# Homework Help: Chromium coated sphere problem

1. Nov 23, 2014

### oxon88

1. The problem statement, all variables and given/known data

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 μm 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–3 respectively

2. Relevant equations

(w=ItA/6F) x (efficiency (%) / 100)

3. The attempt at a solution

Weight deposited, W =
Current, I = 20 amps
Atomic Weight, A = 52 gcm-3
Efficiency = 15%
sphere surface area = 4*π*502 = 31415.92mm2

im not sure where to start here. can anyone advise?

2. Nov 23, 2014

### Staff: Mentor

Calculate volume of the chromium layer.

3. Nov 23, 2014

### oxon88

Ok thanks,

The sphere volume will be 4/3*π*(503) = 523598.7756 mm3

The diameter with the chromium layer will be 100.02mm

So the volume will be 4/3*π*(50.013) = 523912.9977 mm3

523912.9977 mm3 - 523598.7756 mm3 = 314.2221 mm3

4. Nov 23, 2014

### SteamKing

Staff Emeritus
Is the chromium somehow deposited only on one side of the sphere?

Also, since the thickness of the layer of chromium deposited << diameter of the sphere, the volume of chromium is approximately equal to the surface area of the sphere multiplied by the thickness of the chromium layer.

5. Nov 23, 2014

### Staff: Mentor

Once you have the volume, find mass from density. Then, don't expect any more spoonfeeding.

6. Nov 23, 2014

### oxon88

Ah, thank you. I see my mistake. The diameter will be 100.04mm

Therefore the sphere volume will be 4/3*π*(50.023) = 524227.3455

So then the difference in volume will be: 524227.3455 mm3 - 523598.7756 = 628.5677 mm3

Surface area of the sphere will be 4*π*50.022 = 31441.0643 mm3

31441.0643 mm3 * 20μm = 628.82 mm3

7. Nov 23, 2014

### oxon88

Ok, so

Mass = Density * Volume = 7.2 g cm-3 * 625.5677 mm3 = 4504.08744 mg

mass = 4.5 grams

Are my units correct?

8. Nov 23, 2014

### Staff: Mentor

Not clear how you did it, but the mass looks OK.

9. Nov 23, 2014

### SteamKing

Staff Emeritus
The surface area of the sphere has units of mm2 in this case. When you multiply the surface area by the thickness of the coating, then you get mm3

10. Nov 24, 2014

### oxon88

Many thanks.

I guess I can now just use my original formula?

Weight deposited, W = 4.5 grams
Current, I = 20 amps
Atomic Weight, A = 52 gcm-3
Efficiency = 15%

(w=ItA/6F) * (efficiency(%) / 100)

11. Nov 25, 2014

### oxon88

so after plugging the numbers in to the formula, I get:

4.5 = (20 * t * 52) / (6 * 96500)

4.5 = 1040 * t / 579000

4.5 = 0.0017962 * t

t = 4.5 / 0.0017962

t = 2505.2885

t = 2505.2885 seconds = 41.75 minutes.

Does this look like an acceptable value? I realise that I need to take the 15% efficiency into account, but I will need some extra time to think about that.

Last edited: Nov 25, 2014
12. Nov 26, 2014

### oxon88

ok so taking into account the 15% efficiency factor...

4.5 = (20*t*52 / 6 * 96500) * (15 / 100)

4.5 = (13 * t / 48250)

t = 16701.9 seconds

t = 278.365 minutes

13. Nov 26, 2014

### Staff: Mentor

Don't round down intermediate values, only the final result. But yes, 4 hours 40 minutes is what I got.

14. Dec 4, 2014

### oxon88

ok thanks for all the help