To determine the oxidation state of gold in the gold salt, we can use the concept of Faraday's laws of electrolysis.
According to Faraday's laws, the amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The proportionality constant is called the Faraday constant (F) and is equal to 96,485 C/mol.
In this case, the student passed the same quantity of electricity through both electrolytic cells, and she found that 2.64 grams of Ag and 1.61 grams of Au were deposited at the cathodes.
To find the amount of electricity passed (Q) in coulombs, we can use the formula:
Q = (grams of substance deposited) / (gram equivalent weight)
The gram equivalent weight (GEW) is the molar mass of the substance divided by the number of electrons involved in the reaction. For silver (Ag), the GEW is equal to the molar mass of Ag (107.87 g/mol) divided by 1 (as Ag has an oxidation state of +1 in salts), which is 107.87 g/mol. For gold (Au), we need to find the GEW by dividing the molar mass of Au (197.0 g/mol) by the number of electrons involved in the reaction.
Now, let's calculate the number of coulombs (C) passed for each metal:
Q_Ag = 2.64 g / (107.87 g/mol) = 0.0244 mol Ag
Q_Au = 1.61 g / GEW_Au
Since gold can have various oxidation states, we need to find the GEW_Au using the amount of electricity passed for silver:
Q_Ag = Q_Au
0.0244 mol Ag = 1.61 g / GEW_Au
Rearranging the equation, we can solve for GEW_Au:
GEW_Au = (1.61 g) / (0.0244 mol Ag)
Once we have the GEW_Au, we can calculate the oxidation state of gold in the gold salt by considering the individual charges and the conservation of charge in the electrolysis process.
Please provide the value of GEW_Au, and I will help you determine the oxidation state of gold in the gold salt.