Question

A copper rod with a surface area of 5 cm² is coupled with an iron rod with a surface area of 10 cm² to form a galvanic cell. When the circuit is closed, the current flows for 2 hours and the weight loss of the iron rod was found to be 0.58 g. Determine the corrosion rate of the iron rod in millimeters per year (mm/yr) assuming a density of 7.86 g/cm³ for the iron.

Answer 1

First, we need to find the volume loss of the iron rod. We can do this by dividing the weight loss by the density of iron:Volume loss = Weight loss / DensityVolume loss = 0.58 g / 7.86 g/cmVolume loss = 0.07379 cmNow, we need to find the thickness loss of the iron rod. We can do this by dividing the volume loss by the surface area of the iron rod:Thickness loss = Volume loss / Surface areaThickness loss = 0.07379 cm / 10 cmThickness loss = 0.007379 cmSince the corrosion occurred over 2 hours, we need to find the corrosion rate per hour:Corrosion rate per hour = Thickness loss / 2 hoursCorrosion rate per hour = 0.007379 cm / 2Corrosion rate per hour = 0.0036895 cm/hourNow, we need to convert the corrosion rate per hour to millimeters per year:1 cm = 10 mm1 hour = 1/24 day1 day = 1/365.25 yearCorrosion rate per year = 0.0036895 cm/hour * 10 mm/cm * 24 hours/day * 365.25 days/yearCorrosion rate per year = 32.37 mm/yearTherefore, the corrosion rate of the iron rod is approximately 32.37 mm/year.