The time required to deposit 4.5g of copper from CuSO\(_4\) solution by passing a current of 2.5 Amperes is (Cu = 64g ; 1F = 96500C/mol)

AcersofChemistry

5 months ago

C.
Q=It
Q=2.5t
Therefore, Q1=(2.5t)C
Cu ----》Cu²+ + 2e-
64g = 96500×2
4.5g = X C
(193000×4.5) ÷ 64
Q2 =13570C
equating Q1 and Q2
2.5t = 13570
t=13570/2.5
t=5428s

OGUNNIRANADEOLA

2 months ago

how did we get 193000

JTG

4 months ago

how do we have 193000

Teddy45

2 months ago

Can we use
mF/Mi=t
t=4.5x193000/64x2.5
t=868000/160
t=5428.125
t=5428sec. (simpler method)