what is the mass of ag deposited on cathode when 0.1 m 100 ml agno3

To determine the mass of silver (Ag) deposited on the cathode when 0.1 M of 100 mL silver nitrate (AgNO3) solution is electrolyzed, we can follow a systematic approach using Faraday's laws of electrolysis. Let's break this down step by step.

Understanding the Electrolysis Process

During electrolysis, the silver ions (Ag⁺) in the solution are reduced at the cathode, leading to the deposition of metallic silver. The reaction can be represented as:

• Ag⁺ + e^- → Ag (s)

Calculating the Number of Moles

First, we need to calculate the number of moles of AgNO3 in the solution. The molarity (M) is defined as the number of moles of solute per liter of solution. Given that we have a 0.1 M solution and a volume of 100 mL (which is 0.1 L), we can use the formula:

• Number of moles = Molarity × Volume (in liters)

Substituting the values:

• Number of moles = 0.1 mol/L × 0.1 L = 0.01 moles

Determining the Mass of Silver Deposited

Next, we need to find the mass of silver that corresponds to the moles calculated. The molar mass of silver (Ag) is approximately 107.87 g/mol. Using the formula:

• Mass = Number of moles × Molar mass

We can now calculate the mass:

• Mass = 0.01 moles × 107.87 g/mol = 1.0787 g

Final Result

Thus, when 100 mL of 0.1 M AgNO3 is electrolyzed, approximately 1.08 grams of silver will be deposited on the cathode. This process illustrates the principles of electrochemistry and how the amount of substance deposited can be quantitatively determined through careful calculations.