This set of Separation Processes Multiple Choice Questions & Answers (MCQs) focuses on “Electrodialysis”.
1. What principle is used in electrodialysis?
a) Magnetic field and permeable membranes
b) Electric field and cation selective membranes
c) Electric fields and anion selective membranes
d) Electric fields and ion selective membranes
View Answer
Explanation: An electric field and ion selective membranes is used for separating an electrolyte feed in an dilute and concentrated desalted water.
2. How is the arrangement of electrodialysis?
a) 2 membranes arranged in alternating series pattern
b) 4 membranes arranged in alternating series pattern
c) 2 membranes arranged in simultaneous manner
d) 4 membranes arranged back to back
View Answer
Explanation: The electrodialysis consists of 4 membranes of which two are of same type arranged in alternating series pattern.
3. Why are the elctrodes in electrodialysis neither oxidized nor reduced?
a) Because they act as catalyst
b) Because they are chemically inert
c) Because no acid is present
d) There is no acid-base reaction involved in the process
View Answer
Explanation: The electrodes in electrodialysis are chemically inert with the anode being typically stainless steel. So the electrodes are neither oxidized nor reduced.
4. What is true about electrodialysis?
a) Most easily oxidizable species is oxidized at anode
b) Most easily reducible species is oxidized at cathode
c) Most easily oxidizable species is reduced at anode
d) The species don’t get oxidized or reduced since the electrodes are inert.
View Answer
Explanation: The most easily oxidizable species are oxidized at anode and most easily reducible species are reduced at anode.
5. Why is the electrode rinse solution acidic?
a) To avoid the corrosion of electrodes
b) To avoid the formation of salts
c) To neutralize the OH- ions
d) To make the electrodes acidic
View Answer
Explanation: The electrode rinse solution is acidic to neutralize the OH- ions and to precipitate Mg(OH)2 .
6. What is the total area of all cell pairs if:
Volumetric flow rate(Q) = 1
Difference between feed and diluate ion concentration is Δc=10
Current density is i=10
Current efficiency is e=0.8
a) 1.125F
b) 1.5F
c) 2F
d) 3F
View Answer
Explanation: Since Am = FQΔc/ie, Am= 1*10*F/10*8 = 1.125F.
7. What is the total area of all cell pairs if:
Volumetric flow rate(Q) = 2
Difference between feed and diluate ion concentration is Δc=10
Current density is i=10
Current efficiency is e=0.5
a) 1.125F
b) 1.5F
c) 2F
d) 4F
View Answer
Explanation: Since Am = FQΔc/ie, Am= 2*10*F/10*0.5 = 1.125F.
8. What is the total area of all cell pairs if:
Volumetric flow rate(Q) = 1
Difference between feed and diluate ion concentration is Δc=20
Current density is i=10
Current efficiency is e=0.6
a) 1.125F
b) 1.33F
c) 0.33F
d) 0.44F
View Answer
Explanation: Since Am = FQΔc/ie, Am= 1*20*F/10*0.6 = 1.33F.
9. What is the total area of all cell pairs if:
Volumetric flow rate(Q) = 1
Difference between feed and diluate ion concentration is Δc=18
Current density is i=10
Current efficiency is e=0.9
a) 1.125F
b) 1.5F
c) 2F
d) 3F
View Answer
Explanation: Since Am = FQΔc/ie, Am= 1*18*F/10*0.9 = 2F.
10. What is the total area of all cell pairs if:
Volumetric flow rate(Q) = 5
Difference between feed and diluate ion concentration is Δc=15
Current density is i=10
Current efficiency is e=0.75
a) 12.5F
b) 10F
c) 2F
d) 23F
View Answer
Explanation: Since Am = FQΔc/ie, Am= 5*15*F/10*0.75 = 10F.
Sanfoundry Global Education & Learning Series – Separation Processes.
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