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23. When a certain amount of electrical charge flows through an aqueous solution
of silver nitrate,
3.24 g of silver is deposited on the cathode , Calculate :
a. the mass of copper
Answer :
Cathode : Ag+
+ e−
 Ag
Mole of Ag =
G
Ar
=
3.24
108
= 0.003 = mole of electrons
Anode : Cu  Cu+2
+ 2e−
0.0015 mole ~ 0.0015 mole ~ 0.003 mole
The mass of Copper= n.Ar = 0.0015 x 63,5 = 0,953 g
b. the mass of aluminium that will be deposited by the same quantity of
charge
Answer :
Al 𝟑+
+ 3e−
 Al
0.001 mole ~ 0.003 mole ~ 0.001 mole
The mass of Aluminium = n.Ar = 0.001x27 = 0.027 g
24. Calculate the mass of copperdeposited at the cathode when a current of 0.125
A flows through aqueous copper(II) sulphate of concentration 0.80 mol dm−3
for
30 minutes.
Answer :
G =
eit
96500
=
63.5x0.125x1800
2x96500
= 0.0743g

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Kimia sat no. 23 & 24 hal. 46

  • 1. 23. When a certain amount of electrical charge flows through an aqueous solution of silver nitrate, 3.24 g of silver is deposited on the cathode , Calculate : a. the mass of copper Answer : Cathode : Ag+ + e−  Ag Mole of Ag = G Ar = 3.24 108 = 0.003 = mole of electrons Anode : Cu  Cu+2 + 2e− 0.0015 mole ~ 0.0015 mole ~ 0.003 mole The mass of Copper= n.Ar = 0.0015 x 63,5 = 0,953 g b. the mass of aluminium that will be deposited by the same quantity of charge Answer : Al 𝟑+ + 3e−  Al 0.001 mole ~ 0.003 mole ~ 0.001 mole The mass of Aluminium = n.Ar = 0.001x27 = 0.027 g 24. Calculate the mass of copperdeposited at the cathode when a current of 0.125 A flows through aqueous copper(II) sulphate of concentration 0.80 mol dm−3 for 30 minutes. Answer : G = eit 96500 = 63.5x0.125x1800 2x96500 = 0.0743g