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 2 of 3: Maximum service lifetime of the block

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t = \frac{2 n(\ce{Zn}) F}{I} = \frac{2 \times (25.0/65.38) \times 96485}{0.025} = 2.951 \times 10^6\ \text{s} = 819.7\ \text{h}

Analysis

$\ce{Zn -> Zn^{2+} + 2e^-}$ , so $n_e = 2 n(\ce{Zn}) = 2 \times (25/65.38) = 0.7647$ mol electrons; $Q = n_e F = 73787$ C. At $I = 25$ mA = $0.025$ A, $t = Q/I = 2.951 \times 10^6$ s = 819.7 hours (matches the official key: 819.8 h).