Question 20.7: Determining the Direction of Spontaneity from Electrode Pote...
Determining the Direction of Spontaneity from Electrode Potentials
Consider the reaction
\mathrm{Zn}^{2+}(a q)+2 \mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Zn}(s)+2 \mathrm{Fe}^{3+}(a q)
Does the reaction go spontaneously in the direction indicated, under standard conditions?
PROBLEM STRATEGY
Find the oxidizing agents in the equation; one is on the left side and the other on the right side. Locate these oxidizing agents in a table of electrode potentials. (The oxidizing agent is on the left side of the reduction half-reaction.) The stronger oxidizing agent is the one involved in the half-reaction with the more positive standard electrode potential.
In this reaction, \mathrm{Zn}^{2+} is the oxidizing agent on the left side; \mathrm{Fe}^{3+} is the oxidizing agent on the right side. The corresponding standard electrode potentials are
\begin{aligned} & \mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}(s) ; E^{\circ}=-0.76 \mathrm{~V} \\ & \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) ; E^{\circ}=0.77 \mathrm{~V} \end{aligned}
The stronger oxidizing agent is the one involved in the half-reaction with the more positive standard electrode potential, so \mathrm{Fe}^{3+} is the stronger oxidizing agent. The reaction is nonspontaneous as written.