In reducing Ag⁺ ion for plating onto jewelry, an operating cell voltage of 4.00 V is required. If you run a business that does this kind of work, how much will the electrical energy cost to coat 1.0 x 10⁴ necklaces with 0.10 g silver each? Assume that the cost of electricity is $0.10/kW⋅h.
To calculate the electrical energy cost, we need to determine the total amount of electricity used and then multiply it by the cost per kilowatt-hour (kW⋅h).
First, let's calculate the total electrical energy used:
1. Determine the amount of silver required:
- Given: 1.0 x 10⁴ necklaces with 0.10 g silver each
- Total silver required = (1.0 x 10⁴ necklaces) x (0.10 g silver/necklace)
- Total silver required = (1.0 x 10⁴) g
2. Determine the amount of Ag⁺ ions required:
- The molar mass of silver (Ag) is 107.87 g/mol.
- The equation shows 1 mol Ag⁺ = 1 mol Ag.
- Therefore, 1 g Ag⁺ = (1 mol Ag/107.87 g) = 1/107.87 mol Ag⁺
- The amount of Ag⁺ ions required = (1.0 x 10⁴ g) x (1/107.87 mol Ag⁺/g)
3. Determine the total charge used:
- In each reduction of Ag⁺ ion, 1 mol of Ag⁺ ions requires 1 mol of electrons (e⁻).
- Total charge used = (amount of Ag⁺ ions required) x (Faraday's constant)
- Faraday's constant (F) is 96,485 C/mol.
4. Determine the total electrical energy used:
- The operating cell voltage is given as 4.00 V.
- Electrical energy used = (total charge used) x (operating cell voltage)
Now that we have calculated the total electrical energy used, we can proceed to calculate the cost:
5. Determine the electrical energy in kilowatt-hours (kW⋅h):
- 1 kilowatt-hour (kW⋅h) = 3.6 x 10³ megajoules (MJ)
- Electrical energy in kWh = (electrical energy used) / (3.6 x 10³)
6. Calculate the cost of electricity:
- Given that the cost of electricity is $0.10/kW⋅h.
- Cost of electricity = (electrical energy in kWh) x (cost per kWh)
By following these steps, you should be able to determine the cost of electricity to coat 1.0 x 10⁴ necklaces with 0.10 g silver each.