What mass of copper deposited in one hour by a current of 1.62 a?
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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
#Be Brainly❤️
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
#Be Brainly❤️
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