Question

A baby's spoon with an area of \(6.25 \mathrm{~cm}^{2}\) is plated with silver from \(\mathrm{AgNO}_{3}\) using a current of \(2.00 \mathrm{~A}\) for two hours and 25 minutes. (a) If the current efficiency is \(82.0 \%\), how many grams of silver are plated? (b) What is the thickness of the silver plate formed \(\left(d=10.5 \mathrm{~g} / \mathrm{cm}^{3}\right) ?\)

Short answer

Question: Determine the thickness of the silver plate formed during the silver plating process. Answer: 0.243 cm

Step-by-step solution

Convert time to seconds

To get a consistent unit for the time, convert 2 hours and 25 minutes to seconds: 1 hour = 3600 seconds 1 minute = 60 seconds Therefore, 2 hours and 25 minutes = \((2 \times 3600) + (25 \times 60) = 8700\) seconds.

Calculate the charge

To find the charge (Q) in Coulombs passed through the circuit, use the formula: \(Q = I \times t\) Where, I = Current (2.00 A) t = Time (8700 seconds) \(Q = 2.00 \times 8700 = 17400 \mathrm{C}\)

Calculate the mass of plated silver

Using Faraday's law, we can find the amount of silver deposited as follows: \(1 \mathrm{mol~of~Ag} = 1 \mathrm{eq~of~Ag} = 1 \mathrm{ eq} \times F\) Where F is Faraday's constant and equals 96500 C/mol. Thus, \(1 \mathrm{mol~of~Ag} = 96500 \mathrm{C}\) Now, we can find the moles of silver plated: \(n_\mathrm{Ag} = \frac{Q}{F} = \frac{17400}{96500} = 0.1803 \mathrm{mol}\) Next, use the molar mass of silver (M = 108 g/mol) to find the mass of silver plated: \(m = n_\mathrm{Ag} \times M = 0.1803 \times 108 = 19.44 \mathrm{g}\)

Adjust for current efficiency

Now, we take into account the current efficiency (82.0%): \(19.44 \times 0.82 = 15.94 \mathrm{~g}\) So, 15.94 grams of silver have been plated.

Calculate the thickness of the silver plate

We are given the density (d) of silver as 10.5 g/cm³. First, we need to find the volume of the plated silver: \(V = \frac{m}{d} = \frac{15.94}{10.5} = 1.52 \mathrm{~cm}^{3}\) Now, divide the volume by the spoon's area to find the thickness of the plated silver: \(\mathrm{Thickness} = \frac{V}{\mathrm{Area}}\) Given are of the spoon = 6.25 cm² \(\mathrm{Thickness} = \frac{1.52}{6.25} = 0.243 \mathrm{~cm}\) The thickness of the silver plate formed is 0.243 cm.