Chemistry Forum

A metal cup of surface area \(200. cm^2\) needs to be electroplated with silver to a thickness of 0.200 mm. The density of silver is \(1.05 \times 10^4 \)kg m⁻³.The mass of a silver ion is \( 1 .79 \times\) 10⁻²⁵ kg and the charge is the same magnitude as that on an electron. How long does the cup need to be in the electrolytic tank if a current of 12.5 A is being used?

Any chemistry help will be accepted. I am working on this question.

m = I*t*M(Ag)/(n*F)
m = Rho*V
V = th* A
Rho*th*A = I*t*M(Ag)/(n*F)
t= Rho*th*A* n*F/(I*M(Ag))

m= mass, I = current, t= time, M(Ag) molar mass of silver, n= amount of electrons to Transfer, F Faradyconstant, Rho = specific gravity of silver, th = thickness of plated silver, A = surface of cup

The volume of silver that gets electroplated is 0.00000400 m³.
The mass of silver that gets electroplated is 0.00000400 m³ × 1.05 × 10⁴ kg m⁻³=0.042 kg.

The molar mass of silver= 107.8682 g/mol , so moles of silver 42 g/107.87 g = 0.38935756 mole of silver.

Hence, moles of electrons required Ag ⁺ (aq) + e ⁺→ Ag(s), so 0.38935756 mole of silver requires 0.38935756 mole of electrons.

If i plugged in these values in your formula answer is \(t= \frac{42 g \times 0.38935756 mol e^- \times 96485 C/mol} {12. 5 C/ s \times 107.87 g} = 1170.165 \) seconds

But the correct answer given is 3005 seconds ~ 50 mins approx.

How and why does the answer using your formula differ from the given answer?

What is wrong here?

You devided two times by the molar mass.
If you calculated n = m/M then you cannot do the thing twices.

t= n*F/I , n= number of moles of electrons required, F= Faraday constant, I= current in Ampere or Coulombs/second.

t=(0.38935756 Moles* 96485 C)/12.5 C/s = 3005.37 seconds, or 50 minutes approx.

Yes, now its correct.