A teapot with a surface area of 630 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic cell containing silver nitrate (Ag+ NO3−). The cell is powered by a 12.0-V battery and has a resistance of 1.90 Ω. If the density of silver is 1.05 104 kg/m3, over what time interval does a 0.133-mm layer of silver build up on the teapot? You first need to calculate the total number of silver atoms contained in the plated volume of silver. Note that the nitrate ends up on the other electrode. h
Question
A teapot with a surface area of 630 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic cell containing silver nitrate (Ag+ NO3−). The cell is powered by a 12.0-V battery and has a resistance of 1.90 Ω. If the density of silver is 1.05 104 kg/m3, over what time interval does a 0.133-mm layer of silver build up on the teapot? You first need to calculate the total number of silver atoms contained in the plated volume of silver. Note that the nitrate ends up on the other electrode. h
Solution
To solve this problem, we need to follow several steps:
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First, we need to convert the thickness of the silver layer from millimeters to meters, since the density of silver is given in kg/m^3. So, 0.133 mm = 0.133 x 10^-3 m.
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Next, we calculate the volume of the silver layer. The volume is the surface area times the thickness. The surface area is given in cm^2, so we need to convert it to m^2. 1 m^2 = 10,000 cm^2, so 630 cm^2 = 630 x 10^-4 m^2. Therefore, the volume of the silver layer is 630 x 10^-4 m^2 x 0.133 x 10^-3 m = 8.379 x 10^-8 m^3.
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Now we can calculate the mass of the silver layer using the density. The mass is the density times the volume, so the mass of the silver layer is 1.05 x 10^4 kg/m^3 x 8.379 x 10^-8 m^3 = 8.798 x 10^-4 kg.
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The molar mass of silver is 107.87 g/mol, or 0.10787 kg/mol. We can use this to calculate the number of moles of silver in the layer. The number of moles is the mass divided by the molar mass, so the number of moles of silver is 8.798 x 10^-4 kg / 0.10787 kg/mol = 0.00816 mol.
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The number of atoms in a mole is Avogadro's number, which is approximately 6.022 x 10^23. Therefore, the total number of silver atoms in the layer is 0.00816 mol x 6.022 x 10^23 atoms/mol = 4.91 x 10^21 atoms.
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The charge of a silver ion is +1, so one electron is needed to reduce one silver ion to a silver atom. Therefore, the total charge required is 4.91 x 10^21 electrons x 1.602 x 10^-19 C/electron = 0.787 C.
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The current in the circuit is the voltage divided by the resistance, so the current is 12.0 V / 1.90 Ω = 6.32 A.
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The time required is the total charge divided by the current, so the time is 0.787 C / 6.32 A = 0.125 s.
Therefore, it would take approximately 0.125 seconds for a 0.133-mm layer of silver to build up on the teapot.
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