CGP EXAM QUESTIONS: ES Flashcards

1
Q

Two carbon electrodes are placed into molten copper chloride and the power supply is turned on.

What would you see happening at the anode? Explain your answer.

A

At the anode, you would see bubbles of gas appearing
because chloride ions are losing electrons to form chlorine

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2
Q

Two carbon electrodes are placed into molten copper chloride and the power supply is turned on.

What would you see happening at the cathode? Explain your answer.

A

At the cathode, you would see the elctrode being plated
because copper ions are gaining electrons to form copper metal

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3
Q

A student carries out an electrolysis of dilute solution of magnesium chloride, MgCl2(aq)
Write the half-equation for the reaction that occurs at the cathode

A

2H+(aq) + 2e- –> H2(g)

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4
Q

A student carries out an electrolysis of dilute solution of magnesium chloride, MgCl2(aq)
Write the half-equation for the reaction that occurs at the anode

A

4OH-(aq) –> O2(g) + H2O(l) + 4e-

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5
Q

Chlorine gas is produced industrially from the electrolysis of sodium chloride solution

Name one natural source of sodium chloride

A

rock salt, sea water, brine

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6
Q

Chlorine gas is produced industrially from the electrolysis of sodium chloride solution
A student carries out the electrolysis of concentrated sodium chloride solution. Once the electrolysis they add a few drops of universal indicator to the remaining solution.

What would you expect the student to see? Explain your answer.

A

The universal indicator would turn the solution blue/purple
The solution would be alkaline, because of the hydroxide ions left in solution

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7
Q

Chlorine gas is produced industrially from the electrolysis of sodium chloride solution
​A student carries out the electrolysis of concentrated sodium chloride solution. Once the electrolysis they add a few drops of universal indicator to the remaining solution.
Another student then carries out the electrolysis of dilute sodium chloride solution. They also add a few drops of universal indicator to the remaining solution after the electrolysis is complete.

How would the second’s students observations differ from those of the first student. Explain your answer

A

The universal indicator would turn the solution green
because hydrogen ions and hydroxide ions have been discharged, leaving a neutral solution of NaCl

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8
Q

Chlorine gas is produced industrially from the electrolysis of sodium chloride solution
Chlorine gas can be used to displace iodine from brine

State why chlorine displaces iodine from a solution of iodide ions

A

Chlorine is more reactive than bromine

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9
Q

Lead(II) sulfate(VI) can be made by reacting lead(II) oxide with sulfuric acid.

State the oxidation state of lead in lead(II) oxide

A

+2

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10
Q

Lead(II) sulfate(VI) can be made by reacting lead(II) oxide with sulfuric acid.

State the oxidation state of sulfur in lead(II) sulfate(VI)

A

+6

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11
Q

Chromium (Cr) can exist in a range of oxidation states.

Give the full name of the compound C2O3

A

chromium(III) oxide

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12
Q

What is the formula of iron(II) nitrate(V)?

A

Fe(NO3)2

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13
Q

The equation below shows the reaction of chlorine gas with iron(II) ions in solution:
Cl2(g) + 2Fe2+(aq) –> 2Cl-(aq) + 2Fe3+(aq)

Write a half-equation to show the conversion of chlorine, Cl2, into chloride ions, Cl-

A

Cl2(g) + 2e- –> 2Cl-(aq)

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14
Q

The equation below shows the reaction of chlorine gas with iron(II) ions in solution:
Cl2(g) + 2Fe2+(aq) –> 2Cl-(aq) + 2Fe3+(aq)

Write a half-equation to show the conversion of iron(II) ions, Fe2+, into iron(III) ions, Fe3+

A

Fe2+(aq) –> Fe3+(aq) + e-

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15
Q

The equation below shows the reaction of chlorine gas with iron(II) ions in solution:
Cl2(g) + 2Fe2+(aq) –> 2Cl-(aq) + 2Fe3+(aq)

In this reaction, which species is the reducing agent? Explain your answer

A

Iron(II) ions, Fe2+
because during the reaction they donate electrons/are oxidised

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16
Q

Use oxidation states to balance this redox equation:
Ca(s) + Al3+(aq) –> Ca2+(aq) + Al(s)

A

3Ca(s) + 2Al3+(aq) –> 3Ca2+(aq) + 2Al(s)

17
Q

10cm3 of potassium iodate(V) solution was reached with excess acidified potassium iodide solution. All of the resulting solution was titrated with 0.16 mol dm-3 sodium thiosulfate solution.
If fully reacted with 20cm3 of the sodium thiosulfate solution

Balance the following ionic equation, showing how iodine is formed in the reaction between iodate(V) ions and iodide ions in acidic solution: IO3-(aq) + I-(aq) + H+(aq) –> I2(aq) + H2O(l)

A

IO3-(aq) + 5I-(aq) + 6H+(aq) –> 3I2(aq) + 3H2O(l)

18
Q

10cm3 of potassium iodate(V) solution was reached with excess acidified potassium iodide solution. All of the resulting solution was titrated with 0.16 mol dm-3 sodium thiosulfate solution.
If fully reacted with 20cm3 of the sodium thiosulfate solution

How many moles of thiosulfate ions were there in 20 cm3 of the sodium thiosulfate solution?

