Chromium coated sphere problem

[oxon88]

Homework Statement

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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

Homework Equations

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

The Attempt at a Solution

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Weight deposited, W =
Current, I = 20 amps
Atomic Weight, A = 52 gcm^-3
Faraday's constant, F = 96500
Efficiency = 15%
sphere surface area = 4*π*50² = 31415.92mm²im not sure where to start here. can anyone advise?

 

[Borek]

Calculate volume of the chromium layer.

 

[oxon88]

Ok thanks,

The sphere volume will be 4/3*π*(50³) = 523598.7756 mm³The 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

 

[SteamKing]

Ok thanks,

The sphere volume will be 4/3*π*(50³) = 523598.7756 mm³The diameter with the chromium layer will be 100.02mm

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.

 

[Borek]

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

 

[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

 

[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?

 

[Borek]

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

 

[SteamKing]

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

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

 

[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
Faraday's constant, F = 96500
Efficiency = 15%

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

 

[oxon88]

so after plugging the numbers into 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 realize that I need to take the 15% efficiency into account, but I will need some extra time to think about that.

 

[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



Does my answer look feasible?

 

[Borek]

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

 

[oxon88]

ok thanks for all the help