{Al}(s) & \longrightarrow 2 \mathrm{Al}^{3+}(a q)+6 \mathrm{e}^{-} & -E_{\mathrm{Al}}^{\circ}=1.66 \mathrm{~V} \\ 3Fe^{2+}(aq) + 6e^-& \longrightarrow 3Fe(s) & E_{\mathrm{Fe}}^{\circ}= -0.41 \mathrm{~V} \\ \hline 2 \mathrm{Al}(s)+ 3 \mathrm{Fe}^{2+}(a q) & \longrightarrow 2AI^{3+}(aq)+3 \mathrm{Fe}(s) & {\mathbf{E}_{\mathrm{Fe}}^{\circ}-E_{\mathrm{Al}}^{\circ}=1.25 \mathrm{~V}} \end{array}
Alternatively, you can calculate the cell emf from the formula
E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}
to obtain
\begin{aligned} E_{\text {cell }}^{\circ} & =E_{\mathrm{Fe}}^{\circ}-E_{\mathrm{Al}}^{\circ} \\ & =-0.41 \mathrm{~V}-(-1.66 \mathrm{~V}) \\ & =1.25 \mathrm{~V} \end{aligned}