Grade 12

Electrolysis, corrosion and metal protection

Using electricity to force a reaction to run backwards, and the same electrochemistry behind rust.

1 Lower secondary 2 Upper secondary 3 Undergraduate 4 Advanced 5 Research

IntuitionIntuition: a galvanic cell run backwards

A galvanic cell produces electricity from a spontaneous reaction. In electrolysis an external source pushes electrons the other way and forces a non-spontaneous reaction, such as splitting water. Oxidation still happens at the anode and reduction at the cathode, but now the anode is the positive pole and the cathode the negative one.

Change the current and the time. The volume of

\ce{H2}

is always twice that of

\ce{O2}

and grows in proportion to the charge

It

SchoolSchool level: Faraday’s law

n(e^-) = \frac{It}{F},\qquad m = \frac{M\,I\,t}{z\,F},\qquad F = 96485\ \text{C/mol}

Example: Copper plating

A current of 2.0 A flows for 30 min through $\ce{CuSO4}$ solution. What mass of copper is deposited at the cathode?

Solution

$Q = It = 2.0 \times 1800 = 3600$ C, so $n(e^-) = 3600/96485 = 0.0373$ mol. $\ce{Cu^2+ + 2e- -> Cu}$ needs 2 electrons per atom: $n(\ce{Cu}) = 0.0187$ mol and $m = 0.0187 \times 63.55 \approx 1.19$ g.

In water electrolysis, $\ce{2H2O + 2e- -> H2 + 2OH-}$ at the cathode and $\ce{2H2O -> O2 + 4H+ + 4e-}$ at the anode. Four electrons make one $\ce{O2}$ but only two make one $\ce{H2}$ , hence the 2 : 1 volume ratio.

UndergraduateUndergraduate: corrosion as a short-circuited cell

Rusting is a tiny galvanic cell on the iron surface. Where iron is exposed it acts as anode, $\ce{Fe -> Fe^2+ + 2e-}$ ; elsewhere, where oxygen and water are present, the cathode reaction $\ce{O2 + 2H2O + 4e- -> 4OH-}$ takes place. The products form $\ce{Fe(OH)2}$ and finally hydrated iron(III) oxide, rust.

To protect iron, coat it (paint, zinc) or connect it to a more easily oxidized metal such as zinc or magnesium: the “sacrificial anode” corrodes instead. Zinc works because $E^\circ(\ce{Zn^2+/Zn}) = -0.76$ V is lower than $E^\circ(\ce{Fe^2+/Fe}) = -0.44$ V.