Question

how long would it take to electroplate a flute with 28.3 g of silver (107.87 g/mol) at a constant current of 2.0 amps using AgNO3

Answer 1

It would take 211 hours to plate 28.3 g of silver on a flute.

The equation of the reaction;

Ag^+(aq) + e -----> Ag(s)

We know that 1 F of electricity is required to deposit 107.87 g of Ag. Also 1F = 96500 C

Now;

Since  107.87 g is deposited by 96500 C of electricity

28.3 g is deposited by  28.3 g ×  96500 C/ 107.87 g

= 25317 C

Also;

Q = It

I = current

t = time

25317 C =  2.0 amps × t

t = 25317 C /2.0 amps

t = 12658.5 seconds or 211 hours

Learn more about electrolysis: brainly.com/question/12054569

Answer 2

Answer:

• It would take $3.5hrs$ to electroplate the flute

Explanation:

Electrolysis equation is:

$Ag1+ + 1e- ------> Ag$

1 mol of Ag requires 1 mol of electron

1 mol of electron = 96485 C

So,

1 mol of Ag requires 96485 C

let us calculate mol of element deposited:

molar mass of Ag = $107.87 g/mol$

number of mol of Ag, $n = \frac{mass of Ag}{molar mass of Ag}$

$n = \frac{28.3}{107.87}\\\\n = 0.2623 mol$

$total charge = mol of element deposited * charge required for 1 mol\\\\total charge = 0.2623*9.649*10^4\\\\total charge = 2.531*10^4 C$

Therefore,

$time = \frac{Q}{i}\\\\= \frac{2.531*10^4}{2}\\\\= 1.265*10^4 seconds\\\\= 3.515 hr$

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