Electrochemistry: five questions to try now
Real past-paper questions, the answer key from the mark scheme, and the explanation that goes with it. No account needed to answer them.
Question 1
Compound Y is heated with a mild oxidising agent. One of the products of the reaction reacts with hydrogen cyanide forming 2-hydroxybutanenitrile.
What is compound Y?
Answer: C.
Work backwards from the product.
2-hydroxybutanenitrile is CH₃CH₂CH(OH)CN. It has four carbons, but one of them is the nitrile carbon that came from the HCN, so the carbonyl compound it was made from had only three.
HCN adds across C=O to put the OH and the CN on the same carbon. For the OH to end up on carbon 2 with a CH₃CH₂ group beside it, the starting carbonyl must be propanal, CH₃CH₂CHO.
Propanal is what you get by mild oxidation of a primary alcohol, which is propan-1-ol. That is C.
D fails on the position of the OH. Propan-2-ol oxidises to propanone, a ketone, and HCN adds to that to give 2-hydroxy-2-methylpropanenitrile, which has a branch.
A and B both have four carbons, so with the nitrile carbon added the product would have five.
The carbon count is the fastest way in. Adding HCN always adds one carbon, so a four-carbon nitrile comes from a three-carbon carbonyl, and 'mild oxidising agent' then tells you whether to look for a primary or a secondary alcohol.
2-hydroxybutanenitrile is CH₃CH₂CH(OH)CN. It has four carbons, but one of them is the nitrile carbon that came from the HCN, so the carbonyl compound it was made from had only three.
HCN adds across C=O to put the OH and the CN on the same carbon. For the OH to end up on carbon 2 with a CH₃CH₂ group beside it, the starting carbonyl must be propanal, CH₃CH₂CHO.
Propanal is what you get by mild oxidation of a primary alcohol, which is propan-1-ol. That is C.
D fails on the position of the OH. Propan-2-ol oxidises to propanone, a ketone, and HCN adds to that to give 2-hydroxy-2-methylpropanenitrile, which has a branch.
A and B both have four carbons, so with the nitrile carbon added the product would have five.
The carbon count is the fastest way in. Adding HCN always adds one carbon, so a four-carbon nitrile comes from a three-carbon carbonyl, and 'mild oxidising agent' then tells you whether to look for a primary or a secondary alcohol.
Question 2
The equation for the reaction between aqueous copper ions and aqueous iodide ions is as follows.
2Cu2+(aq) + 4I–(aq) →2CuI(s) + I2(aq)
What is the change in oxidation state of copper?
Answer: C.
2Cu²⁺(aq) + 4I⁻(aq) → 2CuI(s) + I₂(aq)
Copper starts as Cu²⁺, so +2. In CuI the iodide is −1 and the compound is neutral, so the copper must be +1. The change is +2 → +1, which is C.
The check is the iodine. Iodide starts at −1 and half of it finishes as I₂ at 0, so it is oxidised, and something must be reduced to match. Copper gaining one electron per ion is that something, and the equation's 4I⁻ against 2Cu²⁺ is what makes the electrons balance: two iodides are oxidised, releasing two electrons, and two copper ions take one each.
B, +2 to 0, would mean copper metal, which the equation does not produce.
A confuses copper's charge with iodide's.
D invents a starting state: copper does not reach +4.
Copper(I) iodide is the reason this works at all. Cu⁺ is normally unstable in water and disproportionates, but CuI is so insoluble that it drops out of solution as a white precipitate as fast as it forms, which is why the mixture ends up as a white solid under a brown iodine solution.
Copper starts as Cu²⁺, so +2. In CuI the iodide is −1 and the compound is neutral, so the copper must be +1. The change is +2 → +1, which is C.
The check is the iodine. Iodide starts at −1 and half of it finishes as I₂ at 0, so it is oxidised, and something must be reduced to match. Copper gaining one electron per ion is that something, and the equation's 4I⁻ against 2Cu²⁺ is what makes the electrons balance: two iodides are oxidised, releasing two electrons, and two copper ions take one each.
B, +2 to 0, would mean copper metal, which the equation does not produce.
A confuses copper's charge with iodide's.
D invents a starting state: copper does not reach +4.
Copper(I) iodide is the reason this works at all. Cu⁺ is normally unstable in water and disproportionates, but CuI is so insoluble that it drops out of solution as a white precipitate as fast as it forms, which is why the mixture ends up as a white solid under a brown iodine solution.
Question 3
NCl3 reacts with H2O. NCl3 + 3H2O → NH3 + 3HCl O The oxidation state of nitrogen does not change in this reaction. Which statement is correct?
Answer: D.
The stem gives you the key: nitrogen does not change. Nitrogen is −3 in NH₃, so it is −3 in NCl₃ too, which makes each chlorine +1 in NCl₃.
Now follow the chlorine into the product. In HClO the hydrogen is +1 and the oxygen is −2, so
Cl + 1 + (−2) = 0, giving Cl = +1
Chlorine is +1 on both sides, so it is neither oxidised nor reduced.
Hydrogen is +1 in H₂O, in NH₃ and in HClO. Oxygen is −2 in H₂O and in HClO. Nothing changes anywhere, so this is not a redox reaction, which is D. It is a hydrolysis.
A and B are the two halves of the trap. NCl₃ looks like it should behave as chlorine does in water, and a reaction producing HClO looks like the disproportionation of chlorine, but here the chlorine arrives already at +1 and stays there.
C is worth checking rather than dismissing, since hydrogen genuinely does both in some reactions, but every hydrogen here is +1 before and after.
The one counter-intuitive step is chlorine being positive in NCl₃. The stem hands that to you by fixing nitrogen, which is why that sentence is in the question.
