Two different electrolytic cells filled with molten Cu(NO3)2 and molten Al(N03)3 respectively are connected in series. When electricity is passed 2. 7 g Al is deposited on electrode. Calculate the weight of Cu deposited on cathode. [Cu = 63 .5; Al= 27 .0 g mol-1 ]

Question

Two different electrolytic cells filled with molten Cu(NO3)2 and molten Al(N03)3 respectively are connected in series. When electricity is passed 2. 7 g Al is deposited on electrode. Calculate the weight of Cu deposited on cathode. [Cu = 63 .5; Al= 27 .0 g mol-1 ] (A) 190.5 g (B) 9.525g (C) 63.5 g (D) 31.75 g

Tardigrade Admin · Accepted Answer

According to Faraday II law (w1/E1) = (w2/E2) (1 for Al and 2 for copper) (2.7 × 3/27) = (w2 × 2/63.5) (EAl =27/3, and Ecu =63.5/2) w2 = 9.525g

Q. Two different electrolytic cells filled with molten $C u (N O_{3})_{2}$ and molten $A l (N 0_{3})_{3}$ respectively are connected in series. When electricity is passed 2. 7 g Al is deposited on electrode. Calculate the weight of Cu deposited on cathode. $[C u = 63.5; A l = 27.0 g m o l^{- 1}]$

190.5 g

9.525g

63.5 g

31.75 g

According to Faraday II law
$\frac{w _{1}}{E _{1}} = \frac{w _{2}}{E _{2}}$ (1 for Al and 2 for copper)
$\frac{2.7 \times 3}{27} = \frac{w _{2} \times 2}{63.5} (E_{A l} = 27/3, an d E_{c u} = 63.5/2)$
$w_{2} = 9.525 g$

Q. Two different electrolytic cells filled with molten Cu(NO3​)2​ and molten Al(N03​)3​ respectively are connected in series. When electricity is passed 2. 7 g Al is deposited on electrode. Calculate the weight of Cu deposited on cathode. [Cu=63.5;Al=27.0gmol−1]