If 3.25x10-3 kg of gold is deposited onto the negative electrode of an electrolytic cell in a period of 2.78 hours, what is the current through this cell in amperes? Assume that the gold ions carry one elementary unit of charge
To find the current through the cell in amperes, you need to use Faraday's law of electrolysis. According to Faraday's law, the amount of substance deposited or released during electrolysis is directly proportional to the electric charge passed through the electrolytic cell.
1. Start by finding the number of moles of gold deposited onto the negative electrode.
- Convert the mass of gold from kilograms to grams: 3.25 x 10^-3 kg * 1000 g/kg = 3.25 g
- Use the molar mass of gold (197 g/mol) to convert grams to moles: 3.25 g / 197 g/mol = 0.016529 moles
2. Determine the total electric charge passed through the electrolytic cell.
- Since gold ions carry one elementary unit of charge, the number of moles of gold deposited is equal to the number of elementary charges passed.
- To convert moles of gold to elementary charges, multiply by Avogadro's number (6.022 x 10^23 elementary charges/mol):
0.016529 moles * (6.022 x 10^23 elementary charges/mol) = 9.9481 x 10^21 elementary charges
- Note that 1 elementary charge is equal to 1 coulomb (C).
3. Calculate the current through the cell.
- The current (I) is defined as the amount of charge (Q) passing through a conductor per unit time (t). It can be expressed as: I = Q / t
- In this case, the time is given as 2.78 hours, so convert it to seconds: 2.78 hours * 3600 seconds/hour = 1.00 x 10^4 seconds
- Finally, divide the total charge (9.9481 x 10^21 elementary charges or 9.9481 x 10^21 C) by the time (1.00 x 10^4 seconds) to find the current:
I = Q / t = 9.9481 x 10^21 C / 1.00 x 10^4 s = 9.9481 x 10^17 A (amperes)
Therefore, the current through the cell is approximately 9.9481 x 10^17 amperes.