Copper–steel joint in a water line: which metal corrodes, and will one 50 g zinc block do?
A municipal water-supply line in Guwahati carries groundwater (25 °C, pH 6.5, [Fe2+] ≈ 1.0 × 10−3 mol L−1, air-saturated) through a mild-steel pipe joined to a copper fitting. After eight months the iron wall beside the joint shows deep pitting, while the copper surface remains bright. A maintenance note proposes either hot-dip galvanising with zinc or electroplating with tin, accepting that any coating will be scratched during handling. Separately, a sacrificial zinc block is to be bolted to a pump casing that will pass 0.020 A of galvanic current for 90 days; the supplier offers 50 g blocks. Data at 25 °C: E°(Fe2+/Fe) = −0.44 V, E°(Cu2+/Cu) = +0.34 V, E°(O2 + 4H+ → 2H2O) = +1.23 V, E°(Zn2+/Zn) = −0.76 V, E°(Sn2+/Sn) = −0.14 V. Decide which metal corrodes, which coating is safe once scratched, and whether one 50 g block is adequate.
1. Identify the anode and cathode at the joint and calculate the cell emf at the stated water composition. Is the attack on iron thermodynamically justified?
Read the worked answer
Electrode assignment. E°(Fe2+/Fe) is far more negative than E°(Cu2+/Cu), so iron is oxidised: anode Fe(s) → Fe2+(aq) + 2e−. The copper fitting is the cathode, where dissolved O2 is reduced.
Cathode potential at pH 6.5 (taking p(O2) ≈ 1 atm for simplicity): E = 1.23 − 0.0591 × pH = 1.23 − 0.384 ≈ +0.846 V.
Anode potential at [Fe2+] = 1.0 × 10−3 M: E = −0.44 + (0.0591/2) log(10−3) = −0.44 − 0.089 = −0.529 V.
Cell emf: Ecell = 0.846 − (−0.529) ≈ +1.38 V, so ΔG < 0 and pitting is spontaneous.
H+ reduction at this pH would give E = −0.0591 × 6.5 = −0.384 V, i.e. Ecell ≈ +0.15 V - possible, but far weaker than O2 reduction. This is why oxygenated near-neutral water still pits iron, and why acidic water is worse. Copper survives not because it is inert but because it is cathodically protected: it supplies electrons instead of losing them.
2. A scratched zinc coating keeps protecting the pipe, but a scratched tin coating makes rusting faster than on bare iron. Explain both outcomes using the E° values.
Read the worked answer
The decisive point is the relative position of the coating and iron in the activity series, not the coating's own corrosion resistance.
Zinc (E° = −0.76 V) is more active than iron (E° = −0.44 V). At a scratch, zinc is the anode and the exposed iron becomes the cathode: Zn → Zn2+ + 2e−, O2 + 2H2O + 4e− → 4OH−. Iron is protected galvanically, and protection continues even where the coating is broken, because the zinc around the scratch keeps dissolving. This is sacrificial protection.
Tin (E° = −0.14 V) is less active than iron. At a scratch, iron becomes the anode and the large tin area becomes the cathode. The result is a small anode coupled to a large cathode: all the corrosion current is concentrated on a tiny iron area, giving rapid, deep pitting - worse than an uncoated pipe, where the current is spread over the whole surface.
So the specification must call for galvanising; tin plating is acceptable only where the coating is guaranteed intact (as in tin cans, where it protects the food from iron rather than protecting iron from rust).
3. Using the sacrificial-anode duty (0.020 A for 90 days), calculate the zinc consumed and judge whether the 50 g block is acceptable.
Read the worked answer
Charge passed: Q = I × t = 0.020 A × (90 × 86400 s) = 1.555 × 10^5 C.
Moles of electrons: n(e−) = Q/F = 1.555 × 10^5 / 96485 ≈ 1.61 mol.
Zinc dissolution: Zn → Zn2+ + 2e−, so n(Zn) = 1.61/2 ≈ 0.806 mol.
Mass: m = 0.806 × 65.4 ≈ 52.7 g.
Judgement. One 50 g block is nominally short of the 52.7 g required even before allowances, so it fails. Real designs also need a safety factor (often 1.5–2) and must allow for the block's efficiency falling as corrosion products and scale cover it. A realistic specification would be two 50 g blocks, or one block of about 80–100 g, with a scheduled inspection interval. Note also that the calculation assumes 100% current efficiency on zinc; if any current leaks to the pipe through a poor mounting, the block lasts longer but protects less, so the electrical bond must be checked.