what mass of silver is deposited when an current of 2.6A is passed through a solution of silversalt for 40 minutes? (AG =108,F=96500C)
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To find the mass of silver deposited, we can use Faraday's law of electrolysis. Faraday's law states that the amount of substance deposited on an electrode during electrolysis is directly proportional to the amount of electricity passed through the electrolyte.
The formula to calculate the mass of the substance deposited is given by:
Mass (in grams) = (Current (in Amperes) × Time (in seconds) × Atomic Mass) / (Faraday's Constant)
Let's break down the given information and calculate the mass of silver deposited step-by-step:
1. Current (I) = 2.6 Amperes
2. Time (t) = 40 minutes = 40 × 60 = 2400 seconds
3. Atomic Mass of silver (Ag) = 108 grams/mol (given)
4. Faraday's Constant (F) = 96500 Coulombs/mol (given)
Now, let's substitute these values into the formula:
Mass = (2.6 A × 2400 s × 108 g/mol) / 96500 C
Calculating further:
Mass ≈ 0.290 g
Therefore, approximately 0.290 grams of silver will be deposited when a current of 2.6 Amperes is passed through the silver salt solution for 40 minutes.