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Answer 1

We first write the electrode equation for deposit of copper at the cathode.

Cu²⁺ + 2e==>Cu₍s₎

This implies that 2 moles of electrons produce one mole of copper.

Given time and current we can calculate the charge of the electrons:

Q =It where Q is the charge in Coulombs, I is the current in amperes and t the time in seconds.

Q is thus: 60 × 60 × 1.62=3600 × 1.62=5832Coulombs.

1mole of an electron = 96500Coulombs

5832/96500=0.060435moles

2 moles of electrons produce 1 mole of copper thus the moles of copper deposited are:

0.060435/2 = 0.030218 moles

Mass of copper deposited = moles of copper × atomic mass of copper.

Atomic mass of copper =63.5g

63.5 × 0.030218=1.9182g

I think option c is correct

Answer 2

heya...!!!

here is ua answer:

______________________________________♥

★Cu²⁺ + 2e==>Cu₍s₎

★This implies that 2 moles of electrons produce one mole of copper.

★Given time and current we can calculate the charge of the electrons:

★Q =It where Q is the charge in Coulombs, I is the current in amperes and t the time in seconds.

★Q is thus: 60 × 60 × 1.62=3600 × 1.62=5832Coulombs.

1mole of an electron = 96500Coulombs

5832/96500=0.060435moles

★2 moles of electrons produce 1 mole of copper thus the moles of copper deposited are:

0.060435/2 = 0.030218 moles

Mass of copper deposited = moles of copper × atomic mass of copper.

Atomic mass of copper =63.5g

63.5 × 0.030218=1.9182g

★hence option C is the correct answer.

hope it helps...!!!♥