Answer:
36,250.03 seconds to deposit 55.00 g of gold over the iron medallion.
Step-by-step explanation:
To determine the amount of time required to deposit 55.00 g of gold over the iron medallion, we can use Faraday's law of electrolysis.
Faraday's law states that the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
The equation to calculate the amount of substance deposited is:
m = (Q * M) / (n * F)
where:
m is the mass of the substance deposited (in grams),
Q is the total charge passed (in Coulombs),
M is the molar mass of the substance (in grams/mol),
n is the number of electrons transferred in the reaction, and
F is the Faraday's constant (96,485 C/mol).
In this case, we want to deposit 55.00 g of gold, so we can rearrange the equation to solve for Q:
Q = (m * n * F) / M
Now we can substitute the given values:
m = 55.00 g
M = 196.97 g/mol (molar mass of gold)
n = 3 (since Au^3+ ions each gain 3 electrons during the reduction)
F = 96,485 C/mol (Faraday's constant)
Q = (55.00 * 3 * 96,485) / 196.97
Q ≈ 80,574.86 C
Since the current is given as 2.225 A (Amperes), we can use the equation:
Q = I * t
where:
Q is the total charge passed (in Coulombs),
I is the current (in Amperes), and
t is the time (in seconds).
Rearranging the equation to solve for t:
t = Q / I
Substituting the values:
Q = 80,574.86 C
I = 2.225 A
t ≈ 80,574.86 / 2.225
t ≈ 36,250.03 seconds
Therefore, it would take approximately 36,250.03 seconds to deposit 55.00 g of gold over the iron medallion using the given conditions.