Calculate the mass of silver deposited from silver nitrate solution by a current of 2 ampere flowing for 30 minutes

To calculate the mass of silver deposited from a silver nitrate solution when a current flows through it, we can use Faraday's laws of electrolysis. The process involves understanding the relationship between electric current, time, and the amount of substance deposited. Let's work through this step-by-step.

The Basics of Electrolysis

When an electric current passes through an electrolyte (like silver nitrate), it causes the ions in the solution to move. In the case of silver nitrate (AgNO3), the silver ions (Ag⁺) are reduced at the cathode, leading to the deposition of silver metal.

Key Variables

• Current (I): 2 amperes
• Time (t): 30 minutes (which we need to convert to seconds)
• Faraday's constant (F): 96500 coulombs per mole, which is the amount of electric charge required to deposit one mole of a monovalent ion.
• Molar mass of silver (Ag): 107.87 g/mol

Convert Time to Seconds

First, we need to convert the time from minutes to seconds since the unit of current is in amperes (coulombs per second).

30 minutes = 30 x 60 = 1800 seconds

Calculate Total Charge (Q)

Using the formula for electric charge:

Q = I x t

Substituting the values:

Q = 2 A x 1800 s = 3600 coulombs

Calculate Moles of Silver Deposited

Next, we can determine how many moles of silver were deposited using Faraday's laws:

Number of moles (n) = Q / (F x z)

Here, z is the number of moles of electrons involved in the reduction. For silver, which is monovalent, z = 1.

So, we have:

n = 3600 C / (96500 C/mol x 1) ≈ 0.0373 moles

Calculate the Mass of Silver Deposited

Finally, we convert moles of silver into grams using its molar mass:

Mass (m) = n x Molar mass

Substituting the values:

m = 0.0373 moles x 107.87 g/mol ≈ 4.02 grams

Summary of Results

Therefore, the mass of silver deposited from the silver nitrate solution by a current of 2 amperes flowing for 30 minutes is approximately 4.02 grams.