Problem 2

Cathodic protection is widely used to prevent metal corrosion by attaching a more active sacrificial metal to the structure. (a) In an experiment on protecting steel in seawater, a 25.0 g zinc block was attached to a steel apparatus. After some time, the block was reweighed at 28.0 g; assume that the only oxidation product is

\ce{Zn(OH)2}

adhering to the block (

M(\ce{Zn}) = 65.38

M(\ce{Zn(OH)2}) = 99.40

g mol

^{-1}

). Calculate the percentage of zinc that has oxidised. (b) Find the maximum service time (in hours) of this 25.0 g zinc block if the average protective current generated is 25 mA. (c) A steel ship hull with an immersed area of 1000 m

^2

requires an average protective current density of 2.5 mA m

^{-2}

. If zinc sacrificial anodes are used and 10 % of the zinc is lost to secondary processes, calculate the mass of zinc required per year (365 days).

Step 3 of 3: Annual zinc mass for hull protection

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I = 1000 \times 2.5 \times 10^{-3} = 2.5\ \text{A};\quad m = \frac{I t M}{2 F \times 0.90} = \frac{2.5 \times 31536000 \times 65.38}{2 \times 96485 \times 0.90} = 29.67\ \text{kg}

Analysis

The total current is $I = 1000\ \text{m}^2 \times 2.5 \times 10^{-3}\ \text{A m}^{-2} = 2.5$ A. In 365 days ( $t = 31536000$ s), $Q = 7.884 \times 10^7$ C. Theoretical zinc required is $Q M/(2F) = 26.70$ kg. Correcting for the 10 % loss gives $m = 26.70 / 0.90 = 29.67$ kg (matches the official key: 29.68 kg).