An industrial chemist is designing a corrosion-prevention system for a buried steel pipeline. Steel consists primarily of iron, which oxidises according to the half-equation:
Fe2+(aq)+2e−⇌Fe(s)Eθ=−0.44 V \text{Fe}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Fe}(\text{s}) \quad E^\theta = -0.44\text{ V} Fe2+(aq)+2e−⇌Fe(s)Eθ=−0.44 VThe pipeline is to be coated with a metal to provide both a physical barrier and sacrificial protection (preventing corrosion of the underlying iron even if the coating is scratched).
Using the standard electrode potentials below, which coating option should the chemist select, and what is the electrochemical justification?
Tin (Sn\text{Sn}Sn), because Eθ(Sn2+/Sn)=−0.14 VE^\theta(\text{Sn}^{2+}/\text{Sn}) = -0.14\text{ V}Eθ(Sn2+/Sn)=−0.14 V is more positive than that of iron, meaning the tin layer is highly resistant to oxidation.
Zinc (Zn\text{Zn}Zn), because Eθ(Zn2+/Zn)=−0.76 VE^\theta(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}Eθ(Zn2+/Zn)=−0.76 V is more negative than that of iron, meaning zinc is a stronger reducing agent and will oxidise preferentially.
Nickel (Ni\text{Ni}Ni), because Eθ(Ni2+/Ni)=−0.25 VE^\theta(\text{Ni}^{2+}/\text{Ni}) = -0.25\text{ V}Eθ(Ni2+/Ni)=−0.25 V is closer to that of iron, minimising the potential difference and preventing galvanic cell formation.
Copper (Cu\text{Cu}Cu), because Eθ(Cu2+/Cu)=+0.34 VE^\theta(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}Eθ(Cu2+/Cu)=+0.34 V has a positive potential, which completely prevents the oxidation of iron.