If 10.8g of silver is deposited in a silver coulometer connected in series with a copper coulometer, the volue of oxygen liberated is

If 10.8g of silver is deposited in a silver coulometer connected in series with a copper coulometer, the volue of oxygen liberated is

A. 0.56 dm^3

B. 5.60 dm^3

C. 11.20 dm^3

D. 22.40 dm^3

Using Faraday’s laws of electrolysis:

Step 1: Moles of silver deposited

Reaction at silver coulometer:
{Ag}^+ + e^- 一>{Ag}

Molar mass of Ag = 108 g mol⁻¹

Moles of Ag = {10.8}/{108} = 0.10  mol

So, 0.10 mol of electrons flowed through the circuit.

Step 2: Relation to oxygen liberated

At the anode (oxygen liberation):
4e^- 一>O2

Moles of O2 = {0.10}/{4} = 0.025 mol

Step 3: Volume of oxygen at STP

1 mol of gas = 22.4 { dm}^3
{Volume of O}_2 = 0.025 * 22.4 = 0.56 dm³

✅ Correct Answer

A. 0.56 dm³

PQN:22