# Calculating time needed to coat an object using the process of electrolysis

Suppose we have an object with a surface area of $$S=\pu{1.5 dm2}$$, and we need to coat it with hard chrome. The thickness of the coat should be $$d=\pu{80 \mu m}$$. The mixture used for coating is of the following mixture: $$\ce{CrO3}$$ $$\pu{290 g/L}$$, $$\ce{H2SO4}$$ $$\pu{2.5 g/L}$$. The current density is $$J = \pu{50 A dm-2}$$. The current utilization factor is $$\mu = 15\%$$. Using Faraday's law of electrolysis: $$\frac{m}{M}=\frac{q}{F}$$ and the fact that $$J_{\mathrm{actual}}=0.15\cdot\frac{q}{S\Delta t}$$ where $$S$$ represents the surface through which the constant current passes through, I got $$\Delta t = \pu{1425 s}$$, however the answer is $$\Delta t = \pu{2 h 22 min}$$.

The trick you're missing here is that you don't account for how many electrons are needed to reduce $\ce{CrO3}$ to $\ce{Cr}$. Cr has an oxidation state of 6 in $\ce{CrO3}$, so it takes 6 electrons to reduce every chromium atom. The full form of Faraday's law is $$\frac{m}{M} = \frac{q}{zF}$$ where $z$ is the number of electrons transferred. Therefore, you need 6 times as much charge as you initially calculated, which works out to 8550 seconds or 2:22:30.