Question

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Answer 1

Verified answer

Electrolysis

21. Ag+ + e- ---> Ag

a.
m = e.Q/96500
1.5 = 108/1 × Q/96500
Q = 1340.3 Coulomb

b.
m = e.F
1.5 = 108 × F
F = 0.0139 faradays

c. I = Q/t
= 1340.3/(1.5×3600)
= 0.248 A

22.
Pb2+ + 2 e- ----> Pb
Pb2+ + 2 H2O ----> PbO2 + 4 H+ + 2e-

m Pb = e.i.t/96500
= 207/2 × 5 × (4×3600)/96500
= 77.22 gram

m PbO2 = 239/2 × 5 × (4×3600)/96500
= 89.16 gram

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