Using the formula u × RAM ÷[F × Z]
What is the mass of silver deposited when a current of 2.6A is pass through a solution of silver salt for 70 minutes



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.