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Suppose we have an object with a surface area of $S=1.5~\mathrm{dm^2}$, and we need to coat it with hard chrome. The thickness of the coat should be $d=80~\mathrm{\mu m}$. The mixture used for coating is of the following mixture: $\ce{CrO3}$ $290~\mathrm{g/L}$, $\ce{H2SO4}$ $2.5~\mathrm{g/L}$. The current density is $J = 50~\mathrm{A~dm^{-2}}$. 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 $\Delta t = 1425~\mathrm{s}$, however the answer is $\Delta t = 2~\mathrm{h}~22~\mathrm{min}$.

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  • $\begingroup$ You are off by a factor of 6. Pay attention to the oxidation state of chromium $\endgroup$ – getafix Sep 6 '16 at 4:44
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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.

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