CGP EXAM QUESTIONS: ES Flashcards
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.
At the anode, you would see bubbles of gas appearing
because chloride ions are losing electrons to form chlorine
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.
At the cathode, you would see the elctrode being plated
because copper ions are gaining electrons to form copper metal
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
2H+(aq) + 2e- –> H2(g)
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
4OH-(aq) –> O2(g) + H2O(l) + 4e-
Chlorine gas is produced industrially from the electrolysis of sodium chloride solution
Name one natural source of sodium chloride
rock salt, sea water, brine
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.
The universal indicator would turn the solution blue/purple
The solution would be alkaline, because of the hydroxide ions left in solution
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
The universal indicator would turn the solution green
because hydrogen ions and hydroxide ions have been discharged, leaving a neutral solution of NaCl
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
Chlorine is more reactive than bromine
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
+2
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)
+6
Chromium (Cr) can exist in a range of oxidation states.
Give the full name of the compound C2O3
chromium(III) oxide
What is the formula of iron(II) nitrate(V)?
Fe(NO3)2
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-
Cl2(g) + 2e- –> 2Cl-(aq)
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+
Fe2+(aq) –> Fe3+(aq) + e-
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
Iron(II) ions, Fe2+
because during the reaction they donate electrons/are oxidised
Use oxidation states to balance this redox equation:
Ca(s) + Al3+(aq) –> Ca2+(aq) + Al(s)
3Ca(s) + 2Al3+(aq) –> 3Ca2+(aq) + 2Al(s)

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

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?
Number of moles of thiosulfate = 0.16 x (20/1000)
= 3.2 x 10-3 moles
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
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
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?
1/3 = 0.33 moles
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
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)
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?
I2 + 2At- –> 2I- + At2
(sodium) astatide
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?
(potassium) iodide
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?
brown