In an electroplating process,it is desired to deposit 40 mg of silver on a metal part by using a current of 2.0 A.How long must the current be allowed to run to deposit this much silver? The silver ions are singly charged,and the atomic mass of silver is 108 g/mol.(e = 1.60 × 10^-19 C,N_A = 6.02 × 10²³ atoms/mol) A)16 s B)18 s C)20 s D)22 s

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of silver required to deposit 40 mg:

1 mol of Ag = 108 g

1 mol of Ag = 6.02 × 10^23 atoms

40 mg of Ag = 40/108 × 6.02 × 10^23 atoms

40 mg of Ag = 2.23 × 10^21 atoms

Now, we can calculate the amount of charge required to deposit 2.23 × 10^21 silver ions:

1 Ag ion = 1 e⁻ (since it is singly charged)

2.23 × 10^21 Ag ions = 2.23 × 10^21 e⁻

Q = ne

Q = (2.23 × 10^21)(1.60 × 10^-19)

Q = 3.57 × 10² C

Finally, we can use the formula for electric charge:

Q = It

where Q is the charge, I is the current, and t is the time.

Solving for t:

t = Q/I

t = 3.57 × 10² / 2.0

t = 178.5 s

Rounding off to the nearest second, we get:

t = 18 s

Therefore, the correct answer is B) 18 s.

In an Electroplating Process,it Is Desired to Deposit 40 Mg