Answer :
Answer:
the mass of silver deposited when a current of 2.6 A is passed through a solution of silver salt for 70 minutes is approximately 0.0029 grams.
Explanation:
To find the mass of silver deposited, we can use the formula provided and the given data. The formula given is:
Mass of substance deposited
=
×
RAM
×
Mass of substance deposited=
F×Z
u×RAM
Where:
u is the current in amperes (A)
RAM is the relative atomic mass of silver (Ag), which is approximately 107.87 g/mol
F is Faraday's constant, which is approximately
96485
C/mol
96485 C/mol
Z is the number of electrons transferred per ion, which for silver ions (Ag⁺) is 1
Given data:
Current,
=
2.6
u=2.6 A
Time,
=
70
t=70 minutes =
70
×
60
70×60 seconds =
4200
4200 seconds
Now, let's calculate step by step:
Calculate the total charge (Q) passed through the circuit:
=
×
Q=u×t
=
2.6
A
×
4200
s
Q=2.6 A×4200 s
=
10920
C
Q=10920 C
Convert charge (Coulombs) to moles of electrons:
Since 1 Faraday (
F) is equal to
96485
96485 C/mol,
Moles of electrons
=
Moles of electrons=
F
Q
Moles of electrons
=
10920
C
96485
C/mol
Moles of electrons=
96485 C/mol
10920 C
Moles of electrons
≈
0.1132
mol
Moles of electrons≈0.1132 mol
Calculate the mass of silver deposited:
Now, using the formula:
Mass of silver deposited
=
×
RAM
×
Mass of silver deposited=
F×Z
u×RAM
Mass of silver deposited
=
2.6
A
×
107.87
g/mol
96485
C/mol
Mass of silver deposited=
96485 C/mol
2.6 A×107.87 g/mol
Mass of silver deposited
≈
280.0622
96485
g
Mass of silver deposited≈
96485
280.0622
g
Mass of silver deposited
≈
0.0029
g
Mass of silver deposited≈0.0029 g
Therefore, the mass of silver deposited when a current of 2.6 A is passed through a solution of silver salt for 70 minutes is approximately
0.0029 grams.
