Question

0.1F of electricity is passed through aqueous solution of copper sulphate using Pt electrodes.If 3.18 grams of copper is deposited at cathode the volume of oxygen liberated at STP) at anode will be (Atomic weight of copper 63.6: Atomic weight of oxygen​

Answer 1

Answer:

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Answer 2

Given info : 0.1 F of electricity is passed through aqueous solution of copper sulphate using Pt electrode. 3.18 gm of copper is deposited at cathode.

To find :  the volume of oxygen liberated at STP at anode will be ..

solution : at cathode, Cu²⁺ + 2e⁻ => Cu

at anode, 4OH⁻  => H₂ + O₂ + 4e⁻

so, no of moles of Cu deposited at cathode is equal to half of no of moles of  oxygen gas liberated from anode.

mass of copper deposited at cathode is 3.18 gm.

so no of moles of copper = 3.18/63.6 = 0.05 mol

so the no of moles of oxygen gas liberated at anode is 0.05/2 = 0.025 mol

now at STP , volume of oxygen liberated at anode, V = no of moles of oxygen gas × 22.4 L

= 0.025 × 22.4 L = 0.56 L

therefore the volume of oxygen gas liberated at STP at anode will be 0.56 L