A current of 4.00 A is passed through a solution of AgNO3 for 1.00 hr. What mass of silver is deposited at the cathode? (In grams)
Use Faraday's constant F = 9.65x10^4 C/mol
0.5556
To find the mass of silver deposited at the cathode, we need to use Faraday's laws of electrolysis. According to Faraday's laws, the amount of substance deposited or liberated at an electrode is directly proportional to the amount of electric charge passing through the electrolytic cell.
The formula to calculate the mass of a substance deposited at the electrode is:
Mass = (Current × Time) / (Number of electrons × Faraday's constant)
In this case, we have the current of 4.00 A passing through the solution of AgNO3 for 1.00 hour (which can be converted to seconds). We also know Faraday's constant, which is F = 9.65 × 10^4 C/mol.
Now, we need to determine the number of electrons involved in the redox reaction for silver. By balancing the half-reaction for the reduction of Ag+ to Ag, we can see that 1 mol of Ag+ requires 1 mol of electrons (Ag+ + e- → Ag). Therefore, the number of electrons is 1 for this reaction.
Next, we can plug the values into the formula:
Mass = (Current × Time) / (Number of electrons × Faraday's constant)
= (4.00 A × 3600 s) / (1 × 9.65 × 10^4 C/mol)
After calculating this expression, you'll get the mass of silver deposited at the cathode in grams.
C = amperes x seconds
C = 4.00 x 1 hr x (60 min/hr) x (60 s/min) = ?
96,500 C will deposit 107.9 g Ag.
107.9 g Ag x (? C from above)/96,500 = ? g Ag deposited.