A

Number of moles of thiosulfate = 0.16 x (20/1000)
= 3.2 x 10-3 moles

19
Q

10cm3 of potassium iodate(V) solution was reached with excess acidified potassium iodide solution. All of the resulting solution was titrated with 0.16 mol dm-3 sodium thiosulfate solution.
If fully reacted with 20cm3 of the sodium thiosulfate solution

In the titration, iodine reacted with sodium thiosulfate according to this equation:
I2(aq) + 2Na2S2O3(aq) –> 2NaI(aq) + Na2S4O6(aq)

Calculate the number of moles of iodine that reacted with the sodium thiosulfate solution

A

2 moles of thiosulfate reacts with 1 mole of iodine, so there were (3.2 x 10-3)/2
= 1.6 x 10-3 moles of iodine

20
Q

10cm3 of potassium iodate(V) solution was reached with excess acidified potassium iodide solution. All of the resulting solution was titrated with 0.16 mol dm-3 sodium thiosulfate solution.
If fully reacted with 20cm3 of the sodium thiosulfate solution

How many moles of iodate(V) ions produce 1 mole of iodine from potassium iodide?

A

1/3 = 0.33 moles

21
Q

10cm3 of potassium iodate(V) solution was reached with excess acidified potassium iodide solution. All of the resulting solution was titrated with 0.16 mol dm-3 sodium thiosulfate solution.
If fully reacted with 20cm3 of the sodium thiosulfate solution

What is the concentration of the potassium iodate(V) solution?
Give your answer to an appropriate number of significant figures

A

There must be (1.6 x 10-3) x 0.33 = 5.3 x 10-4 moles of iodate(V) solution
So concentration of potassium iodate(V) = 5.3 x 10-4 moles x (1000/10) = 0.053 mol dm-3
(to 2 s.f)

22
Q

Astatine is a halogen. It can be found below iodine in Group 7 of the periodic table

Write an ionic equation for the reaction between iodine solution and sodium astatide (NaAt)
Which substance is oxidised?

A

I2 + 2At- –> 2I- + At2

(sodium) astatide

23
Q

A student has a sample of an aqueous potassium halide solution. She knows it contains either chloride, bromide or iodide ions. The student adds a few drops of aqueous bromine solution to the test tube and a reaction takes place.

Which halide ion is present in the potassium halide solution?

A

(potassium) iodide

24
Q

A student has a sample of an aqueous potassium halide solution. She knows it contains either chloride, bromide or iodide ions. The student adds a few drops of aqueous bromine solution to the test tube and a reaction takes place.

What colour will the aqueous solution in the test tube be after the reaction has finished?

A

brown

25
Q

A student wants to make hydrogen iodide using sodium iodide. Which of these reagents should they use?

A) Hydrochloric acid

B) Ammonia

C) Phosphoric acid

D) Sulfuric acid

A

C) Phosphoric acid

26
Q

A sample of hydrogen bromide gas is dissolved in water to form an aqueous solution

State what you would see if universal indicator was added to the solution. Explain your answer.

A

The solution would turn red/pink
because hydrogen bromide solution is strongly acidic

27
Q

A sample of hydrogen bromide gas is dissolved in water to form an aqueous solution

If sulfuric acid is added to the solution, the hydrogen bromide ions reduce the sulfuric acid to sulfur dioxide. Write a full, balanced equation for this reaction

A

2HBr + H2SO4 –> Br2 + SO2 + 2H2O

28
Q

Industrially, chlorine can be produced by electrolysing sodium chloride solution

Describe a test you could perform to confirm that a sample of a solution contained chloride ions. Include any observations that you would see if the result of the test was positive.

A

Add nitric acid to the sample. Then add silver nitrate solution.
If the sample contains chloride ions, white precipitate will form

29
Q

Industrially, chlorine can be produced by electrolysing sodium chloride solution

Explain why chlorine needs to be handled and stored extremely carefully

A

chlorine is toxic and corrosive, so must be kept away from skin and eyes. Chlorine irritates the respiratory system, so must not be breathed in. Chlorine is a strong oxidising agent, so it must be kept away from flammable materials

30
Q

Calculate the atom economy of this reaction. Methanol is the desired product

CH3Br + NaOH –> CH3OH + NaBr

A

% atom economy = (Mr of desired product/sum of Mr of all products) x 100%
= (Mr (CH3OH) / (Mr (CH3OH) + Mr (NaBr)) x 100%
= (12.0 + (3 x 1.0) + 16.0 + 1)/((12.0 + (3 x 1.0) + 16.0 + 1) + (23.0 +79.9)) x 100%
= 32.0/(32.0 + 102.9) x 100%
= 23.7%

31
Q

Dichromate(VI) ions and water were mixed together and allowed to reach equilibrium at a fixed temperature:

Cr2O72-(aq) + H2O(l) ⇌ 2CrO42-(aq) + 2H+(aq)

Write an expression for Kc for this reaction

A

Kc = ([CrO42-]2[H+]2) / ([Cr2O72-][H2O])

32
Q

Dichromate(VI) ions and water were mixed together and allowed to reach equilibrium at a fixed temperature:

Cr2O72-(aq) + H2O(l) ⇌ 2CrO42-(aq) + 2H+(aq)

Calculate the value of the equilibrium constant at RTP, given that the equilibrium concentrations of the reactants and products are:
[Cr2O72-] = 0.20 mol dm-3
[H2O] = 0.60 mol dm-3
[CrO42-] = 0.10 mol dm-3
[H+] = 0.30 mol dm-3

A

Kc = ([0.10]2[0.30]2) / ([0.20][0.60])

33
Q

Dichromate(VI) ions and water were mixed together and allowed to reach equilibrium at a fixed temperature:

Cr2O72-(aq) + H2O(l) ⇌ 2CrO42-(aq) + 2H+(aq)

What would happen to the value of Kc if the concentration of water increased? Explain your answer

A

There would be no change as the concentrations of the other reagents would change to keep Kc constant