Question 7 of 8: Magnesium sacrificial anode in a hot water heater

Question 7: Magnesium sacrificial anode in a hot water heater (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A cathodic-protection anode consumed steadily over its service life:

Find. (a) the half-cell reaction occurring at the anode, and (b) the average current the anode has delivered to protect the steel tank over its fourteen-year life.

Approach. Identify the oxidation half-reaction that consumes the magnesium, convert the mass lost into moles and then into charge through Faraday’s law of electrolysis, and divide by the elapsed time to obtain the average current.

1. Part (a) — write the anode reaction. The anode of a galvanic couple is where oxidation occurs; magnesium is the most active of the common engineering metals and gives up two electrons: $$\boxed{\ \mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2e^{-}\ }$$ with a standard electrode potential of −2.37 V versus the standard hydrogen electrode. That answers part (a), and the valence of 2 quoted by the question is precisely the number of electrons in this equation.
2. Note the balancing cathodic reactions, which are what the anode is protecting against. The electrons released travel through the metal to the steel tank wall and are consumed there by $$\mathrm{O}_2 + 2\,\mathrm{H}_2\mathrm{O} + 4e^{-} \rightarrow 4\,\mathrm{OH}^{-} \qquad\text{(aerated water)}$$ and, in the deaerated hot water typical of a tank in service, $$2\,\mathrm{H}_2\mathrm{O} + 2e^{-} \rightarrow \mathrm{H}_2 + 2\,\mathrm{OH}^{-}$$ Because the steel is now a cathode receiving electrons, its own dissolution reaction Fe → Fe²⁺ + 2e⁻ is suppressed. This is the entire principle of sacrificial protection.
3. Part (b) — convert the mass lost into moles of magnesium. $$n_{Mg} = \frac{m}{M} = \frac{900\ \text{g}}{24.3\ \text{g/mol}} = 37.037\ \text{mol}$$
4. Convert moles into charge using Faraday’s law. Each mole of magnesium releases two moles of electrons, and each mole of electrons carries one Faraday of charge, so $$Q = n\,n_{Mg}\,F = 2(37.037)(96\,485) = 7.147\times 10^{6}\ \text{C}$$ Equivalently, 74.07 mol of electrons have crossed the metal–water interface over the life of the anode.
5. Express the service life in seconds. Taking a 365-day year, $$t = 14 \times 365 \times 24 \times 3600 = 4.4150\times 10^{8}\ \text{s}$$
6. Divide to obtain the average current. $$I = \frac{Q}{t} = \frac{7.147\times 10^{6}}{4.4150\times 10^{8}} = 0.016188\ \text{A}$$ $$\boxed{\ I = 16.19\ \text{mA}\ }$$ Combining the previous steps, the whole calculation is the single expression I = nFm/(Mt), which is Faraday’s law rearranged for current.
7. Cross-check against the theoretical capacity of magnesium. The charge magnesium can deliver per gram is $$\frac{nF}{M} = \frac{2(96\,485)}{24.3} = 7941\ \text{C/g} = 2.206\ \text{A}\cdot\text{h/g}$$ the standard textbook figure for magnesium anodes. Nine hundred grams therefore holds 1985 A·h, and spreading that over the 122 640 h of fourteen years returns 16.19 mA, confirming the answer by an independent route. The anode has been eaten at an average 64.29 g per year, or about 0.18 g a day.