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Today, we are going to take a quick look at how we can solve 3 MCQ questions on electrolysis.

By the end of these 3 examples, I hope you will have a better idea on how to approach electrolysis MCQ questions and how to solve them so that you can score better marks for your MCQ paper!

Question 1 (electrolysis)

electrolysis question 1

What changes would occur when a current is passed through the solution after some time?
A.   Electrode Y will ionise and become smaller in size.
B.   A brown coloured gas is produced at electrode Y
C.   Oxygen is produced at electrode X and hydrogen gas at electrode Y
D.   The colour of the solution around electrode X turns blue, while the colour around electrode Y turns red.

 

Before I give you the answer, let’s take a look at how you may approach this question.

Step 1:
List all ions present – sodium ions, sulfate ions, hydrogen ions, hydroxide ions.

Step 2: 
Determine if electrodes are inert/reactive.

Step 3:
Determine ions discharged – hydrogen ions and hydroxide ions. Although the solution is concentrated, you need to know that sodium ions and sulfate ions are NEVER discharged.
Hydrogen ions are discharged at electrode X forming hydrogen gas. Hydroxide ions are discharged at Y forming oxygen gas and water.

Step 4:
Link back to question.

A is incorrect as platinum electrodes are inert and will NOT take part in the reaction/break down.
B is incorrect as neither hydrogen gas or oxygen gas is brown in colour.
C is incorrect as hydrogen gas should be formed at X and oxygen gas at Y.

D is the CORRECT answer.
As hydrogen ions at X are discharged, the concentration of hydroxide ions will be higher, hence the solution turns alkaline.
As hydroxide ions at Y are discharged, the concentration of hydrogen ions will be higher, hence the solution turns acidic!


 

Question 2 (simple cells)

electrolysis question 2
Which of the following statements about the above setup is correct?
A.   The electrolyte turns from colourless to blue
B.   Hydrogen ions are attracted to the copper electrode
C.   Effervescence is seen at the copper electrode
D.   Mobile electrons will allow the electrolyte to conduct electricity

 

Once again, before I give you the answer, let’s take a look at how you may approach this question.

Step 1:
Identify ions present in electrolye – hydrogen ions, sulfate ions and hydroxide ions

Step 2:
Determine which electrode is more reactive.
Since electrons flow from copper metal to metal T, copper is the MORE reactive metal.

Step 3:
Determine half equations.
@ copper electrode:
Cu(s) –> Cu2+(aq)  +  2e-

@ metal T:
Since metal T is accepting electrons from the copper metal, it becomes NEGATIVELY CHARGED. (**This doesn’t mean that it is the negative electrode! The negative electrode is the electrode that produces electrons – in this case copper!)
2H+(aq) + 2e-  –> H2 (g)

Step 4:
Link back to question

B is incorrect. As seen in step 3, hydrogen ions are attracted toward metal T and discharged at T.
C is incorrect. As seen in step 3, the copper electrode decreases in size as copper atoms form copper ions. NO gas is formed.
D is incorrect. The electrolyte is able to conduct electricity as it contains FREE MOVING IONS.

A is the CORRECT answer.
As copper ions are formed, the solution turns blue.


 

Question 3 (electrolysis)

electrolysis question 3
All 3 cells contain copper (II) sulfate solution as the electrolyte. The switch is closed and the colour of the electrolyte is noted as the reaction progresses. In which cells will the blue colour fade?
A.    Z only
B.    X and Z
C.    Y and Z
D.    X, Y and Z.

 

Once again, before I give you the answer, let’s take a look at how you may approach this question.

Step 1:
Identify ions present in electrolyte – hydrogen ions, copper ions, sulfate ions, hydroxide ions.

Step 2:
Determine which electrodes are inert and reactive.
Copper – reactive
Platinum – inert

Step 3.
Determine ions discharged.
In X,
The anode is a reactive metal, hence it will break down. Cu (s)  –>  Cu2+ (aq)  +  2e-
Although the cathode is reactive, it does NOT break down. Instead, copper ions are discharged. Cu2+(aq) + 2e-  –>  Cu(s)

In Y,
The anode is platinum. It is inert. Hydroxide ions are discharged to form water and oxygen gas.
The cathode, although reactive, does not break down. Copper ions are discharged. Cu2+(aq) + 2e-  –>  Cu(s)

In Z,
Both the electrodes are inert.
@ anode: Hydroxide ions are discharged to form water and oxygen gas.
@ cathode: Copper ions are discharged. Cu2+(aq) + 2e-  –>  Cu(s)

Step 4:
Link back to question

From step 3, we note that the only solution that has copper ions formed is solution X. Hence it will NOT decolourise.
In Y and Z, there are no reactions to replace the copper ions discharged. Hence the solutions will decolourise.

Therefore, the CORRECT answer is C.


 

 

I hope these 3 worked examples have given you a better idea how to approach and solve Electrolysis and Simple Cells MCQ questions!
Do feel free to drop us comments if you have any other questions on the above.

To your success!