Now follow the chlorine into the product. In HClO the hydrogen is +1 and the oxygen is −2, so
Cl + 1 + (−2) = 0, giving Cl = +1
Chlorine is +1 on both sides, so it is neither oxidised nor reduced.
Hydrogen is +1 in H₂O, in NH₃ and in HClO. Oxygen is −2 in H₂O and in HClO. Nothing changes anywhere, so this is not a redox reaction, which is D. It is a hydrolysis.
A and B are the two halves of the trap. NCl₃ looks like it should behave as chlorine does in water, and a reaction producing HClO looks like the disproportionation of chlorine, but here the chlorine arrives already at +1 and stays there.
C is worth checking rather than dismissing, since hydrogen genuinely does both in some reactions, but every hydrogen here is +1 before and after.
The one counter-intuitive step is chlorine being positive in NCl₃. The stem hands that to you by fixing nitrogen, which is why that sentence is in the question.
Question 4
Four equations representing reactions of nitrogen or one of its compounds are given.
Which equation represents a disproportionation reaction?
Answer: D.
Disproportionation is one element being oxidised and reduced in the same reaction, so look for a single element appearing in the products in two oxidation states, one above its starting value and one below.
D. 2NO₂ + H₂O → HNO₃ + HNO₂
Nitrogen starts at +4 in NO₂. In HNO₃ it is +5, and in HNO₂ it is +3. One nitrogen up, one down, from a single starting species. That is disproportionation, so D.
A is neutralisation. Nitrogen stays at +5 throughout, and no oxidation number changes at all.
B is redox, but not disproportionation. Nitrogen is reduced from 0 to −3 and hydrogen is oxidised from 0 to +1, so two different elements change.
C is an acid-base reaction. Nitrogen stays at −3 in both NH₄Cl and NH₃.
The quickest filter is that a disproportionating element must appear twice among the products. Only D has nitrogen in two different products, which narrows it to one option before any numbers are worked out.
D. 2NO₂ + H₂O → HNO₃ + HNO₂
Nitrogen starts at +4 in NO₂. In HNO₃ it is +5, and in HNO₂ it is +3. One nitrogen up, one down, from a single starting species. That is disproportionation, so D.
A is neutralisation. Nitrogen stays at +5 throughout, and no oxidation number changes at all.
B is redox, but not disproportionation. Nitrogen is reduced from 0 to −3 and hydrogen is oxidised from 0 to +1, so two different elements change.
C is an acid-base reaction. Nitrogen stays at −3 in both NH₄Cl and NH₃.
The quickest filter is that a disproportionating element must appear twice among the products. Only D has nitrogen in two different products, which narrows it to one option before any numbers are worked out.
Question 5
The compound potassium bismuthate(V), KBiO3, is a powerful oxidising agent.
What is the significance of the (V) in potassium bismuthate(V)?
Answer: A.
A Roman numeral in a name is always the oxidation number of the element it follows, and here it follows bismuth. So the (V) says the bismuth is +5. That is A.
You can confirm it from the formula KBiO₃. Potassium is +1 and each oxygen is −2:
+1 + Bi + 3(−2) = 0, giving Bi = +5 ✓
C is the trap and worth pinning down. The numeral describes the atom, not the ion. An oxidation number belongs to an element, and the bismuthate ion as a whole has a charge rather than an oxidation number.
B is a different quantity. The bismuthate ion BiO₃⁻ carries a charge of −1, not +5, and a charge is written 1− while an oxidation number is written +1, which is why the two conventions are kept apart.
D is not a real quantity: in a neutral compound the charges must sum to zero.
The same rule reads every such name. Iron(III) chloride has iron at +3, and manganate(VII) has manganese at +7, not a charge of 7 on the MnO₄⁻ ion, which is only −1.
You can confirm it from the formula KBiO₃. Potassium is +1 and each oxygen is −2:
+1 + Bi + 3(−2) = 0, giving Bi = +5 ✓
C is the trap and worth pinning down. The numeral describes the atom, not the ion. An oxidation number belongs to an element, and the bismuthate ion as a whole has a charge rather than an oxidation number.
B is a different quantity. The bismuthate ion BiO₃⁻ carries a charge of −1, not +5, and a charge is written 1− while an oxidation number is written +1, which is why the two conventions are kept apart.
D is not a real quantity: in a neutral compound the charges must sum to zero.
The same rule reads every such name. Iron(III) chloride has iron at +3, and manganate(VII) has manganese at +7, not a charge of 7 on the MnO₄⁻ ion, which is only −1.
What this practice covers
These questions are drawn from past CIE 9701 Chemistry papers. You answer, you find out immediately whether you were right, and you get the reasoning for the correct option and for each distractor. Wrong answers go to a mistakes locker so you can come back to exactly those.
Practice is free. You need an account only so your progress and your mistakes are still there next time.
What examiners see students get wrong here
These are the errors that cost marks on electrochemistry, taken from our own topic notes. Read them before you practise and you will recognise the traps in the questions.
- Saying an oxidising agent is oxidised. It is reduced, because it takes electrons from something else.
- Giving oxygen -2 in hydrogen peroxide. It is -1 there.
- Balancing a redox equation for atoms only. The charge must balance as well.
- Saying the cathode is positive. In electrolysis it is negative, and reduction happens there.
- Calling any reaction where one thing is oxidised and another reduced disproportionation. It has to be the same element doing both.
- Forgetting that a solid ionic compound does not conduct, so it must be molten or in solution.
Revise it first
If any of the above is unfamiliar, work through the notes before practising: Electrochemistry revision notes.