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

Question

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 $\Delta t = \pu{2 h 22 min}$.

Answer 1

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.