AQA A-Level Chemistry Paper 1, 2017: Question 10

Skills & knowledge assessed

• Oxidation state calculation (including ions and neutral molecules).
• Interpreting E° values to rank oxidising/reducing strength.
• Choosing half-equations to match a target EMF.
• Constructing a conventional cell diagram.
• Using E° to justify an oxidising acid, writing the overall redox equation, and calculating EMF.

How marks are won

• Use the table exactly as given: all E° are standard reduction potentials.
• Pick cathode as the more positive E° reduction; anode is the other half reversed.
• State ions correctly (e.g. Cl⁻ not Cl₂).
• When justifying an oxidising agent: explicitly compare E° numbers.

✅ Correct answers (as per mark scheme)

• N in NO₃⁻ = +5 (allow Roman numerals)
• N in NO = +2 (allow Roman numerals)

📐 Calculations (step-by-step)

1. NO₃⁻: let oxidation state of N = x. Oxygen is −2 each.
2. Sum equals ion charge: x + 3(−2) = −1
3. x − 6 = −1 → x = +5
4. NO: neutral molecule, sum = 0. Let N = x, O = −2.
5. x + (−2) = 0 → x = +2

❌ Common errors

• Forgetting the overall charge on NO₃⁻ (using 0 instead of −1).
• Using −1 for oxygen (only true in peroxides like H₂O₂).
• Writing “5+” without sign clarity; always show +5, +2.

💡 Key knowledge

• O is usually −2 in compounds (except peroxides and with fluorine).
• The oxidation states in a species add up to its overall charge.

✅ Correct answer (mark scheme)

Cl⁻ (aq) / chloride ions

💡 Key knowledge (how to decide)

• Table values are reduction half-equations. The species on the right is the reduced form and can act as a reducing agent.
• The weaker the reducing agent, the less willing it is to be oxidised → it corresponds to a half-equation with a very positive E° (its oxidation would be very unfavourable).
• Cl₂ + 2e⁻ → 2Cl⁻ has E° = +1.36 V (very positive), so Cl⁻ is very hard to oxidise → weakest reducing agent here.

🧠 Exam technique

• For “reducing agent” in a reduction table: scan the right-hand species.
• Pick the one paired with the most positive E° (least likely to be oxidised).
• Write the exact ion asked for: Cl⁻, not “chlorine”.

❌ Common errors (explicitly flagged by mark scheme)

• Answering Cl₂: that’s the oxidising agent in the listed reduction half-equation, not the reducing agent.
• Answering “chlorine”/“Cl”: not an ion and not accepted.

💡 Key knowledge (finding the pair)

• E°cell = E°(cathode, reduction) − E°(anode, reduction).
• To get +0.43 V, look for two E° values with a difference of 0.43 V.
• From Table 4: +0.77 (Fe³⁺/Fe²⁺) and +0.34 (Cu²⁺/Cu) differ by 0.43.
• More positive reduction happens at the cathode: Fe³⁺ + e⁻ → Fe²⁺ is the cathode.
• The other half runs in reverse at the anode: Cu(s) → Cu²⁺ + 2e⁻.

✅ Correct cell diagram (what you should write)

One acceptable conventional representation:

Cu(s) | Cu²⁺(aq) || Fe³⁺(aq), Fe²⁺(aq) | Pt(s)

📐 EMF check (quick)

1. Cathode E° = +0.77 V (Fe³⁺/Fe²⁺)
2. Anode E° = +0.34 V (Cu²⁺/Cu)
3. E°cell = 0.77 − 0.34 = +0.43 V

❌ Common errors

• Putting Fe³⁺/Fe²⁺ on the left: would reverse the sign of E°cell.
• Forgetting an inert electrode for Fe³⁺/Fe²⁺ (no solid conductor present) → include Pt(s).
• Using a comma incorrectly: use Fe³⁺(aq), Fe²⁺(aq) to show both ions present in the same half-cell.
• Using a single line instead of || for the salt bridge.

