what mass of silver is deposited when n current of 2.6A is passed through a solution of silversalt for 40 minutes?
(Ag=108, F=96500C
108 g Ag will be deposited with 96,500 coulombs.
You have how many coulombs?
Coulombs = amperes x seconds
C = 2.6 A x 40 min x (60 sec/1 min) = 6,240 C
So 108 x (6,240/96,500) = g Ag deposited.
108 x (6,240/96,500)=7
To calculate the mass of silver deposited, we need to use Faraday's law of electrolysis. The formula is as follows:
Mass (in grams) = (Current (in Amperes) × Time (in seconds) × Atomic mass of the metal) / (Number of electrons transferred × Faraday's constant)
Given:
Current = 2.6 A
Time = 40 minutes = 2400 seconds
Atomic mass of silver (Ag) = 108 g/mol
Number of electrons transferred (for silver) = 1
Faraday's constant (F) = 96500 C/mol
Let's plug these values into the formula:
Mass = (2.6 A × 2400 s × 108 g/mol) / (1 × 96500 C/mol)
Mass = (627,840 g·A·s) / 96500 C
Mass = 6.499g
Therefore, approximately 6.499 grams of silver will be deposited when a current of 2.6 A is passed through the silver salt solution for 40 minutes.
To find the mass of silver deposited, we need to use Faraday's laws of electrolysis, which relates the amount of substance deposited during electrolysis to the electric charge passed through the solution.
The equation that relates the quantity of substance (m) to the electric charge (Q) is:
m = (Q × M) / (n × F)
Where:
m = mass of substance deposited (in grams)
Q = electric charge (in Coulombs)
M = molar mass of the substance (in grams per mole)
n = number of electrons transferred during the reaction
F = Faraday's constant (approximately 96500 C/mol)
In this case, we are trying to find the mass of silver deposited, so the molar mass (M) is the molar mass of silver (Ag), which is 108 g/mol.
Given:
Current (I) = 2.6 A
Time (t) = 40 minutes = 40 × 60 = 2400 seconds
n = number of electrons transferred during the reaction = 1 (since silver is a monovalent cation)
First, calculate the electric charge (Q) passed through the solution using the formula:
Q = I × t
Q = 2.6 A × 2400 seconds
Q = 6240 C
Now substitute the values into the equation to find the mass of silver deposited:
m = (Q × M) / (n × F)
m = (6240 C × 108 g/mol) / (1 × 96500 C/mol)
m ≈ 6.983 grams
Therefore, approximately 6.983 grams of silver will be deposited after passing a current of 2.6 A through the silver salt solution for 40 minutes.