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ananya51
@ananya51
July 2021
2
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When electricity is passed through a solution of AlCl3 13.5 g of Al is deposited. The number of faradays must
be
(1) 1.0 (2) 1.5 (3) 0.5 (4) 2
Answers & Comments
dhathri123
Verified answer
Hey sisy !!!!!!
here is your required answer- (2)
the reaction at cathode is
Al ( 3+) + 3e---------->Al. (n=3)
→applying the equation
w = E×i×t/F
by substituting the values
♦ i × t =
13.5 /(27/3)×F
=13.5/9 ×F
=1.5 F
I hope this will help u ;)
♥dhathri123♥
139 votes
Thanks 192
ananya51
thank u my dear sis
dhathri123
u r welcome ;)
TheRuhanikaDhawan
Verified answer
Al 3+ + 3e ----------> Al
1 mole of Al ---- 27
13.5 is given -- which is 0.5 mole
1 Faraday = 1 mole of electrons
1 mole ----------- 30 faraday
0.5 mole ---------- ?
0.5 x 30
= 1.5 faraday
-----------------------------------------------------------------------------------------
50 votes
Thanks 63
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Answers & Comments
Verified answer
Hey sisy !!!!!!here is your required answer- (2)
the reaction at cathode is
Al ( 3+) + 3e---------->Al. (n=3)
→applying the equation
w = E×i×t/F
by substituting the values
♦ i × t =
13.5 /(27/3)×F
=13.5/9 ×F
=1.5 F
I hope this will help u ;)
♥dhathri123♥
Verified answer
Al 3+ + 3e ----------> Al1 mole of Al ---- 27
13.5 is given -- which is 0.5 mole
1 Faraday = 1 mole of electrons
1 mole ----------- 30 faraday
0.5 mole ---------- ?
0.5 x 30
= 1.5 faraday
-----------------------------------------------------------------------------------------