Q. two electrolyte cells containing silver nitrate solution and dilute sulphuric acid solution were connected in series. A steady current of 2.5 amp was passed through them till 1.078 g of silver was deposited.[Ag= 107.8 g/mol]. A) how much electricity was consumed? B) what was the weight of oxygen gas liberated? I need the procedure pls

To tackle this problem, we need to apply some fundamental concepts from electrochemistry, particularly Faraday's laws of electrolysis. Let's break down the question into two parts: calculating the electricity consumed and determining the weight of oxygen gas liberated.

Calculating Electricity Consumed

First, we need to find out how much electricity was consumed during the deposition of silver. According to Faraday's first law of electrolysis, the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the cell. The formula we use is:

Q = n × F

Where:

• Q = total charge in coulombs (C)
• n = number of moles of the substance deposited
• F = Faraday's constant (approximately 96485 C/mol)

Next, we need to calculate the number of moles of silver deposited. We can use the formula:

n = m / M

Where:

• m = mass of silver deposited (1.078 g)
• M = molar mass of silver (107.8 g/mol)

Plugging in the values:

n = 1.078 g / 107.8 g/mol = 0.01 mol

Now, substituting this value into the first equation:

Q = 0.01 mol × 96485 C/mol = 964.85 C

Determining Weight of Oxygen Gas Liberated

Next, we need to find out how much oxygen gas was liberated during the electrolysis of dilute sulfuric acid. The reaction at the anode can be represented as:

2H₂O → O₂ + 4H⁺ + 4e^-

This indicates that for every 4 moles of electrons (4e^-), 1 mole of oxygen gas (O₂) is produced. To find the number of moles of oxygen gas liberated, we first need to calculate the total charge passed through the cell, which we already found to be 964.85 C.

Using Faraday's constant, we can find the number of moles of electrons:

n_e = Q / F

Substituting the values:

n_e = 964.85 C / 96485 C/mol = 0.01 mol

Now, since 4 moles of electrons produce 1 mole of oxygen gas, we can find the moles of oxygen gas produced:

n_O2 = n_e / 4 = 0.01 mol / 4 = 0.0025 mol

Finally, we can calculate the mass of oxygen gas liberated using the molar mass of oxygen (approximately 32 g/mol):

m_O2 = n_O2 × M_O2

Substituting the values:

m_O2 = 0.0025 mol × 32 g/mol = 0.08 g

Summary of Results

To summarize:

• The total electricity consumed was approximately 964.85 C.
• The weight of oxygen gas liberated was 0.08 g.

This process illustrates the relationship between electricity and chemical reactions in electrolysis, showcasing how we can quantify the outcomes based on the principles of electrochemistry.