🧠 Exam technique (cell diagrams)

• Left = anode (oxidation), right = cathode (reduction) for a spontaneous cell with +EMF.
• Single | = phase boundary; double || = salt bridge.
• If a half-cell has only ions/gases, add an inert electrode: Pt(s) (or C/graphite, but Pt is standard).

✅ Correct answers (mark scheme aligned)

• Acid: nitric acid, HNO₃ (must be nitric acid; “NO₃⁻ ions” alone loses the acid mark per guidance)
• Justification: E°(NO₃⁻/NO) = +0.96 V is more positive than E°(Cu²⁺/Cu) = +0.34 V
• Overall equation: 3Cu + 8H⁺ + 2NO₃⁻ → 3Cu²⁺ + 2NO + 4H₂O
• EMF: 0.62 V

💡 Key knowledge (why nitric acid works)

• Copper is below hydrogen in the reactivity series, so it does not react with dilute HCl/H₂SO₄ to make H₂.
• To oxidise Cu(s) to Cu²⁺, you need an oxidising agent with a more positive reduction potential than Cu²⁺/Cu.
• In nitric acid, the oxidising species is NO₃⁻ in acidic conditions (NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O, E° = +0.96 V).

📐 Calculations (step-by-step): overall equation + EMF

A) Build the overall reaction

1. Oxidation (reverse copper reduction):
   Cu(s) → Cu²⁺(aq) + 2e⁻
2. Reduction (given):
   NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(aq) + 2H₂O(aq)
3. Balance electrons (LCM of 2 and 3 is 6):
   3Cu(s) → 3Cu²⁺(aq) + 6e⁻
   2NO₃⁻(aq) + 8H⁺(aq) + 6e⁻ → 2NO(aq) + 4H₂O(aq)
4. Add and cancel electrons:
   3Cu + 8H⁺ + 2NO₃⁻ → 3Cu²⁺ + 2NO + 4H₂O

B) Calculate EMF for the same overall reaction

1. E°(cathode) = E° for NO₃⁻/NO = +0.96 V
2. E°(anode) = E° for Cu²⁺/Cu = +0.34 V (still use reduction value in the subtraction)
3. E°cell = 0.96 − 0.34 = 0.62 V

🧠 Exam technique (what to write to secure each mark)

• Acid mark: write “nitric acid (HNO₃)”. If you only say “nitrate ions”, you risk losing that mark (mark scheme guidance).
• Justification mark: explicitly compare the E° values: “0.96 V > 0.34 V”. A simple statement like “it has a higher E°” must clearly reference the correct couples.
• Equation mark: show the balanced redox equation in acidic conditions (H⁺ and H₂O included). If you instead write a molecular equation, ensure it matches an allowed form (see below).
• EMF mark: show the subtraction in the correct direction: E°(cathode) − E°(anode).

❌ Common errors (and why they lose marks)

• Choosing HCl / H₂SO₄: these provide H⁺ but not a strong enough oxidising agent to oxidise Cu under standard conditions.
• Wrong comparison: comparing NO₃⁻/NO to H⁺/H₂ (0.00 V) instead of to Cu²⁺/Cu (+0.34 V) misses the point of oxidising copper.
• Sign error in EMF: doing 0.34 − 0.96 = −0.62 V (this would imply a non-spontaneous cell).
• Unbalanced overall equation: especially H and O in acidic solution—use H⁺ and H₂O methodically.
• State confusion: mark scheme shows NO as (aq) in the half-equation; examiners typically accept correct chemistry even if you write NO(g), but don’t change species identity.

🧠 30-second check before moving on

• Oxidation states: sums correct with charge? (NO₃⁻ must sum to −1)
• Reducing agent: chosen from RHS species; is it the weakest (most positive E° pair)?
• Cell diagram: anode left, cathode right, || present, Pt included if only ions.
• EMF: cathode − anode; values copied correctly from table.

✅ Answers recap

• 10.1: N in NO₃⁻ = +5; N in NO = +2
• 10.2: weakest reducing agent = Cl⁻
• 10.3: cell with +0.43 V: Cu(s)|Cu²⁺ || Fe³⁺,Fe²⁺|Pt
• 10.4: acid = HNO₃; equation as given; EMF = 0.62 V