Math Help - Electroplating equation problem

1. Electroplating equation problem

A sphere 100 mm diameter is to be coated with chromium from a solution
containing chromium in the six valent (hexavalent) state. How much time
would be needed to produce a coating 20 micro millimetres thick if;

the current is 20 A

• the cathodic efficiency is 15%

• the atomic weight and density of chromium are 52 and 7.2 gcm respectively

So far Iam working on the basis that the equation to be transposed should be

w=ItA/2F

Any help here would be extremely grateful as I am really struggling with this one. Many thanks.....

2. Re: Electroplating equation problem

The easy part: The volume of a sphere of radius r is given by [tex]\frac{4}{3}\pi r^3[/itex]. So the 100 mm sphere will have volume $\frac{4}{3}\pi (100)^3$. The 20 micro millimeter coating will add .02 millimeters to the radius and a 100.002 mm sphere will have volume [tex]\frac{4}{3}\pi (100.02)^3[/quote].

The difference in the two volumes, $\frac{4}{3}\pi\left((100.002)^3- 100^3)$, is the volume of chromium needed.

Now, the rest of the problem depends upon how fast that chromium can be deposited depends upon information I don't have but I suspect you do. In particular, I have no idea how you would use the equation you give, w= itA/2F, because I have no idea what those letters represent. You should have learned that when you learned the formula.

3. Re: Electroplating equation problem

That was the wrong equation I have since found. All I was given was the info in post one. I got a value of 1 hour 11 minutes but doubt that is correct.
Hope you can help though, been at that question for days and I do mean days....I have no idea how to solve it

4. Re: Electroplating equation problem

HallsofIvy could you help further with my problem please I still haven't got any further with it

5. Re: Electroplating equation problem

Originally Posted by Jock
HallsofIvy could you help further with my problem please I still haven't got any further with it
As HallsofIvy stated, he does not have access to your textbook, nor the formulas in the textbook. It will contain the formulas that will explain how current, cathode current efficiency, and atomic weight/density of the material used affect the rate at which the electroplating coating develops. You should probably ask this question on a chemistry or physics forum. Here are Faraday's laws of electrolysis:

$Q \propto zm/M$

$Q = It = zFn$

where $Q$ is the charge passed, $I$ is the current passed, $z$ is the oxidation state, $m$ and $M$ are the mass (in grams) and molar mass respectively of oxidized or reduced species, $F$ is the Faraday constant ($F = 96,485 \text{ C mol}^{-1}$), and $n$ is the amount of substance oxidized or reduced (in moles).

So, according to the information you gave, $I = 20$, $z = 6$, and $M=52$ so the weight of the material theoretically deposited is given by $m = \dfrac{ItM}{Fz} = \dfrac{104}{57891}t$.

By the density of chromium, I am assuming you are giving it as $7.2\text{ g}/\text{cm}^3$. Next, 100mm = 10cm and 100.02mm = 10.002cm. So, the volume of chromium needed is (as HallsofIvy told you), $\dfrac{4}{3}\pi\left(10.002^3-10^3\right) \text{ cm}^3$. Hence, the weight of the chromium needed is $7.2\dfrac{4}{3}\pi \left(10.002^3-10^3\right) \text{ g}$.

Finally, you are told that the cathodic efficiency is 15%. That means that in this reaction, only 15% of the mass theoretically reduced will actually be reduced. So, you need $0.15m$ to be equal to the amount of chromium you want deposited. Putting everything together, this gives

$t = \dfrac{57891}{104}\dfrac{1}{0.15}\dfrac{7.2\cdot 4\pi}{3}\left(10.002^3-10^3\right)\text{ s} \approx 18.66\text{ hr}$.

I am not a chemist nor a physicist, so I have no idea if I am even in the ballpark. I just tried to plug in the values you gave into the formulas given by Faraday's laws. I assumed that amps was the correct unit of current (since Faraday's constant is coulombs per mole, and amps is coulombs per second). I assumed that $z=6$, although I don't know enough about the reaction to know if that is correct, either. All of this should be in your book. I am guessing.

6. Re: Electroplating equation problem

I have got 8 hours and 4 minutes its not quite right though as that works out at 7.8 grams of chromium

7. Re: Electroplating equation problem

I showed you my calculation. I don't know what numbers you are using to get your calculations, so I have no way of helping you further.

8. Re: Electroplating equation problem

I follow your method but I haven't been taught anything like so I was using the only method I know which is the weight deposited method in post 1. Very very confused with this

9. Re: Electroplating equation problem

The formula you posted in (1) assumes that the substance has oxidation state 2. Your problem gives that the substance is in the six valent (hexavalent) state. So, your formula needs to be modified (as I showed) to:

$w = ItA/6F$ where $I$ is current in amps, $A$ is atomic weight in grams/mole, and $F$ is Faraday's constant. So, plugging in the numbers you are given, your formula becomes $w = \dfrac{20\cdot t\cdot 52}{6\cdot 96485} = \dfrac{104}{57891}t$ (I gave this equation as $m = \dfrac{104}{57891}t$).