If 8A pass for 2minutes at the cathode(where the following reaction takes place, Ag+ + e- --->Ag) how many grams of silver metal are passed onto the cathode??
To determine the amount of silver metal passed onto the cathode, you need to calculate the charge that passes through the circuit and then convert it into grams of silver using the Faraday's law of electrolysis.
First, let's calculate the total charge passed through the cathode using the formula:
Q = I * t
where
Q is the charge (in Coulombs),
I is the current (in Amperes), and
t is the time (in seconds).
In this case, the current is 8A (Amperes) and the time is 2 minutes. However, we need to convert the time to seconds, so 2 minutes = 2 * 60 = 120 seconds.
Q = 8A * 120s = 960 Coulombs
Now, to convert the charge into grams of silver, we can use Faraday's constant and the formula:
1 mole of electrons = 1 Faraday = 96,500 Coulombs
The balanced equation tells us that 1 mole of electrons corresponds to the passage of 1 mole of silver (Ag).
So, we can write the conversion factor:
96,500 C = 1 mole of electrons = 1 mole of Ag = 107.87 grams of Ag
Now, let's calculate the grams of silver using the given charge (Q):
grams of Ag = (Q / 96,500) * molar mass of Ag
grams of Ag = (960 / 96,500) * 107.87 grams of Ag
grams of Ag = 1.00391 grams of Ag
Therefore, approximately 1.004 grams of silver metal are passed onto the cathode.